09243500460  09243500460

Chapter 10: Vector Algebra

Exercise – 10.1

Question 1: Graphically represent a 40 km displacement towards 30 o east of north.

Answer 1:

Vector \(\overrightarrow{OP}\) represent a 40 km displacement towards 30o east of north.

 

Question 2: Categorize the following measures as vectors and scalars.

(a)  20 kg                                          (b)  4 meters north – south           (c) 80o

(d)  70 watt                                      (e)  10– 17 coulomb                         (f)  56 m / s– 2

 

Answer 2:

(a)  In 20 kg, only magnitude is involved. So, it is a scalar quantity.

(b)  In 4 meters north – south, both the direction and magnitude are involved. So, it is a vector quantity

(c) In 80o, only magnitude is involved. So, it is a scalar quantity.

(d)  In 70 watt, only magnitude is involved. So, it is a scalar quantity.

(e)  In 10 – 17 coulombs, only magnitude is involved. So, it is a scalar quantity.

(f)  In 56 m / s– 2, both the direction and magnitude are involved. So, it is a vector quantity

 

 

Question 3: Categorize the following quantities as vector and scalar.

(a)  Time period                              (b) distance                                    (c) force

(d)  Velocity                                      (e) work done

 

Answer 3:

(a) In time period, only magnitude is involved. So, it is a scalar quantity.

(b) In distance, only magnitude is involved. So, it is a scalar quantity.

(c) In force, both the direction and magnitude are involved. So, it is a vector quantity

(d) In velocity, both the direction and magnitude are involved. So, it is a vector quantity

(e) In work done, only magnitude is involved. So, it is a scalar quantity.

 

 

Question 4: In the following diagram, recognize the corresponding vectors

(a) Coinitial

(b) Equal

(c) Collinear but not equal

 

Answer 4:

(a) Coinitial vectors are those vectors which have same initial point.  So, \(\overrightarrow{a} \;and\; \overrightarrow{d}\) vectors are coinitial.

(b) Equal vectors are vectors which have same magnitude and direction. So, \(\overrightarrow{b} \;and\; \overrightarrow{d}\) vectors are equal.

(c) Collinear but not equal are those vectors which are parallel but has different directions. So, \(\overrightarrow{a} \;and\; \overrightarrow{c}\) vectors are collinear but not equal.

 

 

Question 5: Check whether the following statements are true or false.

(a) \(\overrightarrow{b} \;and\; \overrightarrow{- b}\) vectors are collinear

(b) The magnitudes of the two collinear are always equal.

(c) Collinear vectors are the two vectors having same magnitude.

 

Answer 5:

(a). True because the two vectors are parallel .

(b). False because collinear vectors must be parallel.

(c). False.

 

Exercise 10.2

 

Question 1: For the following vectors, calculate the magnitude of the following.

\(\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k}; \;\;\; \overrightarrow{n} = \hat{2i} – \hat{7j} – \hat{3k}; \;\;\; \overrightarrow{o} = \frac{1}{\sqrt{3}} \;\hat{i} + \frac{1}{\sqrt{3}}\; \hat{j} – \frac{1}{\sqrt{3}}\; \hat{k}\)

 

Answer 1:

Given, \(\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k}; \;\;\; \overrightarrow{n} = \hat{2i} – \hat{7j} – \hat{3k}; \;\;\; \overrightarrow{o} = \frac{1}{\sqrt{3}} \;\hat{i} + \frac{1}{\sqrt{3}}\; \hat{j} – \frac{1}{\sqrt{3}}\; \hat{k}\)

\(\left | \overrightarrow{m} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}} = \sqrt{3} \\ \left | \overrightarrow{n} \right | = \sqrt{(2) ^{2} + (- 7) ^{2} + (- 3) ^{2}} = \sqrt{4 + 49 + 9} = \sqrt{62} \\ \left | \overrightarrow{o} \right | = \sqrt{(\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2}} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = 1 \\\)

 

 

Question-2: Mention two dissimilar vectors having similar magnitude?

 

Answer-2:

\(\overrightarrow{m} = (\hat{2i} – \hat{2j} + \hat{3k}); \;and\; \overrightarrow{n} = (\hat{2i} + \hat{2j} – \hat{3k}) \\ It\; can\; be\; observed\; that:\; \\ \left | \overrightarrow{m} \right | = \sqrt{(2) ^{2} + (- 2) ^{2} + (3) ^{2}} = \sqrt{17} \; and \\ \left | \overrightarrow{n} \right | = \sqrt{(2) ^{2} + (2) ^{2} + (3) ^{2}} = \sqrt{17}\)

Thus, the two dissimilar vectors \(\overrightarrow{m} \;and\; \overrightarrow{n}\) having the similar magnitude. Because of different directions the two vectors are dissimilar.

 

 

Question 3: Mention two dissimilar vectors having similar direction?

 

Answer 3:

Consider,  \(\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}; \;\;\;and\; \overrightarrow{b} = \hat{2i} + \hat{2j} + \hat{2k} \\ The\; direction\; cosines\; of\; \overrightarrow{a}\; are\; given\; by, \\ p = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \; q = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \;and\; r = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}\)

\(The\; direction\; cosines\; of\; \overrightarrow{b}\; are\; given\; by, \\ p = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}, \;q = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}\; and\; r = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}\) \(The\; direction\; cosines\; of\; \overrightarrow{a}\; and\; \overrightarrow{b}\; are\; similar.\)

Thus, the direction of the two vectors is similar.

 

 

Question 4: \(4 \hat{i} + 5 \hat{j} \;and\; p \hat{i} + q \hat{j}\) are the vectors and they are equal. Obtain the values of p and q

 

Answer 4:

Given, \(4 \hat{i} + 5 \hat{j} \;and\; p \hat{i} + q \hat{j}\) are equal.

The equivalent components are equal.

So, the value of p = 4 and q = 5.

 

Question 5: The initial point of the vector is (3, 2) and the terminal point of the vector is (- 6, 8). Obtain the vector and scalar components of the given vector.

Answer 5:

Given,

The initial point of the vector A (3, 2) and the terminal point of the vector is B (- 6, 8).

The vector \(\overrightarrow{AB} = (- 6 – 3) \hat{i} + (8 – 2) \hat{j} \\ \overrightarrow{AB} = – 9 \hat{i} + 6 \hat{j}\)

The vector components of the given vector are \(– 9 \hat{i} \;and\; 6 \hat{j}\).

The scalar components of the given vector are – 9 and 6.

 

 

Question 6: The vectors \(\overrightarrow{m} = \hat{i} + \hat{3j} + \hat{k}, \;\;\; \overrightarrow{n} = \hat{- 2i} – \hat{5j} – \hat{3k}, \;\;and \; \overrightarrow{o} = \hat{8i} – \hat{j} – \hat{2k}\). Obtain the sum.

 

Answer 6:

Given:

\(\overrightarrow{m} = \hat{i} + \hat{3j} + \hat{k}, \;\;\; \overrightarrow{n} = \hat{- 2i} – \hat{5j} – \hat{3k}, \;\;and \; \overrightarrow{o} = \hat{8i} – \hat{j} – \hat{2k} \\ \overrightarrow{m} + \overrightarrow{n} + \overrightarrow{o} = (1 – 2 + 8) \hat{i} + (3 – 5 – 1) \hat{j} + (1 – 3 – 2) \hat{k} \\ = 7 \hat{i} – 3 \hat{j} – 4 \hat{k}\)

Question 7: Obtain the unit vector of \(\overrightarrow{p} = \hat{i} + \hat{2j} + \hat{k}\) in the direction of the given vector.

