Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.5

The algebra of vectors is a branch of mathematics that deals with vector quantities—objects defined by both magnitude and direction, unlike scalars, which have only magnitude. This field is essential in physics, engineering, and computer science for analyzing forces, velocities, and other vector quantities.

Question 1. If the position vector of a point (-4,-3) be \overrightarrow{a} , find |\overrightarrow{a}|.

Solution:

We have, 

\overrightarrow{a} = -4\hat i - 3\hat j \\|\overrightarrow{a}| = \sqrt[]{(-4)^2+ (-3)^2} \\ = \sqrt[]{16+9} \\ = \sqrt[]{25} \\= 5 \\|\overrightarrow{a}| = 5 

Question 2. If the position vector  \overrightarrow{a} of a point (12,n) is such that |\overrightarrow{a}|= 13 , find the value(s).

Solution:

We have, 

\overrightarrow{a} = 12\hat i + n \hat j \\ |\overrightarrow{a}| = \sqrt[]{12^2 + n^2} \\ 13 = \sqrt[]{144 + n^2}

On squaring both sides, 

(13)^2 = (\sqrt[]{144 + n^2})^2 \\169 = 144 + n^2 \\n^2 = 169 - 144 \\n^2 = 25 \\n^2 = \pm \sqrt[]{25} \\n = \pm 5

Question 3. Find a vector of magnitude 4 units which is parallel to the vector \sqrt[]{3} \hat i + \hat{j} .

Solution: 

Given, 

\overline{a} = \sqrt{3 \hat i} + \hat{j}

Let \overline{b} is a vector parallel to \overline{a} Therefore, \overline{b} = \lambda{\overline{a}} for any scalar \\ = \lambda(\sqrt{3} \hat i + \lambda \hat j) \\ = \overline{b} = \lambda \sqrt{3} \hat i + \lambda \hat j \\ |\overline{b}| = \sqrt{(\lambda \sqrt{3})^2+(\lambda)^2} \\ = \sqrt{4 \lambda^2} \\ |\overline{b}| = 2\lambda \\ 4 = 2\lambda \\ \lambda = 2 \\ \overline{b}= \lambda \sqrt{3} \hat i + \lambda \hat j \\ \overline{b}= 2 \sqrt{3} \hat i + \lambda \hat j

Question 4. Express \overlinearrow{AB} in terms of unit vectors (i)A = (4,-1),B = (1,3) (ii)A = (-6,3) , B = (-2,-5)

Solution:

(i) We have, A = (4,-1) B = (1,3) Position Vector of A = 4\hat i - \hat j Position Vector of B = \hat i + 3 \hat j Now, \overlinearrow {AB} = Position Vector of B - Position Vector of A \\ ( \hat i + 3 \hat j - 4\hat i + \hat j) \\ \overlinearrow{AB} = -3\hat i + 4 \hat j \\ |\overlinearrow{AB}| = \sqrt{(-3)^2 + (4)^2} \\ = \sqrt{9+16} \\ = \sqrt{25} \\ |\overlinearrow{AB}| = 5 Therefore, \overlinearrow{AB} = -3\hat i + 4 \hat j (ii) We have, A = (-6,3) B = (-2,-5) Position Vector of A = -6\hat i + 3\hat j Position Vector of B = -2\hat i - 5\hat j Now, \overlinearrow {AB} = Position Vector of B - Position Vector of A \\ (-2\hat i -5 \hat j) - (-6\hat i + 3\hat j) \\ \overlinearrow{AB} = 4\hat i - 8\hat j \\ |\overlinearrow{AB}| = \sqrt{(4)^2 + (-8)^2} \\ = \sqrt{16+64} \\ = \sqrt{80} \\ = \sqrt{16*5} \\ |\overlinearrow{AB}| = 4 \sqrt{5} Therefore, \overlinearrow{AB} = 4\hat i - 8\hat j

