If $\overline{a} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\overline{b} = \hat{i} - \hat{j} - \hat{k}$,then the projection of $\overline{b}$ in the direction of $\overline{a}$ is

  • A
    $\frac{1}{\sqrt{29}}$
  • B
    $\frac{2}{\sqrt{3}}$
  • C
    $\frac{5}{\sqrt{3}}$
  • D
    $\frac{3}{\sqrt{29}}$

Explore More

Similar Questions

Find the projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$.

Let $a = \hat{i} + \hat{j} + \hat{k}$,$b = 2\hat{i} + 2\hat{j} + \hat{k}$,and $c = 5\hat{i} + \hat{j} - \hat{k}$ be three vectors. The area of the region formed by the set of points whose position vectors $\vec{r}$ satisfy the equations $\vec{r} \cdot \vec{a} = 5$ and $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ is closest to which integer?

If the position vectors of the vertices of a triangle are $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $3\hat{i} - 4\hat{j} - 4\hat{k}$,then the triangle is:

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

If $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$ are vectors such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then the value of $\lambda$ 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