 

Answer 7:

The unit vector \(\hat{p}\) in the direction of vector \(\overrightarrow{p} = \hat{i} + \hat{2j} + \hat{k}\)

\(\left | \overrightarrow{p} \right | = \sqrt{(1) ^{2} + (2) ^{2} + (1) ^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6} \\ \hat{p} = \frac{\overrightarrow{p}}{\left | \overrightarrow{p} \right |} = \frac{\hat{i} + \hat{2j} + \hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{2j} + \frac{1}{\sqrt{6}} \hat{k} \\ = \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}\)

 

 

Question 8: For a vector \(\overrightarrow{AB}\), obtain the unit vector where the point A (2, 3, 4) and point B (5, 6, 7). The unit vector should be in the direction of given vector.

 

Answer 8:

Given, points A (2, 3, 4) and B (5, 6, 7).

\(\overrightarrow{AB} = (5 – 2) \hat{i} + (6 – 3) \hat{j} + (7 – 4) \hat{k} \\ \overrightarrow{AB} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ \left | \overrightarrow{AB} \right | = \sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3 \sqrt{3} \\ The\; unit\; vector\; in\; the\; direction\; of\; \overrightarrow{AB}\; is \\ \frac{\overrightarrow{AB}}{\left | \overrightarrow{AB} \right |} = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}\)

 

 

Question 9: In the direction of \(\overrightarrow{m} + \overrightarrow{n}\), obtain the unit vector for given vectors \(\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \; and\; \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k}\).

 

Answer 9:

Given,

The vectors \(\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \; and\; \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k}\)

\(\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \\ \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k} \\ \overrightarrow{m} + \overrightarrow{n} = (3 + 2) \hat{i} + (- 1 – 3) \hat{j} + (2 – 1) \hat{k} \\ = 5 \hat{i} – 4 \hat{j} + 1 \hat{k} \\ \left | \overrightarrow{m} + \overrightarrow{n} \right | = \sqrt{(5) ^{2} + (- 4) ^{2} + (1) ^{2}} = \sqrt{25 + 16 + 1} = \sqrt{42}\)

Thus, in the direction of \(\overrightarrow{m} + \overrightarrow{n}\), the vector is,

\(\frac{(\overrightarrow{m} + \overrightarrow{n})}{\left | \overrightarrow{m} + \overrightarrow{n} \right |} = \frac{5 \hat{i} – 4 \hat{j} + 1 \hat{k}}{\sqrt{42}} \\ = \frac{1}{\sqrt{42}} 5 \hat{i} – \frac{1}{\sqrt{42}} 4 \hat{j} + \frac{1}{\sqrt{42}} \hat{k} \\ = \frac{5}{\sqrt{42}} \hat{i} – \frac{4}{\sqrt{42}} \hat{j} + \frac{1}{\sqrt{42}} \hat{k}\)

 

 

Question 10: A vector \(6 \hat{i} – 2 \hat{j} + 3 \hat{k}\) has a magnitude of 8 units. Find the vector in the direction of given vector.

Answer 10:

Suppose, \(\overrightarrow{m} = 6 \hat{i} – 2 \hat{j} + 3 \hat{k} \\ \left | \overrightarrow{m} \right | = \sqrt{6^{2} + (- 2) ^{2} + 3^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \\ \hat{m} = \frac{\overrightarrow{m}}{\left | \overrightarrow{m} \right |} = \frac{6 \hat{i} – 2 \hat{j} + 3 \hat{k}}{7}\).

Thus, the vector in the direction of given vector which has 8 units magnitude is given by,

\(8 \hat{m} = 8 (\frac{6 \hat{i} – 2 \hat{j} + 3 \hat{k}}{7}) = \frac{48}{7} \hat{i} – \frac{16}{7} \hat{j} + \frac{24}{7} \hat{k}\)

 

 

Question 11: Prove whether the vectors \(3 \hat{i} – 4 \hat{j} + 5 \hat{k} \;and\; 9 \hat{i} – 12 \hat{j} + 15 \hat{k}\) are collinear.

 

Answer 11:

Suppose, \(\overrightarrow{p} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \;and\; \overrightarrow{q} = 9 \hat{i} – 12 \hat{j} + 15 \hat{k}\)

The condition for the vectors to be collinear is,

\(\overrightarrow{q} = \lambda \overrightarrow{p}\)

Accordingly,

\(9 \hat{i} – 12 \hat{j} + 15 \hat{k} = 3\; (3 \hat{i} – 4 \hat{j} + 5 \hat{k} )\), which satisfies the condition with \(\lambda = 3\)

Hence, proved

 

 

Question 12: Obtain the direction cosines of the vectors \(2 \hat{i} – 4 \hat{j} + 6 \hat{k}\)

 

Answer 12:

\(\overrightarrow{m} = 2 \hat{i} – 4 \hat{j} + 6 \hat{k} \\ \left | \overrightarrow{m} \right | = \sqrt{(2) ^{2} + (- 4) ^{2} + (6) ^{2}} = \sqrt{4 + 16 + 36} = \sqrt{56} \\ Thus,\; the\; direction\; cosines\; of\; \overrightarrow{m} \;are\; \left ( \frac{2}{\sqrt{56}}, \frac{- 4}{\sqrt{56}}, \frac{6}{\sqrt{56}} \right )\)

 

 

Question 13: P (1, 2, – 3) and Q (- 1, – 2, 1) are the joining points of a vector directed from P to Q. Obtain the direction cosines of the vector.

 

Answer 13:

P (1, 2, – 3) and Q (- 1, – 2, 1) are the joining points of a vector.

\(\overrightarrow{PQ} = (- 1 – 1) \hat{a} + (- 2 – 2) \hat{b} + (1 – (- 3)) \hat{c} \\ \overrightarrow{PQ} = (- 2) \hat{a} + (- 4) \hat{b} + (4) \hat{c} \\ \left | \overrightarrow{PQ} \right | = \sqrt{(- 2) ^{2} + (- 4) ^{2} + 4^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\)

The direction cosines of the vector \(\overrightarrow{PQ}\) are \(\left ( – \frac{2}{6}, – \frac{4}{6}, \frac{4}{6} \right ) = \left ( – \frac{1}{3}, – \frac{2}{3}, \frac{2}{3} \right )\)

 

 

Question 14: Prove that the \(\hat{i} + \hat{j} + \hat{k}\) is evenly tending to the axes OX, OY and  OZ

Answer 14:

Suppose, \(\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k} \\ Then,\; \left | \overrightarrow{m} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}} = \sqrt{3}\)

The direction cosines of the vector \(\overrightarrow{m}\) are \(\left ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right )\)

Now, let α, β, and γ be the angles formed by with the positive directions of x, y, and z axes.

Thus, we obtain,

\(cos \;\alpha = \frac{1}{\sqrt{3}},\; cos \;\beta = \frac{1}{\sqrt{3}} \;and\; cos \;\gamma = \frac{1}{\sqrt{3}}\)

Thus, the vector is evenly tending to the axes OX, OY and OZ

 

 

Question 15: The position vectors of a joining points A and B are \(\hat{i} + 2 \hat{j} – \hat{k} \;and\; – \hat{i} + \hat{j} + \hat{k}\) respectively. Obtain the position vector of C which divides the given points in 2 : 1 ratio.

(a) Internally

(b) Externally

 

Answer 15:

The position vector of C which divides the given points in m : n ratio is written as:

(a) Internally: \(\frac{m \overrightarrow{b} + n \overrightarrow{a}}{m + n}\)

(b) Externally: \(\frac{m \overrightarrow{b} – n \overrightarrow{a}}{m – n}\)

Given,

Position vectors of a joining points A and B,

\(\overrightarrow{OA} = \hat{i} + 2 \hat{j} – \hat{k} \;and\;  \overrightarrow{OB} = \hat{i} + \hat{j} + \hat{k}\)

(a) The position vector of point C which divides the line joining two points A and B internally in the ratio 2:1 is given by,

\(\overrightarrow{OC} = \frac{2 (- \hat{i} + \hat{j} + \hat{k}) + 1 (\hat{i} + 2\hat{j} – \hat{k})}{2 + 1} = \frac{(- 2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + (\hat{i} + 2\hat{j} – \hat{k})}{3} \\ = \frac{- \hat{i} + 4 \hat{j} + \hat{k}}{3} \\ = – \frac{1}{3} \hat{i} + \frac{4}{3} \hat{j} + \frac{1}{3} \hat{k}\)

(b) The position vector of point C which divides the line joining two points A and B externally in the ratio 2:1 is given by,

\(\overrightarrow{OC} = \frac{2 (- \hat{i} + \hat{j} + \hat{k}) – 1 (\hat{i} + 2\hat{j} – \hat{k})}{2 – 1} \\ = (- 2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + (\hat{i} + 2\hat{j} – \hat{k}) \\ = – 3 \hat{i} + 3 \hat{k}\)

 

 

Question 16: A (3, 4, 5) and B (5, 2, – 3) are the joining points of a vector. Obtain the midpoint position vector.

