# Ncert Solutions For Class 12 Maths Ex 10.2

## Ncert Solutions For Class 12 Maths Chapter 10 Ex 10.2

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

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

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?

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

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?

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

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.

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.

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.

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.

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 $$\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \; and\; \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k}$$.

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.

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.

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 $$2 \hat{i} – 4 \hat{j} + 6 \hat{k}$$

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

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.

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

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

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.

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.

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?

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

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.

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

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.