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Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.5
  • Last Updated : 18 Mar, 2021

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

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