 

Answer 16:

The midpoint position vector with the joining points A (3, 4, 5) and B (5, 2, – 3),

Suppose, \(\overrightarrow{OC}\) be the required vector, then,

\(\overrightarrow{OC} = \frac{(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) + (5 \hat{i} + 2 \hat{j} + (- 3) \hat{k})}{2} = \frac{(3 + 5) \hat{i} + (4 + 2) \hat{j} + (5 – 3) \hat{k}}{2} \\ = \frac{8 \hat{i} + 6 \hat{j} + 2 \hat{k}}{2} \\ = 4 \hat{i} + 3 \hat{j} + \hat{k})\)

 

 

Question 17: Prove that the points P, Q and R with position vectors, \(\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}\) respectively from the vertices of a right angled triangle.

 

Answer 17:

Given,

The points P, Q and R with position vectors \(\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}\)

\(\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}\) \(\overrightarrow{PQ} = \overrightarrow{q} – \overrightarrow{p} = (2 – 3) \hat{a} + (-1 + 4) \hat{b} + (1 + 4) \hat{c} = – \hat{a} + 3 \hat{b} + 5 \hat{c} \\ \overrightarrow{QR} = \overrightarrow{r} – \overrightarrow{q} = (1 – 2) \hat{a} + (- 3 + 1) \hat{b} + (- 5 – 1) \hat{c} = – \hat{a} – 2 \hat{b} – 6 \hat{c} \\ \overrightarrow{RP} = \overrightarrow{p} – \overrightarrow{r} = (3 – 1) \hat{a} + (- 4 + 3) \hat{b} + (- 4 + 5) \hat{c} = 2\hat{a} – \hat{b} + \hat{c} \\\) \(\left | \overrightarrow{PQ} \right | ^{2} = (- 1) ^{2} + 3^{2} + 5^{2} = 1 + 9 + 25 = 35 \\ \left | \overrightarrow{QR} \right | ^{2} = (-1) ^{2} + (- 2) ^{2} + (- 6) ^{2} = 1 + 4 + 36 = 41 \\ \left | \overrightarrow{RP} \right | ^{2} = 2^{2} + (- 1)^{2} + 1^{2} = 4 + 1 + 1 = 6 \\ \left | \overrightarrow{PQ} \right | ^{2} + \left | \overrightarrow{QR} \right | ^{2} = 35 + 6 = 41 = \left | \overrightarrow{QR} \right | ^{2}\)

Hence, PQR is a right angled triangle.

 

 

Question 18: Which of the following is incorrect in the triangle PQR?

\((i) \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} = 0 \\ (ii) \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{PR} = 0 \\ (iii) \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{RP} = 0 \\ (iv) \overrightarrow{PQ} – \overrightarrow{RQ} + \overrightarrow{RP} = 0\)

 

Answer 18:

In a given triangle, applying the triangle law of addition, we get,

\(\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR} …. (1) \\ \overrightarrow{PQ} + \overrightarrow{QR} = – \overrightarrow{RP} \\ \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} = \overrightarrow{0} …. (2) \\ Statement\; (i)\; is\; true \\ \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR} \\ \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{PR} = \overrightarrow{0} \\ Statement\; (ii)\; is\; true \\\)

From equation (2), we get:

\(\overrightarrow{PQ} – \overrightarrow{RQ} + \overrightarrow{RP} = \overrightarrow{0} \\ Statement\; (iv)\; is\; true \\ Considering\; statement\; (iii) \\ \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{RP} = \overrightarrow{0} \\ \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{RP} ….. (3)\)

From equations (3) and (1), we get:

\(\overrightarrow{PR} = \overrightarrow{RP} \\ \overrightarrow{PR} = – \overrightarrow{PR} \\ 2 \overrightarrow{PR} = \overrightarrow{0} \\ \overrightarrow{PR} = \overrightarrow{0}, \; is \;not\; true. Statement\; (iii)\; is\; true \\\)

 

 

Question 19: Check whether the corresponding statements are true if the two vectors \(\overrightarrow{p} \;and\; \overrightarrow{q}\) are collinear.

\((i) \overrightarrow{q} = \lambda \overrightarrow{p}, for\; some\; scalar\; \lambda \\ (ii) \overrightarrow{p} = \pm \overrightarrow{q} \\ (iii) The\; components\; of\; \overrightarrow{p} \;and\; \overrightarrow{q}\; are\; proportional \\ (iv) \overrightarrow{p} \;and\; \overrightarrow{q}\; have\; different\; magnitudes\; and\; have\; similar\; direction.\)

 

Answer 19:

The two vectors are said to be collinear when they are parallel to each other.

\(\overrightarrow{p} \;and\; \overrightarrow{q}\) are collinear vectors.

Thus, we have,

\(\overrightarrow{q} = \lambda \overrightarrow{p}, (for\; some\; scalar\; \lambda) \\ Suppose,\; \lambda = \pm 1,\; then\; \overrightarrow{q} = \pm 1 \overrightarrow{p} \\ If,\; \overrightarrow{p} = p _{1} \hat{i} + p _{2} \hat{j} + p _{3} \hat{k} ,\; \overrightarrow{q} = q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k},\; then \;\overrightarrow{q} = \lambda \overrightarrow{p}.\) \(q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k} = \lambda (p _{1} \hat{i} + p _{2} \hat{j} + p _{3} \hat{k}) \\ q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k} = (\lambda p _{1}) \hat{i} + (\lambda p _{2}) \hat{j} + (\lambda p _{3}) \hat{k} \\ q _{1} = \lambda p _{1}, q _{2} = \lambda p _{2}, q _{3} = \lambda p _{3} \\ => \frac{q _{1}}{p _{1}} = \frac{q _{2}}{p _{2}} = \frac{q _{3}}{p _{3}} = \lambda\)

Hence, the components of \(\overrightarrow{p} \;and\; \overrightarrow{q}\) are proportional.

Though, vectors \(\overrightarrow{p} \;and\; \overrightarrow{q}\) can have different directions.

Thus, statement (iv) is incorrect.

