NCERT Solutions Class-12-Chapter-10-Vector Algebra
Excercise-10.3
Note:
Scalar (Dot) Product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.
Projection: The projection of vector \( \vec{a} \) on \( \vec{b} \) is given by \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
Perpendicularity Condition: Two non-zero vectors are perpendicular if \( \vec{a} \cdot \vec{b} = 0 \).
Q1
Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( \sqrt{3} \) and 2, respectively having \( \vec{a} \cdot \vec{b} = \sqrt{6} \).
Solution:
Given \( |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = \sqrt{6} \).
Using the formula \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \):
\( \cos \theta = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{3} \times \sqrt{2}}{\sqrt{3} \times 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \).
Since \( \cos \theta = \frac{1}{\sqrt{2}} \), the angle \( \theta = \frac{\pi}{4} \).
Q2
Find the angle between the vectors \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \).
Solution:
Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \).
\( \vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10 \).
\( |\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \).
\( |\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \).
\( \cos \theta = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7} \).
\( \theta = \cos^{-1}\left(\frac{5}{7}\right) \).
Q3
Find the projection of the vector \( \hat{i} - \hat{j} \) on the vector \( \hat{i} + \hat{j} \).
Solution:
Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \).
Projection of \( \vec{a} \) on \( \vec{b} \) is \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
\( \vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0 \).
Magnitude \( |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).
Projection = \( \frac{0}{\sqrt{2}} = 0 \).
Q4
Find the projection of the vector \( \hat{i} + 3\hat{j} + 7\hat{k} \) on the vector \( 7\hat{i} - \hat{j} + 8\hat{k} \).
Solution:
Let \( \vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} \) and \( \vec{b} = 7\hat{i} - \hat{j} + 8\hat{k} \).
Projection = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
\( \vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60 \).
\( |\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114} \).
Projection = \( \frac{60}{\sqrt{114}} \).
Q5
Show that each of the given three vectors is a unit vector:
\( \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}), \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}), \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \)
Also, show that they are mutually perpendicular to each other.
\( \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}), \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}), \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \)
Also, show that they are mutually perpendicular to each other.
Solution:
Let the vectors be \( \vec{a}, \vec{b}, \vec{c} \).
Magnitude Check:
\( |\vec{a}| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{4+9+36} = \frac{1}{7}\sqrt{49} = 1 \).
Similarly, \( |\vec{b}| = 1 \) and \( |\vec{c}| = 1 \). So, they are unit vectors.
Perpendicularity Check:
\( \vec{a} \cdot \vec{b} = \frac{1}{49}[2(3) + 3(-6) + 6(2)] = \frac{1}{49}[6 - 18 + 12] = 0 \).
\( \vec{b} \cdot \vec{c} = \frac{1}{49}[3(6) + (-6)(2) + 2(-3)] = \frac{1}{49}[18 - 12 - 6] = 0 \).
\( \vec{c} \cdot \vec{a} = \frac{1}{49}[6(2) + 2(3) + (-3)(6)] = \frac{1}{49}[12 + 6 - 18] = 0 \).
Since the dot products are zero, they are mutually perpendicular.
Q6
Find \( |\vec{a}| \) and \( |\vec{b}| \), if \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8 \) and \( |\vec{a}| = 8|\vec{b}| \).
Solution:
Expanding the dot product:
\( \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8 \).
\( |\vec{a}|^2 - |\vec{b}|^2 = 8 \).
Given \( |\vec{a}| = 8|\vec{b}| \), substitute this into the equation:
\( (8|\vec{b}|)^2 - |\vec{b}|^2 = 8 \Rightarrow 64|\vec{b}|^2 - |\vec{b}|^2 = 8 \Rightarrow 63|\vec{b}|^2 = 8 \).
\( |\vec{b}|^2 = \frac{8}{63} \Rightarrow |\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}} \).
\( |\vec{a}| = 8 \times \sqrt{\frac{8}{63}} = \frac{16\sqrt{2}}{3\sqrt{7}} \).
