Exercise 10.3 Practice
Chapter 10: Vector Algebra – NCERT Solutions
Topic: Scalar (Dot) Product of Vectors
Q1
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Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$, respectively having $\vec{a} \cdot \vec{b} = \sqrt{6}$.
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\sqrt{6} = \sqrt{3} \cdot 2 \cos \theta \Rightarrow \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\boxed{\theta = \frac{\pi}{4}}$
Q2
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Find the angle between the vectors $i - 2j + 3k$ and $3i - 2j + k$.
$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$.
$|\vec{a}| = \sqrt{1+4+9} = \sqrt{14}$, $|\vec{b}| = \sqrt{9+4+1} = \sqrt{14}$.
$\cos \theta = \frac{10}{\sqrt{14}\sqrt{14}} = \frac{10}{14} = \frac{5}{7}$.
$\boxed{\theta = \cos^{-1}(\frac{5}{7})}$
Q3
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Find the projection of the vector $i - j$ on the vector $i + j$.
Projection of $\vec{a}$ on $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
$\vec{a} \cdot \vec{b} = 1(1) + (-1)(1) = 0$.
$\boxed{0}$
Q4
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Find the projection of the vector $i + 3j + 7k$ on the vector $7i - j + 8k$.
$\vec{a} \cdot \vec{b} = 1(7) + 3(-1) + 7(8) = 7 - 3 + 56 = 60$.
$|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}$.
$\boxed{\frac{60}{\sqrt{114}}}$
Q5
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Show that each of the given three vectors is a unit vector:
$\frac{1}{7}(2i + 3j + 6k), \frac{1}{7}(3i - 6j + 2k), \frac{1}{7}(6i + 2j - 3k)$.
Also, show that they are mutually perpendicular.
$\frac{1}{7}(2i + 3j + 6k), \frac{1}{7}(3i - 6j + 2k), \frac{1}{7}(6i + 2j - 3k)$.
Also, show that they are mutually perpendicular.
Magnitude of first: $\frac{1}{7}\sqrt{4+9+36} = \frac{1}{7}\sqrt{49} = 1$. Similarly for others.
Dot product of 1st and 2nd: $\frac{1}{49}(6 - 18 + 12) = 0$.
Dot product of 2nd and 3rd: $\frac{1}{49}(18 - 12 - 6) = 0$.
Dot product of 3rd and 1st: $\frac{1}{49}(12 + 6 - 18) = 0$.
$\boxed{\text{Shown}}$
Q6
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Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.
$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 8$.
Substitute $|\vec{a}| = 8|\vec{b}|$: $(8|\vec{b}|)^2 - |\vec{b}|^2 = 8 \Rightarrow 63|\vec{b}|^2 = 8$.
$|\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}}$.
$\boxed{|\vec{a}| = \frac{16\sqrt{2}}{3\sqrt{7}}, |\vec{b}| = \frac{2\sqrt{2}}{3\sqrt{7}}}$
Q7
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Evaluate the product $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.
$= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) - 10(\vec{b} \cdot \vec{a}) - 35|\vec{b}|^2$.
$\boxed{6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2}$
Q8
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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 $\frac{1}{2}$.
Let $|\vec{a}| = |\vec{b}| = x$. $\vec{a} \cdot \vec{b} = x \cdot x \cos 60^\circ = \frac{x^2}{2}$.
Given $\frac{x^2}{2} = \frac{1}{2} \Rightarrow x^2 = 1 \Rightarrow x = 1$.
$\boxed{1}$
Q9
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Find $|\vec{x}|$, if for a unit vector $\vec{a}$, $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$.
$|\vec{x}|^2 - |\vec{a}|^2 = 12$. Since $|\vec{a}| = 1$, $|\vec{x}|^2 - 1 = 12$.
$|\vec{x}|^2 = 13 \Rightarrow |\vec{x}| = \sqrt{13}$.
$\boxed{\sqrt{13}}$
Q10
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If $\vec{a} = 2i + 2j + 3k$, $\vec{b} = -i + 2j + k$ and $\vec{c} = 3i + j$ are such that $\vec{a} + \lambda \vec{b}$ is perpendicular to $\vec{c}$, find the value of $\lambda$.
$(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0$.
$\vec{a} \cdot \vec{c} + \lambda (\vec{b} \cdot \vec{c}) = 0$.
