If the image of the point P(1, -2, 1) with re | vedclass

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

Question

(D) The line passes through $B(1, 1, 2)$ and $C(2, 2, 1)$. The direction ratios of the line are $(2-1, 2-1, 1-2) = (1, 1, -1)$.The equation of the lin

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(D) The line passes through $B(1, 1, 2)$ and $C(2, 2, 1)$. The direction ratios of the line are $(2-1, 2-1, 1-2) = (1, 1, -1)$.The equation of the line is $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{-1} = k$.Any point $Q$ on the line is $(k+1, k+1, -k+2)$.The vector $\vec{PQ} = (k+1-1, k+1-(-2), -k+2-1) = (k, k+3, 1-k)$.Since $\vec{PQ}$ is perpendicular to the line,the dot product of $\vec{PQ}$ and the direction vector $(1, 1, -1)$ is zero:$1(k) + 1(k+3) - 1(1-k) = 0$$k + k + 3 - 1 + k = 0 \Rightarrow 3k + 2 = 0 \Rightarrow k = -\frac{2}{3}$.Substituting $k$ to find $Q$: $x = -\frac{2}{3} + 1 = \frac{1}{3}$,$y = -\frac{2}{3} + 1 = \frac{1}{3}$,$z = -(-\frac{2}{3}) + 2 = \frac{8}{3}$.$Q$ is the midpoint of $PR$,where $R = (l, m, n)$:$\frac{1+l}{2} = \frac{1}{3} \Rightarrow l = -\frac{1}{3}$.$\frac{-2+m}{2} = \frac{1}{3} \Rightarrow m = \frac{8}{3}$.$\frac{1+n}{2} = \frac{8}{3} \Rightarrow n = \frac{13}{3}$.Therefore,$l^2 + m^2 + n^2 = (-\frac{1}{3})^2 + (\frac{8}{3})^2 + (\frac{13}{3})^2 = \frac{1+64+169}{9} = \frac{234}{9} = 26$.

If the image of the point $P(1, -2, 1)$ with respect to the line passing through the points $B(1, 1, 2)$ and $C(2, 2, 1)$ is $R(l, m, n)$,then $l^2 + m^2 + n^2 =$

Similar Questions

The shortest distance between the lines $1+x=2y=-12z$ and $x=y+2=6z-6$ is

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane,then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is

If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :

The equation of the straight line $3x + 2y - z - 4 = 0$ and $4x + y - 2z + 3 = 0$ in the symmetrical form is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module