A perpendicular is drawn from a point P on th | vedclass

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

Question

(A) Let a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1} = \lambda$ be $(2\lambda + 1, -\lambda - 1, \lambda)$.Since the foot

Vedclass · Accepted Answer

$(2, 0, 1)$

Vedclass · Answer

(A) Let a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1} = \lambda$ be $(2\lambda + 1, -\lambda - 1, \lambda)$.Since the foot of the perpendicular $Q$ lies on both planes $x + y + z = 3$ and $x - y + z = 3$,we can find the intersection of these two planes.Adding the two equations: $(x + y + z) + (x - y + z) = 3 + 3 \Rightarrow 2x + 2z = 6 \Rightarrow x + z = 3$.Subtracting the two equations: $(x + y + z) - (x - y + z) = 3 - 3 \Rightarrow 2y = 0 \Rightarrow y = 0$.Since $Q$ lies on the line of intersection of the two planes,its coordinates must satisfy $y = 0$ and $z = 3 - x$.Thus,$Q$ is of the form $(x, 0, 3 - x)$.Since $PQ$ is perpendicular to the plane $x + y + z = 3$,the vector $\vec{PQ}$ must be parallel to the normal vector of the plane,$\vec{n} = (1, 1, 1)$.$vec{PQ} = (x - (2\lambda + 1), 0 - (-\lambda - 1), 3 - x - \lambda) = (x - 2\lambda - 1, \lambda + 1, 3 - x - \lambda)$.Since $\vec{PQ} = k(1, 1, 1)$,we have $x - 2\lambda - 1 = \lambda + 1 = 3 - x - \lambda$.From $\lambda + 1 = 3 - x - \lambda$,we get $2\lambda + x = 2$.From $x - 2\lambda - 1 = \lambda + 1$,we get $x - 3\lambda = 2$.Solving $2\lambda + x = 2$ and $x - 3\lambda = 2$,we subtract the equations: $5\lambda = 0 \Rightarrow \lambda = 0$.Substituting $\lambda = 0$ into $x - 3\lambda = 2$,we get $x = 2$.Thus,the coordinates of $Q$ are $(2, 0, 3 - 2) = (2, 0, 1)$.

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

Similar Questions

In what ratio does the $xy$-plane divide the line segment joining the points $(1, 2, 3)$ and $(4, 2, 1)$?

Find the position vector of the point where the line $r = (i - j + k) + t(i + j + k)$ meets the plane $r \cdot (i + j - k) = 5$.

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$,then a value of $\alpha$ is

If $\theta$ is the angle between the line $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 2}{4}$ and the plane $2x + y - 3z + 4 = 0$,then $64 \csc^2 \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module