Let P be a point in the first octant,whose im | vedclass

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

Question

(A) Let $P \equiv (x_0, y_0, z_0)$.The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \f

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $P \equiv (x_0, y_0, z_0)$.The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = k$.The image $Q$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = -2 \frac{x_0+y_0-3}{1^2+1^2} = -(x_0+y_0-3)$.Thus,$x_Q = x_0 - (x_0+y_0-3) = 3-y_0$ and $y_Q = y_0 - (x_0+y_0-3) = 3-x_0$.Since $Q$ lies on the $z$-axis,$x_Q = 0$ and $y_Q = 0$,which implies $x_0 = 3$ and $y_0 = 3$.The distance of $P(3, 3, z_0)$ from the $x$-axis is $\sqrt{y_0^2 + z_0^2} = 5$.Substituting $y_0 = 3$,we get $\sqrt{3^2 + z_0^2} = 5$,so $9 + z_0^2 = 25$,which gives $z_0^2 = 16$,so $z_0 = 4$ (since $P$ is in the first octant).$P$ is $(3, 3, 4)$. The image $R$ of $P$ in the $xy$-plane is $(3, 3, -4)$.The length of $PR$ is the distance between $(3, 3, 4)$ and $(3, 3, -4)$,which is $|4 - (-4)| = 8$.

Similar Questions

Find the distance of the point $(2, 1, 0)$ from the plane $2x + y + 2z + 5 = 0$.

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

The value of $m$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{2z-m}{3}$ lies in the plane $2x-5y+2z=7$,is

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module