The square of the distance of the image of th | vedclass

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

Question

(C) Let the given line be $L: \frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4} = \lambda$. Any point $M$ on the line is given by $M(3\lambda+1, 2\lambda, 4

Vedclass · Accepted Answer

$62$

Vedclass · Answer

(C) Let the given line be $L: \frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4} = \lambda$. Any point $M$ on the line is given by $M(3\lambda+1, 2\lambda, 4\lambda+2)$.Since $AM$ is perpendicular to the line $L$,the direction vector $\vec{AM} = (3\lambda+1-6, 2\lambda-1, 4\lambda+2-5) = (3\lambda-5, 2\lambda-1, 4\lambda-3)$ must be perpendicular to the direction vector of the line $\vec{b} = (3, 2, 4)$.Thus,$\vec{AM} \cdot \vec{b} = 0 \implies 3(3\lambda-5) + 2(2\lambda-1) + 4(4\lambda-3) = 0$.$9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0 \implies 29\lambda - 29 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$,we get $M(4, 2, 6)$.Let $I(x, y, z)$ be the image of $A(6, 1, 5)$ in the line. Since $M$ is the midpoint of $AI$,we have $\frac{x+6}{2} = 4, \frac{y+1}{2} = 2, \frac{z+5}{2} = 6$.$x+6 = 8 \implies x = 2$; $y+1 = 4 \implies y = 3$; $z+5 = 12 \implies z = 7$.So,the image point is $I(2, 3, 7)$.The square of the distance of $I(2, 3, 7)$ from the origin $(0, 0, 0)$ is $2^2 + 3^2 + 7^2 = 4 + 9 + 49 = 62$.

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin is:

Similar Questions

Lines $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other. Then,the value of $k$ is . . . . . . .

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

$A$ line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ at points $P$ and $Q$ respectively. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$,then $l^2$ is

The length of the perpendicular from the point $A(1, -2, -3)$ on the line $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}$ is (in $\text{ units}$)

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module