If the image of the point P(1, -2, 3) in the | vedclass

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

Question

(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{

Vedclass · Accepted Answer

$2\sqrt{42}$

Vedclass · Answer

(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.Any point $F$ on this line can be represented as $(\lambda+1, 4\lambda-2, 5\lambda+3)$.Since $F$ lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $F$ into the plane equation:$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$$2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$$-6\lambda + 6 = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ back into the coordinates of $F$,we get $F(2, 2, 8)$.Since $F$ is the midpoint of $PQ$,the distance $PQ = 2PF$.The distance $PF = \sqrt{(2-1)^2 + (2-(-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$.Therefore,$PQ = 2PF = 2\sqrt{42}$.

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$,then $PQ$ is equal to:

Similar Questions

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

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

The equation of the plane containing the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a - 6, \beta, \gamma)$. If $a + b + c = 5$,then $7\beta - 9\gamma$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module