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Vector Algebra

Multiple Choice Questions

501.

If $|\vec{a}| = |\vec{b}| = 1 and |\vec{a} + \vec{b}| = \sqrt{3}$ , then the value of $(3 \vec{a} - 4 \vec{b}) . (2 \vec{a} + 5 \vec{b})$ is :

- 21
$- \frac{21}{2}$
21
$\frac{21}{2}$

502.

If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5 and \vec{a}, \vec{b}, \vec{c}$ are such that each is perpendicular to the sum of other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is :

5 $\sqrt{2}$
$\frac{5}{\sqrt{2}}$
$10 \sqrt{2}$
$10 \sqrt{3}$

503.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, then $|2 \vec{a} - \vec{b}, 2 \vec{b} - \vec{c}, 2 c - \vec{a}|$ is equal to :

1
0
$- \sqrt{3}$

504.

If $\vec{u}, \vec{v}, \vec{w}$ be vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$ and $|\vec{u}| = 3, |\vec{v}| = 4, |\vec{w}| = 5$ , then $\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}$ is equal to :

47
- 47
0
- 25

505.

If $\vec{a}$ is perpendicular to $\vec{b} and \vec{c}$ , $|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 4$ and the angle between $\vec{b} and \vec{c}$ is $\frac{2 π}{3}$ , then $[\vec{a} \vec{b} \vec{c}]$ is equal to :

$4 \sqrt{3}$
6 $\sqrt{3}$
$12 \sqrt{3}$
$18 \sqrt{3}$

506.

If $\vec{a}, \vec{b} and \vec{c}$ are perpendicular to $\vec{b} + \vec{c}, \vec{c} + \vec{a} and \vec{a} + \vec{b}$ respectively and, if $|\vec{a} + \vec{b}|$ = 6, $|\vec{b} + \vec{c}|$ = 8 and $|\vec{c} + \vec{a}|$ = 10, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to :

5 $\sqrt{2}$
50
10 $\sqrt{2}$
10

507.

A vector perpendicular to $2 \hat{i} + \hat{j} + \hat{k}$ and coplanar with $\hat{i} + 2 \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + 2 \hat{k}$ is :

$5 (\hat{j} - \hat{k})$
$\hat{i} + 7 \hat{j} - \hat{k}$
$5 (\hat{j} + \hat{k})$
$2 \hat{i} - \hat{j} - \hat{k}$

508.

If $\vec{a} = 2 \hat{i} - 3 \hat{j} + p \hat{k}$ and $\vec{a} \times \vec{b} = \vec{a} = 4 \hat{i} + 2 \hat{j} - 2 \hat{k}$ , then p is :

509.
Let $\vec{a} = \hat{i} - \hat{j}, \vec{b} = \hat{j} - \hat{k}, \vec{c} = \hat{k} - \hat{i}$ . If $\vec{d}$ is a unit vector such that $\vec{a} . \vec{d} = 0 = [\vec{b} \vec{c} \vec{d}]$ , then $\vec{d}$ is (are) :

$\pm \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}$

$\pm \frac{\hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{6}}$

$\pm \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

$\pm \hat{k}$

$\pm \frac{\hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{6}}$

$Let \vec{d} = d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k} \vec{a} . \vec{d} = d_{1} - d_{2} = 0 \Rightarrow d_{1} = d_{2} . . . (i) Also, \vec{d} is a unit vector \Rightarrow {d_{1}}^{2} + {d_{2}}^{2} + {d_{3}}^{2} = 1 . . . (iii) [\vec{b} \vec{c} \vec{d}] = 0 \Rightarrow |\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ d_{1} & d_{2} & d_{3} \end{matrix}| = 0 \Rightarrow - 1 (- d_{3} - d_{1}) - 1 (- d_{2}) = 0 \Rightarrow d_{1} + d_{2} + d_{3} = 0 \Rightarrow 2 d_{1} + d_{3} = 0 \Rightarrow d_{3} = - 2 d_{1} . . . (iii) Using Eqs . (iii) and (i) in Eq . (ii) we get {d_{1}}^{2} + {d_{1}}^{2} + 4 {d_{1}}^{2} = 1 \Rightarrow 6 {d_{1}}^{2} = 1 \Rightarrow d_{1} = \pm \frac{1}{\sqrt{6}} \Rightarrow d_{2} = \pm \frac{1}{\sqrt{6}} and d_{3} = \mp \frac{2}{\sqrt{6}} ∴ Required vector is \pm \frac{1}{\sqrt{6}} (\hat{i} + \hat{j} - 2 \hat{k})$

510.

If $\vec{a} and \vec{b}$ are unit vectors such that $[\vec{a} \vec{b} \vec{a} \times \vec{b}] = \frac{1}{4}$ , then angle between $\vec{a} and \vec{b}$ is :

$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{2}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

509. Let a→ = i^ - j^, b→ = j^ - k^, c→ = k^ - i^. If d→ is a unit vector such that a→ . d→ = 0 = b→ c→ d→, then d→ is (are) : ± i^ + j^ - k^3 ± i^ + j^ - 2k^6 ± i^ + j^ + k^3 ± k^