Let a unit vector which makes an angle of $60^{\circ}$ with $2\hat{i}+2\hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\vec{C}$. Then $\vec{C}+\left(-\frac{1}{2}\hat{i}+\frac{1}{3\sqrt{2}}\hat{j}-\frac{\sqrt{2}}{3}\hat{k}\right)$ is

(D) $\frac{\sqrt{2}}{3}\hat{i}-\frac{1}{2}\hat{k}$

(A) $\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right)\hat{i}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3\sqrt{2}}\right)\hat{j}+\left(\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{3}\right)\hat{k}$

(B) $\frac{\sqrt{2}}{3}\hat{i}+\frac{1}{3\sqrt{2}}\hat{j}-\frac{1}{2}\hat{k}$

(C) $-\frac{\sqrt{2}}{3}\hat{i}+\frac{\sqrt{2}}{3}\hat{j}+\left(\frac{1}{2}+\frac{2\sqrt{2}}{3}\right)\hat{k}$

Vector Algebra — JEE Main · 2024 · 04 Apr 2024 · Shift 1 Question with Solution

Let a unit vector which makes an angle of

6 0^{\circ}

with

2 \hat{i} + 2 \hat{j} - \hat{k}

and an angle of

4 5^{\circ}

with

\hat{i} - \hat{k}

C

. Then

C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})

(A) $(\frac{1}{3} + \frac{1}{2}) \hat{i} + (\frac{1}{3} - \frac{1}{3 2}) \hat{j} + (\frac{1}{3} + \frac{2}{3}) \hat{k}$
(B) $\frac{2}{3} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{1}{2} \hat{k}$
(C) $- \frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + (\frac{1}{2} + \frac{2 2}{3}) \hat{k}$
(D) $\frac{2}{3} \hat{i} - \frac{1}{2} \hat{k}$ ✓ Correct

Answer: (D)

Solution

Let

C = (a, b, c)

with

a^{2} + b^{2} + c^{2} = 1

$6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ (magnitude 3): $2 a + 2 b - c = 3 cos 6 0^{\circ} = \frac{3}{2}$ .
$4 5^{\circ}$ with $\hat{i} - \hat{k}$ (magnitude $2$ ): $a - c = 2 cos 4 5^{\circ} = 1$ .

From these,

a = c + 1

and

c + 2 b = - \frac{1}{2}

. Substituting into

a^{2} + b^{2} + c^{2} = 1

gives

9 b^{2} = \frac{1}{2}

, so

b = \pm \frac{1}{3 2}

.

Taking

b = - \frac{1}{3 2}

gives

a = \frac{1}{2} + \frac{2}{3}

c = - \frac{1}{2} + \frac{2}{3}

. Then

C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k}) = \frac{2}{3} \hat{i} - \frac{1}{2} \hat{k} .

Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Vector Algebra

Solution

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