The coordinates of the foot of the perpendicu | vedclass

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

Question

(D) Let the given point be $P(2, -1, 5)$ and the line be $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k}

Vedclass · Accepted Answer

$(1, 2, 3)$

Vedclass · Answer

(D) Let the given point be $P(2, -1, 5)$ and the line be $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$.Any point $M$ on the line can be represented as $(10\lambda + 11, -4\lambda - 2, -11\lambda - 8)$.The direction ratios of the line $PM$ are $(10\lambda + 11 - 2, -4\lambda - 2 - (-1), -11\lambda - 8 - 5) = (10\lambda + 9, -4\lambda - 1, -11\lambda - 13)$.Since $PM$ is perpendicular to the given line,the dot product of the direction ratios of $PM$ and the line vector $(10, -4, -11)$ must be zero:$10(10\lambda + 9) - 4(-4\lambda - 1) - 11(-11\lambda - 13) = 0$.$100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0$.$237\lambda + 237 = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into the coordinates of $M$:$M = (10(-1) + 11, -4(-1) - 2, -11(-1) - 8) = (1, 2, 3)$.

The coordinates of the foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ are

Similar Questions

The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1}$ and $\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular to each other,then $k$ is

Let the shortest distance between the lines $L : \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}, \lambda \geq 0$ and $L_1 : x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,then which of the following is $NOT$ possible?

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$,$B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes,then $\alpha + \beta =$

The perimeter of a square whose two sides lie along the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module