If the position vectors of three points $A, B, C$ are $\hat{i} + \hat{j} + \hat{k}$,$2\hat{i} + 3\hat{j} - 4\hat{k}$,and $7\hat{i} + 4\hat{j} + 9\hat{k}$ respectively,then find the unit vector perpendicular to the plane of triangle $ABC$.

  • A
    $31\hat{i} - 18\hat{j} - 9\hat{k}$
  • B
    $\frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}}$
  • C
    $\frac{31\hat{i} + 38\hat{j} + 9\hat{k}}{\sqrt{2486}}$
  • D
    None of these

Explore More

Similar Questions

Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :

Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to

$(2a + 3b) \times (5a + 7b) = $

The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is

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