If the shortest distance between the straight | vedclass

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

Question

(C) The given lines are:$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The given lines are:$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \hat{j} + 3\hat{k}$.$L_{2}: \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$. Point $P_{2} = (2, \lambda, 3)$,direction vector $\vec{v}_{2} = \hat{i} + 2\hat{j} + 4\hat{k}$.Let $\vec{a} = P_{2} - P_{1} = (2-1)\hat{i} + (\lambda-2)\hat{j} + (3-1)\hat{k} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$.The shortest distance $d$ is given by $d = \frac{|\vec{a} \cdot (\vec{v}_{1} 	imes \vec{v}_{2})|}{|\vec{v}_{1} 	imes \vec{v}_{2}|}$.First,calculate $\vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 1 & 2 & 4 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(8-3) + \hat{k}(4-1) = -2\hat{i} - 5\hat{j} + 3\hat{k}$.Magnitude $|\vec{v}_{1} 	imes \vec{v}_{2}| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.Now,$\vec{a} \cdot (\vec{v}_{1} 	imes \vec{v}_{2}) = (1)(-2) + (\lambda-2)(-5) + (2)(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$.Given $d = \frac{1}{\sqrt{38}}$,so $\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$.$|14 - 5\lambda| = 1$.Case $1$: $14 - 5\lambda = 1 \Rightarrow 5\lambda = 13 \Rightarrow \lambda = 2.6$.Case $2$: $14 - 5\lambda = -1 \Rightarrow 5\lambda = 15 \Rightarrow \lambda = 3$.Since we need the integral value of $\lambda$,the answer is $3$.

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$,$\lambda \in R$ is $\frac{1}{\sqrt{38}}$,then the integral value of $\lambda$ is equal to :

Similar Questions

Find the distance of the point $(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

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

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

The equation of the line,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module