Question 5. Find the coordinates of the tip of the position vector which is equivalent to \overrightarrow{AB}, where the coordinates of A and B are (-1,3) and (-2,1)

Solution:

We have, 

A = (-1,3)

B = (-2,1)

Now, 

Position Vector of A = -\hat i + 3 \hat j 

Position Vector of -2 \hat i + 1\hat j

Therefore, 

\overrightarrow{AB} = Position Vector of B - Position Vector of A \\ = (-2\hat i + \hat j) - (-\hat i + 3 \hat j) \\ = -2 \hat i + \hat j + \hat i -3 \hat j \\ = -\hat i - 2 \hat j

Coordinate of the position vector \overrightarrow{AB} =  -\hat i - 2 \hat j

Question 6. ABCD is a parallelogram. If the coordinates of A,B,C are (-2,-1), (3,0),(1,-2) respectively, find the coordinates of D.

Solution:

Here, A = (-2,-1)

B = (3,0)

C = (1,-2)

Let us assume D be (x , y).

Computing Position Vector of AB, we have,

= Position Vector of B - Position Vector of A

= (3 \hat i) - (-2 \hat i - \hat j) \\ \overrightarrow{AB} = 5 \hat i + \hat j \\ \overrightarrow{DC} = Position Vector of C - Position Vector of D \\ = ( \hat i - 2 \hat j) - (x \hat i + y \hat j) \\ = \hat i - 2 \hat j - x \hat i - y \hat j \\ \overrightarrow{DC} = (1-x)\hat i + (-2-y)\hat j

Comparing LHS and RHS of both, 

5 = 1-x 

x = -4 

And, 

1 = -2-y

y = -3

So, coordinates of D = (-4,-3).

Question 7. If the position vectors of the points A(3,4), B(5,-6) and C(4,-1) are \vec{a}, \vec{b}, \vec{c} respectively, compute the value of \vec{a} + 2\vec{b} - 3\vec{c}.

Solution:

Computing the position vectors of all the points we have, 

\vec{a} = 3\hat{i} + 4\hat{j} \\ \vec{b} = 5\hat{i} -6\hat{j} \\ \vec{c} = 4\hat{i} - \hat{j}

Now, 

Computing the final value after substituting the values, 

\vec{a} + 2\vec{b}-3\vec{c} = (3\hat i + 4\hat j ) + 2(5 \hat i - 6 \hat j) - 3(4 \hat i - \hat j) \\ = 3 \hat i + 4 \hat j + 10 \hat i - 12 \hat j -12 \hat i + 3 \hat j \\ = \hat i - 5 \hat j \\Therefore, \\ \vec{a} + 2\vec{b}-3\vec{c} = \hat i - 5 \hat j

Question 8. If \vec {a} be the position vector whose tip is (-5,3), find the coordinates of a point B such that \overrightarrow{AB} = \vec {a} , the coordinates of A being (-4,1).

Solution:

Given, Coordinate of A = (4,-1) Position vector of A = 4\hat i - \hat j Position vector of \vec{a} = 5\hat i - 3\hat j Let coordinate of point B = (x, y) Position vector of B = x\hat i + y\hat j Given that, \overrightarrow{AB} = \vec{a} Position vector of B - Position vector of A = \vec{a} (x\hat i + y\hat j) - (4\hat i - \hat j) = 5\hat i - 3\hat j \\ (x - 4) \hat i + (y+1)\hat j = 5 \hat i - 3 \hat j Comparing the coefficients of LHS and RHS x - y = 5 x = 9 Also, y + 1 = 3 y = -1 So, coordinate of B = (9,-4)

Question 9. Show that the points 2 \hat i , -\hat i -4 \hat j and -\hat i+4 \hat j form an isosceles triangle. 

