Let a unit vector which makes an angle of 60^ with 2 i +2 j - k and angle 45^ with i - k be C . Then C + (-(1)/(2) i + 1 3 2 j - 2 3 k ) is :

Correct answer: A. 2 3 i -(1)/(2) k C =C_1 i +C_2 j +C_3 k C_1 ^2+C_2 ^2+C_3 ^2=1 C ·(2 i +2 j - k )=|C| 9 60^ 2 C_1+2 C_2-C_3=(3)/(2) C_1-C_3=1 C_1+2 C_2=(1)/(2) C_1= 2 3 +(1)/(2) C_2= -1 3 2 C_3= 2 3 -(1)/(2)

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Vector Algebra

Let a unit vector which makes an angle of 60^ with 2 i +2 j - k and...

Let a unit vector which makes an angle of

60^{\circ}

with

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

45^{\circ}

with

\hat{i}-\hat{k}

be

\overrightarrow{\mathrm{C}}

. Then

\overrightarrow{\mathrm{C}}+

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

is :

Options

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

Correct answer

A. $\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}$

Step-by-step solution

\overrightarrow{\mathrm{C}}=\mathrm{C}_1 \hat{\mathrm{i}}+\mathrm{C}_2 \hat{\mathrm{j}}+\mathrm{C}_3 \hat{\mathrm{k}}

\mathrm{C}_1{ }^2+\mathrm{C}_2{ }^2+\mathrm{C}_3{ }^2=1

\overrightarrow{\mathrm{C}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=|\mathrm{C}| \sqrt{9} \cos 60^{\circ}

2 \mathrm{C}_1+2 \mathrm{C}_2-\mathrm{C}_3=\frac{3}{2}

\mathrm{C}_1-\mathrm{C}_3=1

\mathrm{C}_1+2 \mathrm{C}_2=\frac{1}{2}

\mathrm{C}_1=\frac{\sqrt{2}}{3}+\frac{1}{2}

\mathrm{C}_2=\frac{-1}{3 \sqrt{2}}

\mathrm{C}_3=\frac{\sqrt{2}}{3}-\frac{1}{2}

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