If the lines $\vec{r} = 2\hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = \hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{j} + 2\hat{k})$ intersect each other,then $(\lambda + \mu)$ is equal to

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

Explore More

Similar Questions

Let $\vec{a} = \lambda \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$, and $\vec{c} = \hat{i} + \mu \hat{j} + \hat{k}$. If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda - \mu$.

Let $b = 4i + 3j$ and $c$ be two vectors perpendicular to each other in the $xy$-plane. All vectors in the same plane having projections $1$ and $2$ along $b$ and $c$ respectively,are given by

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}$,then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = $ . . . . . . .

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