The distance of the point (-1, 2, -2) from th | vedclass

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

Question

(D) Given planes are $P_{1}: 2x + 3y + 2z = 0$ and $P_{2}: x - 2y + z = 0$.The normal vectors are $\vec{n}_{1} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $

Vedclass · Accepted Answer

$\frac{\sqrt{34}}{2}$

Vedclass · Answer

(D) Given planes are $P_{1}: 2x + 3y + 2z = 0$ and $P_{2}: x - 2y + z = 0$.The normal vectors are $\vec{n}_{1} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $\vec{n}_{2} = \hat{i} - 2\hat{j} + \hat{k}$.The direction vector $\vec{v}$ of the line of intersection $L$ is given by $\vec{v} = \vec{n}_{1} 	imes \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 2 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(3 + 4) - \hat{j}(2 - 2) + \hat{k}(-4 - 3) = 7\hat{i} - 7\hat{k}$.Simplifying the direction ratios,we get $(1, 0, -1)$.Since the planes pass through the origin $(0, 0, 0)$,the line $L$ passes through the origin. Thus,the equation of line $L$ is $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1} = \lambda$.Any point $Q$ on the line $L$ is $(\lambda, 0, -\lambda)$.Let $P = (-1, 2, -2)$. The vector $\vec{PQ} = (\lambda + 1, -2, -\lambda + 2)$.Since $\vec{PQ}$ is perpendicular to the line $L$ (direction vector $\vec{v} = (1, 0, -1)$),their dot product is zero:$(\lambda + 1)(1) + (-2)(0) + (-\lambda + 2)(-1) = 0$$\lambda + 1 + \lambda - 2 = 0 \Rightarrow 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$.Thus,$Q = (\frac{1}{2}, 0, -\frac{1}{2})$.The distance $PQ = \sqrt{(\frac{1}{2} - (-1))^2 + (0 - 2)^2 + (-\frac{1}{2} - (-2))^2} = \sqrt{(\frac{3}{2})^2 + (-2)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 4 + \frac{9}{4}} = \sqrt{\frac{18}{4} + 4} = \sqrt{\frac{9}{2} + 4} = \sqrt{\frac{17}{2}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2}$.

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is:

Similar Questions

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5,$ and the point $(1,1,1).$

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x + 4y + 12z + 23 = 0$ is:

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

The vector equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$,which is perpendicular to the plane $x-y+z=0$,is

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the $XY$ plane and the $YZ$ plane at points $A$ and $B$ respectively. The equation of the line passing through the points $A$ and $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module