The equation of the line of shortest distance | vedclass

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

Question

(D) Let the points on the two lines be $P(3\lambda + 6, -\lambda + 7, \lambda + 4)$ and $Q(-3\mu, 2\mu - 9, 4\mu + 2)$.The vector $\vec{PQ} = (-3\mu -

Vedclass · Accepted Answer

$\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{-1}$

Vedclass · Answer

(D) Let the points on the two lines be $P(3\lambda + 6, -\lambda + 7, \lambda + 4)$ and $Q(-3\mu, 2\mu - 9, 4\mu + 2)$.The vector $\vec{PQ} = (-3\mu - 3\lambda - 6)\hat{i} + (2\mu + \lambda - 16)\hat{j} + (4\mu - \lambda - 2)\hat{k}$.Since $\vec{PQ}$ is perpendicular to both lines,its dot product with the direction vectors of the lines must be zero.Direction vectors are $\vec{d_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{d_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.$\vec{PQ} \cdot \vec{d_1} = 0 \Rightarrow 3(-3\mu - 3\lambda - 6) - 1(2\mu + \lambda - 16) + 1(4\mu - \lambda - 2) = 0 \Rightarrow -7\mu - 11\lambda = 4$.$\vec{PQ} \cdot \vec{d_2} = 0 \Rightarrow -3(-3\mu - 3\lambda - 6) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0 \Rightarrow 29\mu + 7\lambda = 22$.Solving the system of equations,we get $\lambda = -1$ and $\mu = 1$.Substituting these values,we get $P(3, 8, 3)$ and $Q(-3, -7, 6)$.The direction ratios of line $PQ$ are $(-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3)$,which simplifies to $(2, 5, -1)$.The equation of the line passing through $P(3, 8, 3)$ with direction ratios $(2, 5, -1)$ is $\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{-1}$.

The equation of the line of shortest distance between the lines $\frac{x - 6}{3} = \frac{y - 7}{-1} = \frac{z - 4}{1}$ and $\frac{x}{-3} = \frac{y + 9}{2} = \frac{z - 2}{4}$ is:

Similar Questions

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

If the lines $\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}$ and $\frac{x+1}{4}=\frac{1-3y}{15}=z+1$ are perpendicular to each other,then $\lambda=$

Let $A(2,3,-1), B(4,1,0), C(-1,-1,11)$ be the vertices of a triangle $ABC$. Let $D$ be the point where the bisector of $\angle BAC$ meets the side $BC$. Then the direction ratios of $AD$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module