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Vector Algebra

Multiple Choice Questions

301.

$(a . \hat{i}) \hat{i} + (a . \hat{j}) \hat{j} + (a . \hat{k}) \hat{k}$ is equal to

302.

If $|a| = 3, |b| = 4$ , then a value of $λ$ for which a + $λb$ is perpendicular to a - $λb$ is

$\frac{9}{16}$
$\frac{3}{4}$
$\frac{3}{2}$
$\frac{4}{3}$

303.

The projection of $a = 2 \hat{i} + 3 \hat{j} - 2 \hat{k}$ on $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$ is

$\frac{1}{\sqrt{14}}$
$\frac{2}{\sqrt{14}}$
$\sqrt{14}$
$\frac{- 2}{\sqrt{14}}$

304.

If the vectors $4 \hat{i} + 11 \hat{j} + m \hat{k}$ , $7 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $\hat{i} + 5 \hat{j} + 4 \hat{k}$ are coplanar, then m is equal to

0
38
- 10
10

305.

If $|a \times b| = 4 and |a . b| = 2, then {|a|}^{2} {|b|}^{2}$ is equal to

306.
The angle between the vectors a + b and a - b when a = (1, 1, 4) and b = (1, - 1, 4) is

45°

90°

15°

30°

90°

$Here, \vec{a} = \hat{i} + \hat{j} + 4 \hat{k} \vec{b} = \hat{i} - \hat{j} + 4 \hat{k} We get, \vec{a} + \vec{b} = 2 \hat{i} + 8 \hat{k} and \vec{a} - \vec{b} = 2 \hat{j} ∴ \cos (θ) = \frac{(2 \hat{i} + 8 \hat{k}) . (2 \hat{j})}{|2 \hat{i} + 8 \hat{k}| \times |2 \hat{j}|} = 0 ∴ θ = 90 ° Alternate method : We have, |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \Rightarrow {|\vec{a} + \vec{b}|}^{2} = {|\vec{a} - \vec{b}|}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} \Rightarrow \vec{a} . \vec{b} = 0 ∴ \vec{a} ⊥ \vec{b} \Rightarrow θ = 90 °$

307.

If a, b, c are mutually perpendicular unit vectors, then $|a + b + c|$ is equal to

3
$\sqrt{3}$
0
1

308.

If ABCDEF is a regular hexagon, then AD + EB + FC equals

0
2 AB
4 AB
3 AB

309.

The projection of the vector $2 \hat{i} + \hat{j} - 3 \hat{k}$ on the vector $\hat{i} - 2 \hat{j} + \hat{k}$ is

$- \frac{3}{\sqrt{14}}$
$\frac{3}{\sqrt{14}}$
$- \sqrt{\frac{3}{2}}$
$\frac{3}{\sqrt{2}}$

310.

An unit vector perpendicular to the plane containing the vectors $\hat{i} - \hat{j} + \hat{k}$ and $- \hat{i} + \hat{j} + \hat{k}$ is

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

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

306. The angle between the vectors a + b and a - b when a = (1, 1, 4) and b = (1, - 1, 4) is 45° 90° 15° 30°

306.
The angle between the vectors a + b and a - b when a = (1, 1, 4) and b = (1, - 1, 4) is

45°

90°

15°

30°