Exercise 10.2 Practice
Chapter 10: Vector Algebra – NCERT Solutions
Q1
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Compute the magnitude of the following vectors:
$\vec{a} = i + j + k$; $\vec{b} = 2i - 7j - 3k$; $\vec{c} = \frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j - \frac{1}{\sqrt{3}}k$
$\vec{a} = i + j + k$; $\vec{b} = 2i - 7j - 3k$; $\vec{c} = \frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j - \frac{1}{\sqrt{3}}k$
$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$.
$|\vec{c}| = \sqrt{(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 + (-\frac{1}{\sqrt{3}})^2} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$.
$\boxed{\sqrt{3}, \sqrt{62}, 1}$
Q2
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Write two different vectors having same magnitude.
Let $\vec{a} = i + 2j + 3k$. Magnitude $|\vec{a}| = \sqrt{1+4+9} = \sqrt{14}$.
Let $\vec{b} = 2i + j + 3k$. Magnitude $|\vec{b}| = \sqrt{4+1+9} = \sqrt{14}$.
Vectors are different (different components) but magnitudes are same.
$\boxed{\text{Example: } i + 2j + 3k \text{ and } 2i + j + 3k}$
Q3
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Write two different vectors having same direction.
Let $\vec{a} = i + j + k$.
Let $\vec{b} = 2(i + j + k) = 2i + 2j + 2k$.
Direction cosines are same, but magnitudes are different.
$\boxed{\text{Example: } i + j + k \text{ and } 2i + 2j + 2k}$
Q4
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Find the values of $x$ and $y$ so that the vectors $2i + 3j$ and $xi + yj$ are equal.
Two vectors are equal if their corresponding components are equal.
$2 = x$ and $3 = y$.
$\boxed{x = 2, y = 3}$
Q5
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Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point $(-5, 7)$.
Vector $\vec{v} = (x_2 - x_1)i + (y_2 - y_1)j$.
$\vec{v} = (-5 - 2)i + (7 - 1)j = -7i + 6j$.
Scalar components: $-7, 6$. Vector components: $-7i, 6j$.
$\boxed{\text{Scalar: } -7, 6; \text{ Vector: } -7i, 6j}$
Q6
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Find the sum of the vectors $\vec{a} = i - 2j + k$, $\vec{b} = -2i + 4j + 5k$ and $\vec{c} = i - 6j - 7k$.
$\vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)i + (-2 + 4 - 6)j + (1 + 5 - 7)k$.
$= 0i - 4j - 1k$.
$\boxed{-4j - k}$
Q7
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Find the unit vector in the direction of the vector $\vec{a} = i + j + 2k$.
$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$.
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{6}}i + \frac{1}{\sqrt{6}}j + \frac{2}{\sqrt{6}}k$.
$\boxed{\frac{1}{\sqrt{6}}i + \frac{1}{\sqrt{6}}j + \frac{2}{\sqrt{6}}k}$
Q8
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Find the unit vector in the direction of vector $\vec{PQ}$, where P and Q are the points $(1, 2, 3)$ and $(4, 5, 6)$, respectively.
$\vec{PQ} = (4-1)i + (5-2)j + (6-3)k = 3i + 3j + 3k$.
$|\vec{PQ}| = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$.
Unit vector = $\frac{3i + 3j + 3k}{3\sqrt{3}} = \frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j + \frac{1}{\sqrt{3}}k$.
$\boxed{\frac{1}{\sqrt{3}}(i + j + k)}$
Q9
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For given vectors, $\vec{a} = 2i - j + 2k$ and $\vec{b} = -i + j - k$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.
$\vec{c} = \vec{a} + \vec{b} = (2-1)i + (-1+1)j + (2-1)k = i + k$.
$|\vec{c}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$.
$\boxed{\frac{1}{\sqrt{2}}i + \frac{1}{\sqrt{2}}k}$
Q10
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Find a vector in the direction of vector $5i - j + 2k$ which has magnitude 8 units.
Let $\vec{a} = 5i - j + 2k$. $|\vec{a}| = \sqrt{25+1+4} = \sqrt{30}$.
Unit vector $\hat{a} = \frac{1}{\sqrt{30}}(5i - j + 2k)$.
Required vector $= 8\hat{a}$.
$\boxed{\frac{40}{\sqrt{30}}i - \frac{8}{\sqrt{30}}j + \frac{16}{\sqrt{30}}k}$
Q11
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Show that the vectors $2i - 3j + 4k$ and $-4i + 6j - 8k$ are collinear.
Let $\vec{b} = -4i + 6j - 8k$.
