Let the foot of the perpendicular from the po | vedclass

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

Question

(C) The point $(1, \mu, 2)$ lies on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$.Setting $x=1$, we get $\frac{1-4}{1} = -3$, so $\frac{\mu

Vedclass · Accepted Answer

$\frac{\sqrt{146}}{7}$

Vedclass · Answer

(C) The point $(1, \mu, 2)$ lies on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$.Setting $x=1$, we get $\frac{1-4}{1} = -3$, so $\frac{\mu-9}{2} = -3 \Rightarrow \mu = 3$ and $\frac{2-5}{1} = -3$. Thus, the foot is $(1, 3, 2)$.The vector from $(\lambda, 2, 3)$ to $(1, 3, 2)$ is $(1-\lambda, 3-2, 2-3) = (1-\lambda, 1, -1)$.Since this vector is perpendicular to the line direction $(1, 2, 1)$, we have $(1-\lambda)(1) + (1)(2) + (-1)(1) = 0$, which gives $1-\lambda + 1 = 0 \Rightarrow \lambda = 2$.The lines are $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $L_2: \frac{x-2}{2} = \frac{y-3}{3} = \frac{z+5}{6}$.These are parallel lines with direction vector $\vec{v} = (2, 3, 6)$.The distance between parallel lines is $d = \frac{|(\vec{r_2}-\vec{r_1}) 	imes \vec{v}|}{|\vec{v}|}$.Here, $\vec{r_1} = (1, 2, -4)$ and $\vec{r_2} = (2, 3, -5)$, so $\vec{r_2}-\vec{r_1} = (1, 1, -1)$.The cross product is $(1, 1, -1) 	imes (2, 3, 6) = (6 - (-3), -(6 - (-2)), 3-2) = (9, -8, 1)$.The magnitude is $\sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81+64+1} = \sqrt{146}$.The magnitude of $\vec{v}$ is $\sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.Thus, the distance is $\frac{\sqrt{146}}{7}$.

Similar Questions

If lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are mutually perpendicular to each other,then $p = $ . . . . . . .

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

The equation of the line passing through the points $(3, 4, -7)$ and $(6, -1, 1)$ is

The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0, 2x-y+z+3=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module