$A$ unit vector which is perpendicular to $i + 2j - 2k$ and $-i + 2j + 2k$ is

  • A
    $\frac{1}{\sqrt{5}}(2i - k)$
  • B
    $\frac{1}{\sqrt{5}}(-2i + k)$
  • C
    $\frac{1}{\sqrt{5}}(2i + j + k)$
  • D
    $\frac{1}{\sqrt{5}}(2i + k)$

Explore More

Similar Questions

Consider $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$. Let $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y$,and $A_z$ are the projections of the area of triangle $PQR$ on the $yz, zx$,and $xy$ planes,respectively. Then,the value of $|\vec{B}|^2$ is:

If $a=\hat{i}+\hat{j}$ and $b=3 \hat{i}-2 \hat{j}$, then the vector $r$ satisfying the equations $r \times a=b \times a$ and $r \times b=a \times b$ is

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to

If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo