The distance of the point Q(0, 2, -2) from th | vedclass

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

Question

(D) The direction vector of the required line is the cross product of the direction vectors of the two given lines:$\vec{v} = \begin{vmatrix} \hat{i}

Vedclass · Accepted Answer

$\sqrt{74}$

Vedclass · Answer

(D) The direction vector of the required line is the cross product of the direction vectors of the two given lines:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ -1 & 3 & 2 \end{vmatrix} = \hat{i}(6 - 15) - \hat{j}(4 + 5) + \hat{k}(6 + 3) = -9\hat{i} - 9\hat{j} + 9\hat{k}$.We can take the direction vector as $\vec{d} = \hat{i} + \hat{j} - \hat{k}$.The equation of the line passing through $P(5, -4, 3)$ is $\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$.Any point $M$ on this line is $(5+\lambda, -4+\lambda, 3-\lambda)$.Let $Q$ be $(0, 2, -2)$. The vector $\vec{QM} = (5+\lambda - 0)\hat{i} + (-4+\lambda - 2)\hat{j} + (3-\lambda + 2)\hat{k} = (5+\lambda)\hat{i} + (\lambda-6)\hat{j} + (5-\lambda)\hat{k}$.Since $QM \perp$ line,$\vec{QM} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$.$(5+\lambda)(1) + (\lambda-6)(1) + (5-\lambda)(-1) = 0 \implies 5+\lambda + \lambda-6 - 5+\lambda = 0 \implies 3\lambda - 6 = 0 \implies \lambda = 2$.The point $M$ is $(5+2, -4+2, 3-2) = (7, -2, 1)$.The distance $QM = \sqrt{(7-0)^2 + (-2-2)^2 + (1 - (-2))^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$.

Similar Questions

The coordinates of the foot of the perpendicular from the point $(1, 0, 0)$ to the line $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}$ are

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ will be perpendicular if...

The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ is:

The perpendicular distance of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $P(2,-10,1)$ is:

Find the Cartesian equation of the line passing through the point $\hat{i} + 2\hat{j} + 2\hat{k}$ and parallel to the line joining the points $2\hat{i} - \hat{j} + \hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module