 

 

Exercise 10.3

 

 

Q.1 : Find the angle between two vectors \(\vec{ m }\) and \(\vec{ n }\) with magnitude \(\sqrt{ 3 }\)and 2 , respectively having \(\vec{ m }.\vec{ n } = \sqrt{ 6 }\)

Solution 1:

It is given that,

\(\left | \vec{ m } \right |= \sqrt{ 3 }\) ,

\(\left | \vec{ n } \right |\) = 2

And \(\vec{ m }.\vec{ n } = \sqrt{ 6 }\)

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Now, we know that

Therefore,

\(\boldsymbol{\Rightarrow }\)    \(\sqrt{ 6 } = \sqrt{ 3 } \times 2 \times cos\theta\)

\(\boldsymbol{\Rightarrow }\)   \(\cos \theta = \frac{ \sqrt{ 6 }}{ \sqrt{ 3 } \times 2 }\)

\(\boldsymbol{\Rightarrow }\)   \(\cos \theta = \frac{ 1 }{ \sqrt{ 2 }}\)

\(\boldsymbol{\Rightarrow }\)   \(\theta = \frac{ \pi }{ 4 }\)

Therefore, the angle between the given vectors \(\vec{ m }\) and \(\vec{ n }\) is \(\frac{ \pi }{ 4 }\)

 

 

Q 2 : Find the angle between the vectors \(\hat{ a } – 2 \hat{ b } + 3 \hat{ c } and 3 \hat{ a } – 2 \hat{ b } + \hat{ c }\)

 

Solution 2:

The given vectors are:

\(\vec{ m } = \hat{ a } – 2 \hat{ b } + 3 \hat{ c } and \vec{ n } = 3 \hat{ a } – 2 \hat{ b } + \hat{ c }\) \(\left |\vec{ m } \right | = \sqrt{ 1 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 3 ^{ 2 }}\) \(\left |\vec{ m } \right | = \sqrt{ 1 + 4 + 9 }\) \(\left |\vec{ m } \right | = \sqrt{ 14 }\) \(\left |\vec{ n } \right | = \sqrt{ 3 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 1 ^{ 2 }}\) \(\left |\vec{ n } \right | = \sqrt{ 9 + 4 + 1 }\) \(\left |\vec{ n } \right | = \sqrt{ 14 }\) \(Now, \vec{ m }.\vec{ n } = \left ( \hat{ a } – 2 \hat{ b } + 3 \hat{ c } \right )\left ( 3 \hat{ a } – 2 \hat{ b } + \hat{ c } \right )\) \(Now, \vec{ m }.\vec{ n } = 1.3 + \left ( – 2 \right )\left ( – 2 \right ) + 3.1\) \(Now, \vec{ m }.\vec{ n } = 3 + 4 + 3\) \(Now, \vec{ m }.\vec{ n } = 10\)

Also, we know that

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Therefore,

\(10 = \sqrt{ 14 } \sqrt{ 14 }\cos \theta\) \(\cos \theta = \frac{ 10 }{ 14 }\) \(\theta = \cos^{-1} \frac{ 5 }{ 7 }\)

 

 

Q 3. Find the projection of the vector \(\hat{ a } – \hat{ b }\) on the vector \(\hat{ a } + \hat{ b }\).

 

Solution 3:

Let, \(\hat{ i } = \hat{ a } – \hat{ b }\)

And \(\hat{ j } = \hat{ a } + \hat{ b }\)

Now, projection of vector \(\vec{ i }\) on \(\vec{ j }\) is given by,

\(\frac{ 1 }{ \left | \vec{ j} \right |} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 1 + 1 }} \left \{ 1.1 + \left ( – 1 \right ) \left ( 1 \right ) \right \} = \frac{ 1 }{ \sqrt{ 2 }}\left ( 1 – 1 \right ) = 0\)

Hence the projection of vector \(\vec{ i }\) on \(\vec{ j }\) is 0

 

 

Q 4. Find the projection of the vector \(\hat{ a } + 3 \hat{ b } + 7 \hat{ c }\) on the vector \(7\hat{ a } –  \hat{ b } + 8 \hat{ c }\)

 

Solution 4:

Let \(\hat{ i } = \hat{ a } + 3 \hat{ b } + 7 \hat{ c }\) and \(\hat{ j } = 7\hat{ a } –  \hat{ b } + 8 \hat{ c }\)

Now, projection of vector \(\vec{ i }\) on \(\vec{ j }\) is given by,

\(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 7^{ 2 } + \left ( – 1 \right )^{ 2 } + 8 ^{ 2 }}} \left \{ 1 \left ( 7 \right ) + 3 \left ( – 1 \right ) + 7 \left ( 8 \right )\right \}\) \(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 7 – 3 + 56 }{ \sqrt{ 49 + 1 + 64 }}\) \(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 60 }{ \sqrt{ 114 }}\)

 

Q 5: Show that each of the given three vectors is a unit vector :

\(\frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right )\)

Also, show that they are mutually perpendicular to each other.

 

Solution 5:

\(Let \vec{ i } = \frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) = \frac{ 2 }{ 7 } \hat{ a } + \frac{ 3 }{ 7 } \hat{ b } + \frac{ 6 }{ 7 } \hat{ c } \\ Let \vec{ j } = \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) = \frac{ 3 }{ 7 } \hat{ a } – \frac{ 6 }{ 7 } \hat{ b } + \frac{ 2 }{ 7 } \hat{ c } \\ Let \vec{ k } = \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right ) = \frac{ 6 }{ 7 } \hat{ a } + \frac{ 2 }{ 7 } \hat{ b } – \frac{ 3 }{ 7 } \hat{ c }\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{ i } \right | = \sqrt{ \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ 6 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ i } \right | = \sqrt{ \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 }}\) \(\left | \vec{ i } \right | = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{ j } \right | = \sqrt{ \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ – 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ j } \right | = \sqrt{ \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 }}\) \(\left | \vec{ j } \right | = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{ k } \right | = \sqrt{ \left (\frac{ 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ – 3 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ k } \right | = \sqrt{ \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 }}\) \(\left | \vec{ k } \right | = 1\)

Thus, each of the given three vectors is a unit vector.

\(\boldsymbol{\Rightarrow }\)   \(\vec{ i }.\vec{ j } = \frac{ 2 }{ 7 }\times \frac{ 3 }{ 7 } + \frac{ 3 }{ 7 } \times \left ( \frac{ -6 }{ 7 } \right ) + \frac{ 6 }{ 7 } \times \frac{2}{7}\)

\(\vec{ i }.\vec{ j } = \frac{6}{49} – \frac{18}{49} + \frac{12}{49}\) \(\vec{ i }.\vec{ j } = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\vec{ j }.\vec{ k } = \frac{3}{ 7 }\times \frac{ 6 }{ 7 } + \frac{ -6 }{ 7 } \times \left ( \frac{ 2 }{ 7 } \right ) + \frac{ 2 }{ 7 } \times \frac{ -3}{7}\)

\(\vec{ j }.\vec{ k } = \frac{18}{49} – \frac{12}{49} – \frac{6}{49}\) \(\vec{ j }.\vec{ k } = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\vec{ k }.\vec{ i } = \frac{6}{ 7 }\times \frac{ 2 }{ 7 } + \frac{ 2 }{ 7 } \times \left ( \frac{ 3 }{ 7 } \right ) + \frac{ -3 }{ 7 } \times \frac{ 6}{7}\)

\(\vec{ k }.\vec{ i } = \frac{12}{49} – \frac{6}{49} – \frac{18}{49}\) \(\vec{ k }.\vec{ i } = 0\)

Hence, the given three vectors are mutually perpendicular to each other.

 

Q 6: Find:

\(\left | \vec{m} \right | and \left | \vec{n} \right |\), if  \(\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} + \vec{n} \right ) = 8\) and \(\left | \vec{m} \right | = 8\left | \vec{n} \right |\)

 

Solution 6:

\(\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} – \vec{n} \right ) = 8\)

\(\boldsymbol{\Rightarrow }\)    \(\vec{ m}.\vec{ m} – \vec{ m}\vec{n} + \vec{ n}\vec{ m} – \vec{n}.\vec{ n} = 8\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{ m} \right |^{ 2} – \left | \vec{ n} \right |^{ 2} = 8\)

\(\boldsymbol{\Rightarrow }\)    \(8 \left | \vec{ n} \right |^{2} – \left | n \right |^{ 2} = 8\)  . . . . . . . . .  \(\left [ \left | \vec{ m } = 8\left | \vec{ n} \right | \right | \right ]\)

\(\boldsymbol{\Rightarrow }\)   \(64 \left | \vec{n} \right |^{2} – \left | \vec{n} \right |^{2} = 8\)

\(\boldsymbol{\Rightarrow }\)   \(63 \left | \vec{ n } \right |^{2} = 8\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{n} \right |^{2} = \frac{ 8}{ 63}\)

\(\boldsymbol{\Rightarrow }\)    \(\left | \vec{n} \right |^{2} = \sqrt{\frac{ 8 }{ 63 }}\) [ magnitude of a vector is non-negative]

