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

If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$,then the sum of all possible values of $\lambda$ is:

If the lines with direction cosines $(l, m, n)$ satisfying $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular,then $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = .........$

If the shortest distance between the lines $\vec{r}=(-\hat{i}+3\hat{k})+\lambda(\hat{i}-a\hat{j})$ and $\vec{r}=(-\hat{j}+2\hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{2}{3}}$,then the integral value of $a$ is equal to

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module