Q7
Evaluate the product \( (3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) \).
Solution:
\[ \begin{aligned} &= 3\vec{a} \cdot 2\vec{a} + 3\vec{a} \cdot 7\vec{b} - 5\vec{b} \cdot 2\vec{a} - 5\vec{b} \cdot 7\vec{b} \\ &= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) - 10(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2 \\ &= 6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2 \end{aligned} \]
Q8
Find the magnitude of two vectors \( \vec{a} \) and \( \vec{b} \), having the same magnitude and such that the angle between them is \( 60^\circ \) and their scalar product is \( \frac{1}{2} \).
Solution:
Given \( |\vec{a}| = |\vec{b}| \), \( \theta = 60^\circ \), and \( \vec{a} \cdot \vec{b} = \frac{1}{2} \).
Using formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \):
\( \frac{1}{2} = |\vec{a}| |\vec{a}| \cos 60^\circ \).
\( \frac{1}{2} = |\vec{a}|^2 \cdot \frac{1}{2} \).
\( |\vec{a}|^2 = 1 \Rightarrow |\vec{a}| = 1 \).
So, \( |\vec{a}| = |\vec{b}| = 1 \).
Q9
Find \( |\vec{x}| \), if for a unit vector \( \vec{a} \), \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \).
Solution:
Given \( |\vec{a}| = 1 \).
Expanding the expression:
\( \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12 \).
\( |\vec{x}|^2 - |\vec{a}|^2 = 12 \).
\( |\vec{x}|^2 - 1 = 12 \Rightarrow |\vec{x}|^2 = 13 \).
\( |\vec{x}| = \sqrt{13} \).
Q10
If \( \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} \), \( \vec{b} = -\hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{c} = 3\hat{i} + \hat{j} \) are such that \( \vec{a} + \lambda \vec{b} \) is perpendicular to \( \vec{c} \), then find the value of \( \lambda \).
Solution:
\( \vec{a} + \lambda \vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k}) \).
\( = (2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k} \).
Since this vector is perpendicular to \( \vec{c} = 3\hat{i} + \hat{j} \), their dot product is 0.
\( (2-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0 \).
\( 6 - 3\lambda + 2 + 2\lambda = 0 \).
\( 8 - \lambda = 0 \Rightarrow \lambda = 8 \).
Q11
Show that \( |\vec{a}|\vec{b} + |\vec{b}|\vec{a} \) is perpendicular to \( |\vec{a}|\vec{b} - |\vec{b}|\vec{a} \), for any two nonzero vectors \( \vec{a} \) and \( \vec{b} \).
Solution:
Calculate the dot product:
\[ \begin{aligned} &= (|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} - |\vec{b}|\vec{a}) \\ &= (|\vec{a}|\vec{b})^2 - (|\vec{b}|\vec{a})^2 \\ &= |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2 \\ &= 0 \end{aligned} \]
Since the dot product is 0, the vectors are perpendicular.
Q12
If \( \vec{a} \cdot \vec{a} = 0 \) and \( \vec{a} \cdot \vec{b} = 0 \), then what can be concluded about the vector \( \vec{b} \)?
Solution:
\( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 0 \implies |\vec{a}| = 0 \implies \vec{a} = \vec{0} \).
If \( \vec{a} \) is a zero vector, then \( \vec{a} \cdot \vec{b} \) will always be 0 irrespective of what \( \vec{b} \) is.
Therefore, \( \vec{b} \) can be any vector.
Q13
If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \).
Solution:
Given \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \) and \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \).
Squaring the sum equation:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \).
\( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \).
\( 1 + 1 + 1 + 2S = 0 \Rightarrow 3 + 2S = 0 \).
\( S = -\frac{3}{2} \).
Q14
If either vector \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \cdot \vec{b} = 0 \). But the converse need not be true. Justify your answer with an example.
Solution:
The converse states: "If \( \vec{a} \cdot \vec{b} = 0 \), then either \( \vec{a} = 0 \) or \( \vec{b} = 0 \)".