Solution:

|\overrightarrow{AB}| = 5 units \\ |\overrightarrow{BC}| = \sqrt[]{8^2} \\ |\overrightarrow{BC}| = 8 units \\ |\overrightarrow{AC}| = \sqrt[]{(-3)^2 + (8)^2} \\ = \sqrt[]{9+16} \\ = \sqrt[]{25} \\Here, \\|\overrightarrow{AB}| = |\overrightarrow{AC}|

So, the two sides AB and AC of the triangle ABC are equal. 

Therefore, ABC is an isosceles triangle. 

Question 10. Find a unit vector parallel to the vector \hat i + \sqrt{3} \hat j .

Solution:

We have, 

Let \vec {a} = \hat i + \sqrt{3} \hat j

Suppose \vec{a} is any vector parallel to \vec{a}

\vec{b} = λ \vec{a} , where λ is any scalar.

= λ(\hat i + \sqrt{3} \hat j)

\vec{b} = λ(\hat i + \sqrt{3} \hat j)

Unit vector of

\vec{b}= \frac{\vec{b}}{|\vec{b}|} \\ \hat{b} = \frac{\hat i + \sqrt[]{3} \hat j}{2} \\ \hat{b} = \frac{1}{2}(\hat i + \sqrt[]{3} \hat j) Therefore, \\ \hat{b} = \frac{1}{2}\hat i + \frac{\sqrt{3}}{2} \hat j

Question 11. Find the components along the coordinate axes of the position vector of each of the following points : 

(i) P(3,2)

(ii) Q(-5,1)

(iii) R(-11,-9)

(iv) S(4,-3)

Solution:

(i) Given, P = (3,2)

Position vector of P = 3\hat i + 2\hat j

Component of P along x-axis = 3 \hat i

Component of P along y-axis = 2 \hat j

(ii) Given, Q = (-5,1)

Position vector of Q = -5\hat i + \hat j

Component of Q along x-axis = -5 \hat i

Component of Q along y-axis =\hat j

(iii) Given, R = (-11,-9)

Position vector of R =-11\hat i -9 \hat j

Component of R along x-axis =-11 \hat i

Component of R along y-axis = -9 \hat j

(iv) Given, S = (4,-3)

Position vector of S =4\hat i -3 \hat j

Component of S along x-axis =4 \hat i

Component of S along y-axis = -3 \hat j

Related Article:

Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.6

Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.4

Summary

Exercise 23.5 in Chapter 23 typically covers the topic of vector triple product. The vector triple product is a vector operation that involves three vectors and two cross product operations. It's usually denoted as a × (b × c), where a, b, and c are vectors.

The vector triple product is not associative: a × (b × c) ≠ (a × b) × c

The expansion formula for vector triple product is:

a × (b × c) = (a • c)b - (a • b)c

This formula is also known as Lagrange's formula or the BAC-CAB rule

The vector triple product can be used to solve various problems in physics and engineering

Practice Questions

1). Given vectors a = 2i + 3j + k, b = i + 2j - k, and c = 3i - j + 2k, calculate a × (b × c).

2). Prove that a × (b × c) + b × (c × a) + c × (a × b) = 0 for any vectors a, b, and c.

3). If a • (b × c) = 0, what can you conclude about the vectors a, b, and c?

4). Show that (a × b) • (c × d) = (a • c)(b • d) - (a • d)(b • c).

5). Prove that a × (b × c) = (a • c)b - (a • b)c using the properties of dot and cross products.

6). If a, b, and c are mutually perpendicular unit vectors, evaluate a × (b × c).

7). Given that a • b = 2, b • c = 3, c • a = 4, and |a| = 2, |b| = 3, |c| = 4, calculate the scalar triple product a • (b × c).

8). Prove that a × (b × c) = (a × b) × c if and only if a, b, and c are coplanar.

9). If a × (b × c) = 0, what can you conclude about vectors a, b, and c?

10). Given vectors a = i - 2j + 3k, b = 2i + j - k, and c = -i + 3j + 2k, calculate (a × b) • (b × c).