Find the position vector of the image of the | vedclass

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

Question

(A) Let the given point be $P(2, 1, 3)$ and the line be $L: \vec{r} = (0, 1, -2) + \lambda(1, 1, -1)$.Any point $Q$ on the line is given by $Q(\lambda

Vedclass · Accepted Answer

$-4\hat{i} - \hat{j} - 5\hat{k}$

Vedclass · Answer

(A) Let the given point be $P(2, 1, 3)$ and the line be $L: \vec{r} = (0, 1, -2) + \lambda(1, 1, -1)$.Any point $Q$ on the line is given by $Q(\lambda, 1+\lambda, -2-\lambda)$.The vector $\vec{PQ} = (\lambda-2, \lambda, -5-\lambda)$.Since $PQ$ is perpendicular to the line direction $\vec{v} = (1, 1, -1)$,their dot product is zero:$(\lambda-2)(1) + (\lambda)(1) + (-5-\lambda)(-1) = 0$$\lambda - 2 + \lambda + 5 + \lambda = 0$$3\lambda + 3 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into $Q$,we get the foot of the perpendicular $M(-1, 0, -1)$.Let the image of $P$ be $P'(x, y, z)$. Since $M$ is the midpoint of $PP'$,we have:$M = \frac{P + P'}{2} \implies (-1, 0, -1) = \frac{(2, 1, 3) + (x, y, z)}{2}$$-2 = 2 + x \implies x = -4$$0 = 1 + y \implies y = -1$$-2 = 3 + z \implies z = -5$Thus,the position vector of the image is $-4\hat{i} - \hat{j} - 5\hat{k}$.

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

Similar Questions

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

$A$ line drawn from a point $A(-2,-2,3)$ and parallel to the line $\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$ meets the $YOZ-$ plane in point $P$. Then,the coordinates of the point $P$ are:

The Cartesian equation of the line which passes through the point $(5, -2, 4)$ and is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module