Vector Algebra - Exemplar Solutions

 CBSE Class–12 Mathematics

NCERT Exemplar
Chapter - 10
Vector Algebra - Short Answer Questions

 

---

1. Find the unit vector in the direction of the sum of the vectors and 

Sol. Let denote the sum of and . We have

Now 

Thus, the required unit vector is 

2. Find a vector of magnitude 11 in the direction opposite to that of , where are the points respectively.

Sol. The vector with initial point and terminal point is given by

Thus, 

Therefore, unit vector in the direction of is given by

Hence, the required vector of magnitude 11 in direction of is

3. Find the position vector of a point R which divides the line joining the two points P and Q with position vectors and , respectively, in the ratio , (i) internally and (ii) externally.

Sol. (i) The position vector of the point R dividing the join of P and Q internally in the ratio 1:2 is given by

(ii) The position vector of the point R’ dividing the join of P and Q in the ratio externally is given by

4. If the points are collinear, find the value of .

Sol. Let the given points be . Then

And 

Since are collinear, we have 

(solving equations)

Therefore, .

5. Find a vector of magnitude units which makes an angle of and with respectively.

Sol. Here and 

Therefore, gives

Hence, the required vector is given by

6. If and find such that is perpendicular to 

Sol. We have

Since 

7. Find all vectors of magnitude that are perpendicular to the plane of and 

Sol. Let and Then

a→×b→=∣∣∣∣∣i^1−1j^23k^14∣∣∣∣∣=i^(8−3)−j^(4+2)+k^(3+2)=5i^−5j^+5k^$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ -1& 3& 4\end{array}\right|=\hat{i}(8-3)-\hat{j}(4+2)+\hat{k}(3+2)=5\hat{i}-5\hat{j}+5\hat{k}$

Therefore, unit vector perpendicular to the plane of and is given by

Hence, vectors of magnitude of that are perpendicular to plane of and are 

 

Long Answer Questions

 

---

8. Using vectors, prove that .

Sol. Let and be unit vectors making angles A and B, respectively, with positive direction of . Then [Fig. 10.1]

We know and OQ−→−=ON−→−+NQ−→−=i^cosB+j^sinB$\overrightarrow{OQ}=\overrightarrow{ON}+\overrightarrow{NQ}=\hat{i}\cos B+\hat{j}\sin B$

By definition (A.−→B→=ABcosθ)$\overrightarrow{(A.}\overrightarrow{B}=AB\cos \theta )$

In terms of components, we have

From (1) and (2), we get

.

9. Prove that in a where represents the magnitude of the sides opposite to vertices respectively.

Sol. Let the three sides of the triangle BC, CA and AB be represented by , respectively [Fig. 10.2]

We have 

Which pre-cross multiplying by , and post cross multiplying by , gives

respectively. Therefore,

⇒∣∣∣a→×b→∣∣∣=∣∣b→×c→∣∣=∣∣∣c→×b→∣∣∣$\Rightarrow \left|a^{\to }\times \overrightarrow{b}\right|=\left|b^{\to }\times \overrightarrow{c}\right|=\left|c^{\to }\times \overrightarrow{b}\right|$

Dividing by abc, we get

 

Objective Type Questions

 

---

Choose the correct answer from the given the four options in each of the Examples 10 to 21.

10. The magnitude of the vector is

(A) 5

(B) 7

(C) 12

(D) 1

Sol. (B) is the correct answer.

11 The position vector of the point which divides the join of points with position vectors and in the ratio 1: 2 is

(A) 

(B) 

(C) 

(D) 

Sol. (D) is the correct answer. Applying section formula, the position vector of the required point is 

12. The vector with initial point and terminal point is

(A) 

(B) 

(C) 

(D) None of these

Sol. (A) is the correct answer.

13. The angle between the vectors and is

(A) 

(B) 

(C) 

(D) 

Sol. (B) is the correct answer. Apply in formula 

14. The value of for which the two vectors and are perpendicular is

(A) 2

(B) 4

(C) 6

(D) 8

Sol. (D) is the correct answer.

15. The area of the parallelogram whose adjacent sides are and is

(A) 

(B) 

(C) 3

(D) 4

Sol. (B) is the correct answer. Area of the parallelogram whose adjacent sides are and is.

16. If and then value of is

(A) 

(B) 

(C) 

(D) None of these

Sol. (C) is the correct answer. Using the formula we get 

Therefore, 

17. The 2 vectors and represents the two sides AB and AC, respectively of a . The length of the Median through A is

(A) 

(B) 

(C) 

(D) None of these

Sol. (A) is the correct answer. Median is given by

18. The projection of vector along is

(A) 

(B) 

(C) 2

(D) 

Sol. (A) is the correct answer. Projection of a vector on is

19. If and are unit vector, then what is the angle between and for be a unit vector?

(A) 

(B) 

(C) 

(D) 

Sol. (A) is the correct answer. We have

20. The unit vector perpendicular to the vectors and forming a right-handed system is

(A) 

(B) 

(C) 

(D) 

Sol. (A) is the correct answer. Required unit vector is 

21. If and then lies in the interval

(A) 

(B) 

(C) 

(D) 

Sol. (A) is the correct answer. The smallest value of will exist at numerically smallest value of k, i.e., at k = 0, which gives 

The numerically greatest value of is 2 at which