Let $a, b$ and $c$ be three non-coplanar vectors. The vector equation of a line which passes through the point of intersection of two lines,one joining the points $a+2b-5c$ and $-a-2b-3c$,and the other joining the points $-4c$ and $6a-4b+4c$,is

  • A
    $r=2a-4b+3c+\mu(a-6b+4c)$
  • B
    $r=3a+6b-c+\mu(a+2b+c)$
  • C
    $r=2a+3b-c+\mu(a+b-c)$
  • D
    $r=-2b+3c+\mu(a-4b+3c)$

Explore More

Similar Questions

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$,then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle,for $K$ equal to:

If $G(\vec{g}), H(\vec{h})$ and $P(\vec{p})$ are the centroid,orthocenter,and circumcenter of a triangle respectively,and $x \vec{p} + y \vec{h} + z \vec{g} = 0$,then $(x, y, z) = $

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other,then $\lambda$ is equal to

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$,$\vec{b}$ and $\vec{c}$ is $\beta$,and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

If $a, b, c$ are the position vectors of the points $A, B, C$ respectively,then match the items of List-$I$ with those of List-$II$.
List-$I$List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$$IV$. $A, B, C$ are collinear

The correct match 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