NCERT Solutions For Class 12 Maths Chapter 10

NCERT Solutions Class 12 Maths Vector Algebra

NCERT Solutions for class 12 maths chapter 10 Vector Algebra covers topics such as how to differentiate vectors and scalars, functions on vectors, applying vectors to figures and more. For students who are preparing for competitive exams also need have a good idea about the topic of vector algebra, thus NCERT Solutions for class 12 maths chapter 10 pdf will be extremely helpful. With the help of NCERT Solutions for class 12 maths chapter 10, students will also be able to understand the topic in future references if they hope to take an engineering degree as well.

NCERT Solutions Class 12 Maths Chapter 10 Exercises

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.