If a plane passes through (1, -2, 1) and is p | vedclass

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

Question

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$ of

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$ of the given planes.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 0)$.The equation of the plane passing through $(1, -2, 1)$ is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.The distance of the plane $x + y + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

If a plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,then the distance of that plane from the point $(1, 2, 2)$ is

Similar Questions

Find the intercepts made by the plane $\bar{r} \cdot (2, -3, 4) = 12$ on the coordinate axes.

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes at points $A, B,$ and $C$. Find the area of $\Delta ABC$.

$A$ 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?

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is

$A$ plane passes through the point $(3, 5, 7)$. If the direction ratios of its normal are equal to the intercepts made by the plane $x+3y+2z=9$ with the coordinate axes,then the equation of that plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module