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{n} \right |^{2} = \frac{ 2 \sqrt{2}}{ 3 \sqrt{7}}\)

\(\left | \vec{m} \right | = 8 \left | \vec{n} \right |\) \(\left | \vec{m} \right | = \frac{ 8 \times 2\sqrt{2}}{ 3\sqrt{7}}\) \(\left | \vec{m} \right | = \frac{ 16 \sqrt{2}}{ 3 \sqrt{7}}\)

 

 

Q 7: Find the product of the following : \(\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )\)

 

Solution 7:

\(\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )\) \(= 3 \vec{m}.2 \vec{m} + 3 \vec{m}.7 \vec{n} – 5 \vec{n}.2 \vec{m} – 5 \vec{n}.7 \vec{n}\) \(= 6 \vec{m}. \vec{m} + 21 \vec{m}. \vec{n} – 10 \vec{n}. \vec{m} – 35 \vec{n}. \vec{n}\) \(= 6 \left | \vec{m} \right |^{2} + 11 \vec{m}.\vec{n} – 35 \left | \vec{n} \right |^{2}\)

 

 

Q 8: Find the magnitude of two vectors \(\vec{m} and \vec{n}\), having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{ 1 }{ 2 }\)

 

Solution 8:

Let θ be the angle between the vectors \(\vec{m} and \vec{n}\).

As given in the question:

 

\(\left | \vec{m} \right | = \left | \vec{n} \right |, \vec{m}.\vec{n} = \frac{1}{2} and \theta = 60 ^{\circ}\). . . . . . . . . . . . . . . . . . . . . . .  ( 1 )

We know that:

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Therefore,

\(\frac{1}{2} = \left | \vec{m} \right |\left | \vec{m} \right | cos 60 ^{\circ}\)  . . . . . . . . . . . . . . . . . . . [using ( 1 )]

\(\boldsymbol{\Rightarrow }\)   \(\frac{1}{2} = \left | \vec{m} \right |^{2} \times \frac{1}{2}\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{m} \right |^{2} = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\left | \vec{m} \right |^{2} = \left | \vec{n} \right |^{2} = 1\)

 

 

Q 9: Find:

\(\left | \vec{y} \right |\) , if for a unit vector \(\vec{b}\) , \(\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12\)

 

Solution 9:

\(\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12\)

\(\boldsymbol{\Rightarrow }\)   \(\vec{y}.\vec{y} + \vec{y}.\vec{b} – \vec{b}.\vec{y} – \vec{b}.\vec{b} = 12\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{y} \right |^{ 2 } – \left |\vec{b} \right |^{ 2 } = 12\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{y} \right |^{ 2 } – 1 = 12 \left [ \left |\vec{b} \right | = 1 as \vec{b} is a unit vector \right ]\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{y} \right |^{ 2 } = 13\)

Therefore,

\(\left | \vec{ y } \right | = \sqrt{ 13 }\)

 

 

Q 10: If \(\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}\) ,

 \(\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}\)

and \(\vec{k} = 3\hat{a} + \hat{b}\) are such that  \(\vec{i} + \lambda \vec{j}\) is perpendicular to \(\vec{k}\) then find the value of λ

 

Solution 10:

The given vectors are \(\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}\) , \(\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}\) and \(\vec{k} = 3\hat{a} + \hat{b}\)

Now,

\(\vec{i} + \lambda \vec{j} = \left ( 2 \hat{a} + 2 \hat{b} + 3 \hat{c} \right ) + \lambda \left ( – \hat{a} + 2 \hat{b} + \hat{c} \right ) = ( 2 – \lambda ) \hat{a} + ( 2 + 2 \lambda )\hat{b} + \left ( 3 + \lambda \right )\hat{c}\)

If \(\left (\vec{i} + \lambda \vec{j} \right )\) is perpendicular to \(\vec{k}\) ,then

\(\left (\vec{i} + \lambda \vec{j} \right ).\vec{k} = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\left [\left ( 2 – \lambda \right )\hat{a} + \left ( 2 + 2\lambda \right )\hat{b} + \left ( 3 + \lambda \right )\hat{c} \right ]\left ( 3\hat{a} + \hat{b} \right ) = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\left [\left ( 2 – \lambda \right )3 + \left ( 2 + 2\lambda \right )1 + \left ( 3 + \lambda \right )0 \right ] = 0\)

\(\boldsymbol{\Rightarrow }\)   \(6 – 3\lambda + 2 + 2\lambda = 0\)

\(\boldsymbol{\Rightarrow }\)   \(– \lambda + 8 = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\lambda = 8\)

Hence, the required value of λ is 8.

 

 

Q 11: Show that:

\(\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}\) is perpendicular to \(\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}\) , for any two non zero vectors \(\vec{a} and \vec{b}\)

 

Solution 11:

\(\boldsymbol{\Rightarrow }\)   \(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\)

\(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\) = \(\left |\vec{a} \right |^{2} \vec{b}.\vec{b} – \left |\vec{a} \right |\left |\vec{b} \right |\vec{b}.\vec{a} + \left |\vec{b} \right |\left |\vec{a} \right |\vec{a}.\vec{b} – \left |\vec{b} \right |^{2} \vec{a}.\vec{a}\)

\(= \left | \vec{a} \right |^{ 2 } \left |\vec{b} \right |^{ 2 }\)

\(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\) = 0

Hence, \(\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}\) is perpendicular to \(\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}\)

 

 

Q 12: If \(\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0\), then what can be concluded about the vector \(\vec{b}\) ?

 

Solution:

It is given that \(\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0\)

Now,

that \(\vec{a}.\vec{a} = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{a} \right |^{ 2 } = 0\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{a} \right |= 0\)

Therefore, \(\vec{a}\) is a zero vector

Hence, vector \(\vec{b}\) satisfying \(\vec{a}.\vec{b} = 0\) can be any vector

 

 

Q 13:

If \(\vec{a} , \vec{b} , \vec{c}\) are unit vectors such that \(\vec{a} + \vec{b} + \vec{c}\) = 0, find the value of \(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\)

 

Solution 13:

\(\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 } = \left (\vec{a} + \vec{b} + \vec{c} \right )\left (\vec{a} + \vec{b} + \vec{c} \right )\)

\(\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 }\) = \(\left |\vec{a} \right | ^{2}+ \left |\vec{b} \right |^{ 2 } +\left | \vec{c} \right |^{ 2 } + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )\)

\(\boldsymbol{\Rightarrow }\)   \(0 = 1 + 1 + 1 + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )\)

\(\boldsymbol{\Rightarrow }\)   \(\left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right ) = \frac{ – 3 }{ 2 }\)

 

 

Q 14: If either vector \(\vec{a} = \vec{0} or \vec{b} = \vec{0 }\) ,  then \(\vec{a}.\vec{b} = 0\). But the converse need not be true. Justify your answer with an example.

