If a plane cuts off intercepts -6, 3, 4 from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$

Vedclass · Accepted Answer

$\frac{12}{\sqrt{29}}$

Vedclass · Answer

(C) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.Given intercepts are $a = -6, b = 3, c = 4$.Substituting these values,we get $\frac{x}{-6} + \frac{y}{3} + \frac{z}{4} = 1$.To simplify,multiply by the least common multiple of $6, 3, 4$,which is $12$:$-2x + 4y + 3z = 12$,or $-2x + 4y + 3z - 12 = 0$.The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = -2, B = 4, C = 3, D = -12$.$d = \frac{|-12|}{\sqrt{(-2)^2 + 4^2 + 3^2}} = \frac{12}{\sqrt{4 + 16 + 9}} = \frac{12}{\sqrt{29}}$.

If a plane cuts off intercepts $-6, 3, 4$ from the coordinate axes,then the length of the perpendicular from the origin to the plane is

Similar Questions

$A$ plane meets the coordinate axes at points $A, B$,and $C$ such that the centroid of $\Delta ABC$ is $(1, 2, 3)$. The equation of the plane is

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

The equation of the plane bisecting the line segment joining the points $(2,0,6)$ and $(-6,2,4)$ and perpendicular to it,is

The equation of the plane containing the lines $r = a_1 + \lambda a_2$ and $r = a_2 + \lambda a_1$ is

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$,and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module