This is false. Two non-zero vectors can have a zero dot product if they are perpendicular.
Example: Let \( \vec{a} = \hat{i} \) and \( \vec{b} = \hat{j} \).
\( \vec{a} \cdot \vec{b} = (1)(0) + (0)(1) + (0)(0) = 0 \).
Here, \( \vec{a} \cdot \vec{b} = 0 \), but neither \( \vec{a} \) nor \( \vec{b} \) is a zero vector.
Q15
If the vertices A, B, C of a triangle ABC are \( (1, 2, 3), (-1, 0, 0), (0, 1, 2) \), respectively, then find \( \angle ABC \).
Solution:
\( \angle ABC \) is the angle between vectors \( \vec{BA} \) and \( \vec{BC} \).
\( \vec{BA} = \vec{A} - \vec{B} = (1 - (-1))\hat{i} + (2 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k} \).
\( \vec{BC} = \vec{C} - \vec{B} = (0 - (-1))\hat{i} + (1 - 0)\hat{j} + (2 - 0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k} \).
\( \vec{BA} \cdot \vec{BC} = 2(1) + 2(1) + 3(2) = 2 + 2 + 6 = 10 \).
\( |\vec{BA}| = \sqrt{4+4+9} = \sqrt{17} \).
\( |\vec{BC}| = \sqrt{1+1+4} = \sqrt{6} \).
\( \cos(\angle ABC) = \frac{10}{\sqrt{17} \sqrt{6}} = \frac{10}{\sqrt{102}} \).
\( \angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right) \).
Q16
Show that the points \( A(1, 2, 7), B(2, 6, 3) \) and \( C(3, 10, -1) \) are collinear.
Solution:
\( \vec{AB} = \hat{i} + 4\hat{j} - 4\hat{k} \). Magnitude: \( \sqrt{1+16+16} = \sqrt{33} \).
\( \vec{BC} = \hat{i} + 4\hat{j} - 4\hat{k} \). Magnitude: \( \sqrt{1+16+16} = \sqrt{33} \).
\( \vec{AC} = 2\hat{i} + 8\hat{j} - 8\hat{k} \). Magnitude: \( \sqrt{4+64+64} = \sqrt{132} = 2\sqrt{33} \).
Since \( |\vec{AB}| + |\vec{BC}| = \sqrt{33} + \sqrt{33} = 2\sqrt{33} = |\vec{AC}| \), the points are collinear.
Q17
Show that the vectors \( 2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k} \) and \( 3\hat{i} - 4\hat{j} - 4\hat{k} \) form the vertices of a right angled triangle.
Solution:
Let vectors be \( \vec{a}, \vec{b}, \vec{c} \).
\( |\vec{a}|^2 = 4+1+1 = 6 \).
\( |\vec{b}|^2 = 1+9+25 = 35 \).
\( |\vec{c}|^2 = 9+16+16 = 41 \).
Since \( 6 + 35 = 41 \), i.e., \( |\vec{a}|^2 + |\vec{b}|^2 = |\vec{c}|^2 \), they form a right-angled triangle.
Q18
If \( \vec{a} \) is a nonzero vector of magnitude '\( a \)' and \( \lambda \) a nonzero scalar, then \( \lambda \vec{a} \) is unit vector if
(A) \( \lambda = 1 \) (B) \( \lambda = -1 \) (C) \( a = |\lambda| \) (D) \( a = 1/|\lambda| \)
(A) \( \lambda = 1 \) (B) \( \lambda = -1 \) (C) \( a = |\lambda| \) (D) \( a = 1/|\lambda| \)
Answer: (D)
Solution:
For \( \lambda \vec{a} \) to be a unit vector, its magnitude must be 1.
\( |\lambda \vec{a}| = 1 \)
\( |\lambda| |\vec{a}| = 1 \)
\( |\lambda| a = 1 \)
\( a = \frac{1}{|\lambda|} \)