 

Solution 14:

Consider \(\vec{a} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} + 3 \hat{j } – 6 \hat{k}\)

Then,

\(\vec{a}.\vec{b} = 2.3 + 4.3 + 3 \left ( – 6 \right ) = 6 + 12 – 18 = 0\)

We now observe that :

\(\left |\vec{a} \right | = \sqrt{ 2 ^{2} + 4 ^{ 2 } + 3 ^{2}} = \sqrt{ 29 }\)

Therefore,

\(\vec{a} \neq \vec{0}\) \(\left |\vec{ b } \right | = \sqrt{ 3 ^{2} + 3 ^{ 2 } + ( -6 ) ^{2}} = \sqrt{ 54 }\)

Therefore,

\(\vec{b} \neq \vec{0}\)

Hence, the converse of the given statement need not be true

 

 

Q 15: If the vertices A, B, C of a triangle ABC are ( 1 , 2 , 3 ) , ( – 1 , 0 , 0 ) , ( 0 , 1 , 2 ) , respectively, then find ABC. [  ABC is the angle between the vectors \(\vec{BA} and \vec{BC}\) ]

 

Solution 15:

The vertices of ∆ABC are given as A ( 1 , 2 , 3 ) , B ( – 1 , 0 , 0 ) , and C ( 0 , 1 , 2 ). Also, it is given that ∠ ABC is the angle between the vectors \(\vec{BA} and \vec{BC}\)

\(\vec{BA} = \left \{ 1 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 2 – 0 \right ) \hat{j} + \left ( 3 – 0 \right ) \hat{k} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}\) \(\vec{BC} = \left \{ 0 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 1 – 0 \right ) \hat{j} + \left ( 2 – 0 \right ) \hat{k} =  \hat{i} +  \hat{j} + 2 \hat{k}\)

Therefore,

\(\vec{BA }. \vec{BC} = \left ( 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \right ).\left ( \hat{i} + \hat{j} + 2 \hat{k} \right ) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10\) \(\left |\vec{BA } \right | = \sqrt{ 2 ^{ 2 } + 2 ^{2} + 3^{ 2 }} = \sqrt{ 4 + 4 + 9 } = \sqrt{ 17 }\) \(\left |\vec{BC } \right | = \sqrt{ 1 + 1 + 2 ^{ 2 }} = \sqrt{ 6 }\)

Now, it is known that :

\(\vec{BA} . \vec{BC} = \left |\vec{BA} \right |\left |\vec{BC} \right | cos \left ( \angle ABC \right )\)

Therefore,

\(10 = \sqrt{ 17 } \times \sqrt{ 6 } cos \left ( \angle ABC \right )\)

\(\boldsymbol{\Rightarrow }\)   \(cos \left ( \angle ABC \right ) = \frac{ 10 }{ \sqrt{ 17 } \times \sqrt{6}}\)

\(\boldsymbol{\Rightarrow }\)   \(\angle ABC = \cos^{-1}\left ( \frac{ 10 }{ \sqrt{ 102 }} \right )\)

 

 

Q 16: Show that the points A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) and C ( 3 , 10 , – 1 ) are collinear.

 

Solution 16:

The given points are A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) , and C ( 3 , 10 , – 1 ).

Therefore,

\(\vec{AB} = \left ( 2 – 1 \right ) \hat{i} + \left ( 6 – 2 \right ) \hat{j} + \left ( 3 – 7 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}\) \(\vec{BC} = \left ( 3 – 2 \right ) \hat{i} + \left ( 10 – 6 \right ) \hat{j} + \left ( – 1 – 3 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}\) \(\vec{AC} = \left ( 3 – 1 \right ) \hat{i} + \left ( 10 – 2 \right ) \hat{j} + \left ( – 1 – 7 \right ) \hat{k} = 2 \hat{i} + 8 \hat{j} – 8 \hat{k}\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{AB} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{BC} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{AC} \right | = \sqrt{ 2 ^{2} + 8 ^{2} + 8 ^{2} } = \sqrt{ 4 + 64 + 64 } = \sqrt{ 132 }\) = \(2 \sqrt{33}\)

Therefore,

\(\left |\vec{AC} \right | = \left |\vec{AB} \right | + \left |\vec{BC} \right |\)

Hence, the given points A, B, and C are collinear.

 

 

Q 17: Show that the vectors \(2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}\) form the vertices of a right angled triangle.

 

Solution 17:

Let vectors \(2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}\) be position vectors of points A, B, and C respectively.

i.e , \( \vec{OA}  = 2 \hat{i} – \hat{j} + \hat{k} , OB = \hat{i} – 3 \hat{j} – 5 \hat{k} and OC = 3 \hat{i} – 4 \hat{j } – 4 \hat{k}\)

Therefore,

\(\vec{AB} = \left ( 1 – 2 \right ) \hat{i} + \left ( – 3 + 1 \right ) \hat{j} + \left ( – 5 – 1 \right ) \hat{k} = – \hat{i} – 2 \hat{j} – 6 \hat{k}\) \(\vec{BC} = \left ( 3 – 1 \right ) \hat{i} + \left ( – 4 + 3 \right ) \hat{j} + \left ( – 4 + 5 \right ) \hat{k} = – 2\hat{i} – \hat{j} –  \hat{k}\) \(\vec{AC} = \left ( 2 – 3 \right ) \hat{i} + \left ( – 1 + 4 \right ) \hat{j} + \left ( 1 + 4 \right ) \hat{k} = – \hat{i} + 3 \hat{j} + 5 \hat{k}\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{AB} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( – 2 \right )^{ 2 } + \left ( – 6 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 36 } = \sqrt{ 41 }\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{BC} \right | = \sqrt{ \left ( 2 \right )^{ 2 } + \left ( – 1 \right )^{ 2 } + \left ( 1 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 1 } = \sqrt{ 6 }\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{AC} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( 3 \right )^{ 2 } + \left ( 5 \right ) ^{ 2 }} = \sqrt{ 1 + 9 + 25 } = \sqrt{ 35 }\)

Therefore,

\(\left |\vec{BC} \right |^{ 2 } + \left |\vec{AC} \right | ^{2} = 6 + 35 = 41 = \left |\vec{AB} \right | ^{ 2 }\)

Hence, ∆ ABC is a right – angled triangle.

 

 

Q 18:

If \(\vec{a}\) is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ \(\vec{a}\) is unit vector if

(A) λ = 1

(B) λ = – 1

(C) \(a = \left |\lambda \right |\)

(D) \(a = \frac{ 1 }{ \left | \lambda \right |}\)

 

 

Solution 18:

Vector \(\lambda \vec{a}\) is a unit vector if \(\left |\lambda \vec{a} \right | = 1\)

Now,

\(\left |\lambda \vec{a} \right | = 1\)

\(\boldsymbol{\Rightarrow }\)    \(\left |\lambda \right | \left |\vec{a} \right | = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\left |\vec{a} \right | = \frac{ 1 }{ \left |\lambda \right | }\)  . . . . . . . . . . . . [ λ ≠ 0 ]

\(\boldsymbol{\Rightarrow }\)   \(a = \frac{ 1 }{ \left |\lambda \right | }\)              . . . . . . . . . . . . . . . . . .  [\(\left |\vec{a} \right | = a\) ]

Therefore,  vector \(\lambda \vec{a}\) is a unit vector if \(a = \frac{ 1 }{ \left | \lambda \right |}\)

The correct answer is D.

 

 

Exercise 10.4

 

 

Q 1: Find \(\left | \vec{a }\times \vec{b} \right |\) , if \(\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}\) and \(\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}\)

 

Solution 1:

We have,

\(\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}\) and \(\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}\)

\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 7 & 7 \\ 3 & – 2 & 2 \end{vmatrix}\)

\(\vec{a} \times \vec{b}\) = \(\hat{i} \left ( – 14 + 14 \right ) – \hat{j} \left ( 2 – 21 \right ) + \hat{k} \left ( – 2 + 21 \right ) = 19 \hat{j} + 19 \hat{k}\)

Therefore,

\(\left |\vec{a} \times \vec{b} \right | = \sqrt{\left ( 19 \right ) ^{ 2} + \left ( 19 \right ) ^{2}} = \sqrt{ 2 \times \left ( 19 \right ) ^{ 2 }} = 19 \sqrt{2}\)

 

 

Q 2: Find a unit vector perpendicular to each of the vector \(\vec{a} + \vec{b} and \vec{a} – \vec{b}\), where \(\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}\)

 

Solution 2:

We have,

\(\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}\)

Therefore,

\(\vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j} , \vec{a} – \vec{b} = 2 \hat{i} + 4 \hat{k}\) \(\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}\) \(\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \hat{i} \left ( 16 \right ) – \hat{j} \left ( 16 \right ) + \hat{k} \left ( – 8 \right ) = 16 \hat{i} – 16 \hat{j} – 8 \hat{k}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{16 ^{2} + \left ( – 16 \right )^{2} + \left ( – 8 \right ) ^{ 2 }}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{ 2 ^{2} \times 8 ^{ 2} + 2 ^{2} \times 8 ^{2 } + 8 ^{2}}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = 8 \sqrt{ 2^{2} + 2 ^{2} + 1 } = 8 \sqrt{9} = 8 \times 3 = 24\)

