RD Sharma Class 12 Ex 23.5 Solutions Chapter 23 Algebra of Vectors

Here we provide RD Sharma Class 12 Ex 23.5 Solutions Chapter 23 Algebra of Vectors for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 23.5 Solutions Chapter 23 Algebra of Vectors book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter23
Exercise23.5
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 23.5 Solutions Chapter 23 Algebra of Vectors

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  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  be the position vector whose tip is (-5,3), find the coordinates of a point B such that  , 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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