$\vec{b} = -2(2i - 3j + 4k) = -2\vec{a}$.
Since $\vec{b} = \lambda \vec{a}$ with $\lambda = -2$, they are collinear.
$\boxed{\text{Shown}}$
Q12
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Find the direction cosines of the vector $i + 2j + 3k$.
Magnitude $r = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
Direction cosines are $l = \frac{x}{r}, m = \frac{y}{r}, n = \frac{z}{r}$.
$\boxed{\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}}$
Q13
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Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$, directed from A to B.
$\vec{AB} = (-1-1)i + (-2-2)j + (1-(-3))k = -2i - 4j + 4k$.
$|\vec{AB}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
DCs: $-\frac{2}{6}, -\frac{4}{6}, \frac{4}{6}$.
$\boxed{-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}}$
Q14
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Show that the vector $i + j + k$ is equally inclined to the axes OX, OY and OZ.
Magnitude $r = \sqrt{1+1+1} = \sqrt{3}$.
DCs are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.
Since $\cos \alpha = \cos \beta = \cos \gamma$, the angles are equal.
$\boxed{\text{Shown}}$
Q15
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Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $i + 2j - k$ and $-i + j + k$ respectively, in the ratio 2:1 (i) internally (ii) externally.
$\vec{a} = i + 2j - k$, $\vec{b} = -i + j + k$, $m=2, n=1$.
(i) Internal: $\frac{m\vec{b} + n\vec{a}}{m+n} = \frac{2(-i + j + k) + 1(i + 2j - k)}{3} = \frac{-i + 4j + k}{3}$.
(ii) External: $\frac{m\vec{b} - n\vec{a}}{m-n} = \frac{2(-i + j + k) - 1(i + 2j - k)}{1} = -3i + 3k$.
$\boxed{\text{(i) } -\frac{1}{3}i + \frac{4}{3}j + \frac{1}{3}k \text{ (ii) } -3i + 3k}$
Q16
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Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.
Midpoint formula: $\frac{\vec{P} + \vec{Q}}{2}$.
$\frac{(2+4)i + (3+1)j + (4-2)k}{2} = \frac{6i + 4j + 2k}{2}$.
$\boxed{3i + 2j + k}$
Q17
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Show that the points A, B and C with position vectors $\vec{a} = 3i - 4j - 4k$, $\vec{b} = 2i - j + k$ and $\vec{c} = i - 3j - 5k$, respectively form the vertices of a right angled triangle.
$\vec{AB} = \vec{b} - \vec{a} = -i + 3j + 5k$. $|\vec{AB}|^2 = 1+9+25 = 35$.
$\vec{BC} = \vec{c} - \vec{b} = -i - 2j - 6k$. $|\vec{BC}|^2 = 1+4+36 = 41$.
$\vec{CA} = \vec{a} - \vec{c} = 2i - j + k$. $|\vec{CA}|^2 = 4+1+1 = 6$.
$|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 = |\vec{BC}|^2$.
$\boxed{\text{Shown (Right angled at A)}}$
Q18
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In triangle ABC, which of the following is not true:
(A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$
(B) $\vec{AB} + \vec{BC} - \vec{AC} = \vec{0}$
(C) $\vec{AB} + \vec{BC} - \vec{CA} = \vec{0}$
(D) $\vec{AB} - \vec{CB} + \vec{CA} = \vec{0}$
(A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$
(B) $\vec{AB} + \vec{BC} - \vec{AC} = \vec{0}$
(C) $\vec{AB} + \vec{BC} - \vec{CA} = \vec{0}$
(D) $\vec{AB} - \vec{CB} + \vec{CA} = \vec{0}$
Triangle law: $\vec{AB} + \vec{BC} = \vec{AC}$.
(C) $\vec{AB} + \vec{BC} - \vec{CA} = \vec{AC} - \vec{CA} = \vec{AC} + \vec{AC} = 2\vec{AC} \ne \vec{0}$.
$\boxed{\text{(C)}}$
Q19
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If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:
(A) $\vec{b} = \lambda \vec{a}$
(B) $\vec{a} = \pm \vec{b}$
(C) the respective components of $\vec{a}$ and $\vec{b}$ are proportional
(D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes
(A) $\vec{b} = \lambda \vec{a}$
(B) $\vec{a} = \pm \vec{b}$
(C) the respective components of $\vec{a}$ and $\vec{b}$ are proportional
(D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes
Collinear vectors can have opposite directions. Statement (D) implies they must have the same direction (or describes a specific case as the general rule), which is incorrect in the general sense.
$\boxed{\text{(D)}}$