Therefore, the unit vector perpendicular to each of the vectors \(\vec{a} + \vec{b} and \vec{a} – \vec{b}\) is given by the relation,

= \(\pm \frac{ \left ( \vec{a} + \vec{b} \right ) \times \left ( \vec{a} – \vec{b} \right )}{\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | } = \pm \frac{ 16 \hat{i} – 16 \hat{j} – 8 \hat{k}}{ 24 }\)

= \(\pm \frac{ 2 \hat{i} – 2 \hat{j} – \hat{k} }{ 3 } = \pm \frac{ 2 }{ 3 } \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{ 3} \hat{k}\)

 

 

Q 3: If a unit vector \(\vec{a}\) makes angle \(\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}\) and an acute angle θ with \(\vec{k}\) , then find θ and hence, the compounds of \(\vec{a}\)

 

Solution 3:

Let unit vector \(\vec{a}\) have ( a 1 , a 2 , a 3 ) componenets

\(\vec{a} = a _{1} \hat{i} + a _{2} \hat{ j} + a _{ 3 } \hat{ k }\)

Since, \(\vec{ a }\) is a unit vector , \(\left |\vec{a} \right | = 1\)

Also, it is given that \(\vec{a}\) makes angles \(\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}\) and an acute angle θ with \(\vec{k}\)

Then, we have :

\(\cos \frac{ \pi }{ 3} = \frac{ a _{1}}{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\)   \(\frac{ 1 }{ 2 } = a _{1} . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]\)

\(\cos \frac{ \pi }{ 4 } = \frac{ a _{2 } }{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\)   \(\frac{ 1 }{ \sqrt{ 2 }} = a _{ 2 } . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]\)

Also , \(\cos \theta = \frac{ a _{ 3 }}{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\)   \(a _{ 3 } = \cos \theta\)

\(\left |a \right | = 1\)

\(\boldsymbol{\Rightarrow }\)    \(\sqrt{ {a_{ 1 }}^{ 2 } + {a_{ 2 }}^{ 2 } + {a_{ 3 }}^{ 2 } } = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\frac{ 1 }{ 2 }^{ 2 } + \frac{ 1 }{ \sqrt{ 2 }}^{ 2 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\frac{ 1 }{ 4 } + \frac{ 1 }{ 2 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\frac{ 3 }{ 4 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\)   \(\cos ^{ 2 } \theta = 1 – \frac{ 3 }{ 4 } = \frac{ 1 }{ 4 }\)

\(\boldsymbol{\Rightarrow }\)   \(\cos \theta = \frac{ 1 }{ 2 } \\ \theta = \cos^{-1} \left ( \frac{ 1 }{ 2 } \right ) \\ \theta = \frac{ \pi }{ 3 }\)

Therefore,

\(a_{ 3 } = \cos \frac{ \pi }{ 3 } = \frac{1}{ 2 }\)

Therefore, \(\theta = \frac{ \pi }{ 3} and the components of \vec{a} are \left ( \frac{ 1 }{ 2} , \frac{ 1 }{ \sqrt{ 2}} , \frac{ 1 }{2} \right )\)

 

 

Q 4: Show that:

\(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )\)

 

Solution:

To prove:

\(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )\)

= \(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = \left ( \vec{a} -\vec{ b} \right ) \times \vec{a} + \left ( \vec{a} – \vec{b} \right ) \times \vec{b}\)  . . . . . . . . . . . . [ By distributivity of vector product over addition ]

= \(\vec{a} \times \vec{a} – \vec{b} \times \vec{a} + \vec{a} \times \vec{b} – \vec{b} \times \vec{b}\) . . . . . . . . . . . . . . . [ again, by distributivity of vector product over addition ]

= \(\vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} – \vec{0}\)

= \(2 \vec{a} \times \vec{b}\)

 

 

Q 5: Find \(\lambda and \mu if \left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}\)

 

Solution:

\(\left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}\)

\(\boldsymbol{\Rightarrow }\)    \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda \mu & \end{vmatrix}\) = \(0 \hat{i} + 0 \hat{j} + 0 \hat{k}\)

\(\boldsymbol{\Rightarrow }\)   \(\hat{i} \left ( 6 \mu – 27 \mu \right ) – \hat{j} \left ( 2 \mu – 27 \right ) + \hat{k} \left ( 2 \lambda – 6 \right ) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}\)

On comparing the corresponding components, we have :

\(6 \mu – 27 \lambda = 0 \\ 2 \mu – 27 = 0 \\ 2 \lambda – 6 = 0\)

Now,

\(2 \lambda – 6 = 0 \Rightarrow \lambda = 3\)

\(\boldsymbol{\Rightarrow }\)   \(2 \mu – 27 = 0 \\ \Rightarrow \mu = \frac{ 27 }{ 2 }\)

Therefore, \(\lambda = 3 and \mu = \frac{ 27 }{ 2 }\)

 

 

Q 6: Given that:

\(\vec{a}.\vec{b} = 0 and \vec{a} \times \vec{b} = \vec{0}\)

What can you conclude about the vectors \(\vec{a} and \vec{b}\) ?

 

Solution:

\(\vec{a}.\vec{b} = 0\)

Then ,

  1. i) either \(\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \perp \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0\)
  2. ii) Either \(\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \parallel \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0\)

But, \(\vec{a} and \vec{b}\) cannot be perpendicular and parallel simultaneously.

Therefore, \(\left |\vec{a} \right | = 0 or \left |\vec{b} \right | = 0\)

 

 

Q 7: Let the vectors \(\vec{a }, \vec{b} ,\vec{ c} given as a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}\)

Then show that = \(\vec{a} \times \left ( \vec{ b } + \vec{ c } \right ) = \vec{a} \times \vec{b} + \vec{a }\times \vec{ c }\)

 

Solution:

We have,

\(a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}\) \(\left ( \vec{b} + \vec{c} \right ) = \left ( b_{ 1 } + c _{1} \right ) \hat{i} + \left ( b_{ 2 } + c_{ 2 } \right ) \hat{j} + \left ( b_{ 3 } + c_{ 3 } \right ) \hat{k}\)

Now, \(\vec{a} \times \left ( \vec{b} + \vec{c} \right )\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{ 1 } & a_{ 2} & a_{ 3 } \\ b_{1} + c_{1} & b_{2} + c_{ 2 } & b_{ 3 } + c_{ 3 } \end{vmatrix}\)

\(= \hat{i} \left [ a_{ 2} \left ( b_{ 3 } + c_{3 } \right ) – a_{ 3 } \left ( b_{ 2 } + c_{ 2 } \right ) \right ] – \hat{j} \left [ a_{ 1 } \left ( b_{ 3 } + c_{ 3 } \right ) – a_{ 3 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ] + \hat{k} \left [ a_{ 1 } \left ( b_{ 2 } + c _{ 2 }\right ) – a_{ 2 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ]\)

\(= \hat{i} \left [ a_{ 2 } b_{3} + a_{2} c_{3} – a_{3} b_{2} – a_{3} c_{2} \right ] + \hat{j} \left [ – a_{1} b_{3} – a_{1} c_{3} + a_{3} b_{1} + a_{3} c_{1} \right ] + \hat{k} \left [ a_{1} b_{2} + a_{1} c_{2} – a_{2}b _{1} – a_{2} c_{1} \right ]\) . . . . . . . . . . . . . . . . .  ( 1 )

\(\boldsymbol{\Rightarrow }\)   \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1 } & a_{2} & a_{3} \\ b_{ 1} & b_{ 2} & b_{3} \end{vmatrix}\)