$\vec{a} \cdot \vec{c} = 6 + 2 + 0 = 8$. $\vec{b} \cdot \vec{c} = -3 + 2 + 0 = -1$.
$8 + \lambda(-1) = 0 \Rightarrow \lambda = 8$.
$\boxed{8}$
Q11
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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 $\vec{b}$.
Dot product: $(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$.
$= |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2 = 0$.
$\boxed{\text{Shown}}$
Q12
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If $\vec{a} \cdot \vec{a} = 0$ and $\vec{a} \cdot \vec{b} = 0$, then what can be concluded about the vector $\vec{b}$?
$\vec{a} \cdot \vec{a} = 0 \Rightarrow |\vec{a}|^2 = 0 \Rightarrow \vec{a} = \vec{0}$.
If $\vec{a} = \vec{0}$, then $\vec{a} \cdot \vec{b} = 0$ for any vector $\vec{b}$.
$\boxed{\text{Any vector}}$
Q13
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If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.
$|\vec{a} + \vec{b} + \vec{c}|^2 = 0$.
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$1 + 1 + 1 + 2(\text{Sum}) = 0 \Rightarrow 2(\text{Sum}) = -3$.
$\boxed{-\frac{3}{2}}$
Q14
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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.
Let $\vec{a} = \hat{\imath}$ and $\vec{b} = \hat{\jmath}$. Both are non-zero.
$\vec{a} \cdot \vec{b} = 1(0) + 0(1) + 0(0) = 0$.
Here dot product is zero but vectors are non-zero (they are perpendicular).
$\boxed{\text{Example: } \vec{a} = i, \vec{b} = j}$
Q15
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If the vertices A, B, C of a triangle ABC are $(1, 2, 3), (-1, 0, 0), (0, 1, 2)$, respectively, then find $\angle ABC$.
Angle is between $\vec{BA}$ and $\vec{BC}$.
$\vec{BA} = (1 - (-1))i + (2-0)j + (3-0)k = 2i + 2j + 3k$.
$\vec{BC} = (0 - (-1))i + (1-0)j + (2-0)k = i + j + 2k$.
$\vec{BA} \cdot \vec{BC} = 2(1) + 2(1) + 3(2) = 10$.
$|\vec{BA}| = \sqrt{4+4+9} = \sqrt{17}$, $|\vec{BC}| = \sqrt{1+1+4} = \sqrt{6}$.
$\cos \theta = \frac{10}{\sqrt{17}\sqrt{6}} = \frac{10}{\sqrt{102}}$.
$\boxed{\cos^{-1}(\frac{10}{\sqrt{102}})}$
Q16
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Show that the points $A(1, 2, 7), B(2, 6, 3)$ and $C(3, 10, -1)$ are collinear.
$\vec{AB} = i + 4j - 4k$. $|\vec{AB}| = \sqrt{1+16+16} = \sqrt{33}$.
$\vec{BC} = i + 4j - 4k$. $|\vec{BC}| = \sqrt{33}$.
$\vec{AC} = 2i + 8j - 8k$. $|\vec{AC}| = \sqrt{4+64+64} = \sqrt{132} = 2\sqrt{33}$.
$|\vec{AB}| + |\vec{BC}| = |\vec{AC}|$.
$\boxed{\text{Shown}}$
Q17
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Show that the vectors $2i - j + k$, $i - 3j - 5k$ and $3i - 4j - 4k$ form the vertices of a right angled triangle.
Let vectors be position vectors $\vec{a}, \vec{b}, \vec{c}$.
$\vec{AB} = \vec{b} - \vec{a} = -i - 2j - 6k$. $|\vec{AB}|^2 = 1+4+36 = 41$.
$\vec{BC} = \vec{c} - \vec{b} = 2i - j + k$. $|\vec{BC}|^2 = 4+1+1 = 6$.
$\vec{AC} = \vec{c} - \vec{a} = i - 3j - 5k$. $|\vec{AC}|^2 = 1+9+25 = 35$.
$|\vec{AC}|^2 + |\vec{BC}|^2 = 35 + 6 = 41 = |\vec{AB}|^2$.
$\boxed{\text{Shown (Right angled at C)}}$
Q18
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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) $a = 1/|\lambda|$
$|\lambda \vec{a}| = 1 \Rightarrow |\lambda| |\vec{a}| = 1$.
$|\lambda| a = 1 \Rightarrow a = \frac{1}{|\lambda|}$.
$\boxed{\text{(D)}}$