The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i} + \hat{j} + \hat{k}$,$\hat{i} + \mu \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + \mu \hat{k}$ are coplanar,is

  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • D
    $2$

Explore More

Similar Questions

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$,$\hat{i}+\hat{k}$,and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda$ is equal to

$[(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})] \cdot \vec{d} = \dots$

The volume of the tetrahedron,whose vertices are given by the vectors $-i + j + k$,$i - j + k$,and $i + j - k$ with reference to the fourth vertex as the origin,is

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar vectors such that $\overrightarrow{P} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$,$\overrightarrow{Q} = 4\overrightarrow{a} + 3\overrightarrow{b} + 4\overrightarrow{c}$,and $\overrightarrow{R} = \overrightarrow{a} + \alpha\overrightarrow{b} + \beta\overrightarrow{c}$ are linearly dependent vectors. Then,the number of possible values of $\alpha$ is:

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$

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