The length of the perpendicular drawn from th | vedclass

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

Question

(C) Let the foot of the perpendicular drawn from the point $P(3, -1, 11)$ to the line be $L$. Since $L$ lies on the line $\frac{x}{2} = \frac{y-2}{3}

Vedclass · Accepted Answer

$\sqrt{53}$

Vedclass · Answer

(C) Let the foot of the perpendicular drawn from the point $P(3, -1, 11)$ to the line be $L$. Since $L$ lies on the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} = t$,the coordinates of $L$ are $(2t, 3t+2, 4t+3)$. The direction ratios of the line $PL$ are $(2t-3, 3t+2-(-1), 4t+3-11)$,which simplifies to $(2t-3, 3t+3, 4t-8)$. Since $PL$ is perpendicular to the line with direction ratios $(2, 3, 4)$,their dot product must be zero: $2(2t-3) + 3(3t+3) + 4(4t-8) = 0$. $4t - 6 + 9t + 9 + 16t - 32 = 0$. $29t - 29 = 0 \implies t = 1$. Substituting $t=1$ into the coordinates of $L$,we get $L(2, 5, 7)$. The length of the perpendicular $PL$ is the distance between $P(3, -1, 11)$ and $L(2, 5, 7)$: $PL = \sqrt{(2-3)^2 + (5-(-1))^2 + (7-11)^2} = \sqrt{(-1)^2 + 6^2 + (-4)^2} = \sqrt{1 + 36 + 16} = \sqrt{53}$.

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

Similar Questions

The distance of the point $P(7, 10, 11)$ from the line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ is

The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ 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 :

If the Cartesian equation of the line is $x-1=2y+3=3-z$,then its vector equation is

The ratio in which the $YZ$-plane divides the line segment joining the points $(2, 4, 5)$ and $(3, 5, -4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module