The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

  • A
    $2 x-3 y+4 z=6$
  • B
    $2 x+3 y-4 z=6$
  • C
    $-2 x-3 y+4 z=6$
  • D
    $2 x+3 y+4 z=-6$

Explore More

Similar Questions

$A$ plane passes through $(2,1,2)$ and $(1,2,1)$ and is parallel to the line $2x = 3y$ and $z = 1$. Then the plane also passes through which of the following points?

Let $P(\lambda, 2, 1)$ be a point on the plane which passes through the point $Q(4, -2, 2)$. If the plane is perpendicular to the line joining the points $A(-2, -21, 29)$ and $B(-1, -16, 33)$,then find the value of $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, 5)$. Then,the Cartesian equation of the plane is:

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

If a variable plane,at a distance of $3 \ units$ from the origin,intersects the coordinate axes at $A, B$,and $C$,then the locus of the centroid of $\Delta ABC$ 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