= \(\hat{i} \left [ a_{ 2 } b_{ 3 } – a_{ 3 } b_{2} \right ] + \hat{j} \left [ a_{3} b_{1} – a_{1} b_{3} \right ] + \hat{k} \left [ a_{1} b_{2} – a_{2} b_{1} \right ]\)     . . . . . . . . . . . . . . . . . . . . . . . . . .  ( 2 )

\(\boldsymbol{\Rightarrow }\)   \(\vec{a} \times \vec{ c } = \begin{vmatrix} \hat{ I } & \hat{ j } & \hat{ k } \\ a_{1 } & a_{2} & a_{3} \\ c_{ 1} & c_{ 2 } & c_{3 } \end{vmatrix}\)

\(\hat{i} \left [ a_{ 2 } c_{ 3 } – a_{ 3 } c_{2} \right ] + \hat{j} \left [ a_{3} c_{1} – a_{1} c_{3} \right ] + \hat{k} \left [ a_{1} c_{2} – a_{2} c_{1} \right ]\)  . . . . . . . . . . . . . . . . . . . . . . . . . .  ( 3 )

On adding (2) and (3), we get:

\(\left ( \vec{a} \times \vec{b} \right ) + \left ( \vec{a} \times \vec{ c} \right ) = \hat{i} \left [ a_{ 2 } b_{ 3 } + a_{ 2 } c_{ 3 } + – a _{ 3 } b_{ 2 } – a_{ 3 } c_{ 2 } \right ] + \hat{j} \left [ a_{ 3 } b_{ 1 } + a_{ 3 } c_{ 1 } + – a_{ 1 } b_{ 3 } – a_{ 1 } c_{ 3 } \right ] + \hat{ k } \left [ a_{ 1 } b_{ 2 } + a_{ 1 } c_{ 2 } + – a_{ 2 } b_{ 1 } – a_{ 2 } c_{ 1 } \right ]\) . . . . . . . . . . . . . . . . . . . . . . . . . .  ( 4 )

Now, from (1) and (4), we have:

\(\vec{a} \times \left ( \vec{b} + \vec{c} \right ) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\)

Hence, the given result is proved.

 

 

Question 8: Suppose, any of the vector \(\overrightarrow{m} \;or\; \overrightarrow{n} = \overrightarrow{0}\), then \(\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}\).

Is the above given statement true?

Give reason in support to your answer with an example.

 

Answer 8:

By taking any two non – zero vectors, for the condition \(\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}\).

Suppose, \(\overrightarrow{m} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \\ Then,\)

\(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 4 & 6 \end{vmatrix}\) \(\hat{i} (4 – 4) – \hat{j} (6 – 6) + \hat{k} (4 – 4) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{0} \\ Now,\; we\; find\; that, \\ \left [ \overrightarrow{m} \right ] = \sqrt{(1) ^{2} + (2) ^{2} + (3) ^{2}} = \sqrt{14} \\ So,\; \overrightarrow{m} \neq \overrightarrow{0} \\ \left [ \overrightarrow{n} \right ] = \sqrt{(2) ^{2} + (4) ^{2} + (6) ^{2}} = \sqrt{56} \\ So,\; \overrightarrow{n} \neq \overrightarrow{0}\)

Thus, the above given statement is not true.

 

 

Question 9: P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5) are the vertices of a triangle. Obtain the area.

 

Answer 9:

Given:

Vertices of a triangle P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5)

The contiguous sides \(\overrightarrow{PQ} \;and\; \overrightarrow{QR}\) of the triangle is given as,

\(\overrightarrow{PQ} = (2 – 1) \hat{i} + (3 – 1) \hat{j} + (5 – 2) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{QR} = (1 – 2) \hat{i} + (5 – 3) \hat{j} + (5 – 5) \hat{k} = – \hat{i} + 2 \hat{j} \\ Area\; of\; \Delta \;triangle = \frac{1}{2} \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right |\) \(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ – 1 & 2 & 0 \end{vmatrix} \\ \hat{i} (- 6) – \hat{j} (3) + \hat{k} (2 + 2) = – 6 \hat{i} -3 \hat{j} + 4 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 6) ^{2} + (- 3) ^{2} + 4 ^{2}} = \sqrt{36 + 9 + 16} = \sqrt{61}\)

Thus, \(\frac{\sqrt{61}}{2}\) is the area of triangle ABC.

 

 

Question 10: \(\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}\) are the adjacent sides of a vector. Obtain the area of the parallelogram.

Answer 10:

The area of the parallelogram whose contiguous sides are \(\overrightarrow{m} \;and\; \overrightarrow{n} is \left | \overrightarrow{m} \times \overrightarrow{n} \right |\)

Given,

\(\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}\) are the adjacent sides of a vector.

\(\overrightarrow{m} \times \overrightarrow{n} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 1 & 3 \\ 2 & – 7 & 1 \end{vmatrix} \\ \hat{i} (- 1 + 21) – \hat{j} (1 – 6) + (- 7 + 2) \hat{k} = 20 \hat{i} + 5 \hat{j} – 5 \hat{k} \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = \sqrt{20 ^{2} + 5 ^{2} + 5 ^{2}} = \sqrt{400 + 25 + 25} = 15 \sqrt{2}\)

\(15 \sqrt{2}\) is the area of the given parallelogram.

 

 

Question 11: Suppose, vectors \(\overrightarrow{m} \;and\; \overrightarrow{n} \; in\; such\; a\; way\; that \left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; \overrightarrow{m} \times \overrightarrow{n}\) is a unit vector, suppose the angle between the two vectors is

\((i)\; \frac{\pi }{6}, \;(ii)\; \frac{\pi }{4},\; (iii)\; \frac{\pi }{3} \;(iv) \frac{\pi }{2}\)

 

Answer 11:

Given, \(\left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; as\; we\; know\; \overrightarrow{m} \times \overrightarrow{n} = 1 \\ \overrightarrow{m} \times \overrightarrow{n} = \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | \;sin \theta \hat{s}\), where s is a unit vector perpendicular to both \(\overrightarrow{m} \;and\; \overrightarrow{n}\) and \(\Theta\) is the angle between \(\overrightarrow{m} \;and\; \overrightarrow{n}\)

Now, \(\overrightarrow{m} \;and\; \overrightarrow{n} \;is\; a \;unit \; vector \;if\; \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \hat{s} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \right | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sin\; \Theta = 1 \\ sin\; \Theta = \frac{1}{\sqrt{2}} \\ \Theta = \frac{\pi }{4}\)

The correct answer is option (ii)

 

 

Question 12: The area of the rectangle with P, Q, R and S as the vertices with positive vectors \(– \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \; respectively\; is \\ (i) \frac{1}{2}, \;\;\;\; (ii) 1 \\ (iii) 2, \;\;\;\; (iv) 4\)

 

Answer 12:

Given,

The area of the rectangle PQRS with P, Q, R and S as the vertices with positive vectors such as:

\(\overrightarrow{OP} = – \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k},\; \overrightarrow{OQ} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \;\overrightarrow{OR} = \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; \overrightarrow{OS} – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k}\)

The contiguous sides \(\overrightarrow{PQ} \;and\; \overrightarrow{QR}\) of the rectangle is given as,

\(\overrightarrow{PQ} = (1 + 1) \hat{i} + (\frac{1}{2} – \frac{1}{2}) \hat{j} + (4 – 4) \hat{k} = 2 \hat{i} \;and\; \overrightarrow{QR} = (1 – 1) \hat{i} + \left (- \frac{1}{2} – \frac{1}{2} \right ) \hat{j} + (4 – 4) \hat{k} = – \hat{j}\) \(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & – 1 & 0 \end{vmatrix} \\ \hat{k} (- 2) = – 2 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 2) ^{2}} = 2\)

Area of a rectangle is 2 square units.

Option (iii) is the correct answer.

 

Join BYJU’S NCERT Learning Program

NCERT Solutions

NCERT Solutions Maths

NCERT Solutions