Let L_1 (respectively L_2) be the line passin | vedclass

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

Question

(D) Given that line $L_1$ passes through $\vec{a}_1 = 2 \hat{i}-\hat{k}$ and is parallel to $\vec{b}_1 = 3 \hat{i}-\hat{j}+2 \hat{k}$.Line $L_2$ passe

Vedclass · Accepted Answer

$\frac{9}{\sqrt{35}}$

Vedclass · Answer

(D) Given that line $L_1$ passes through $\vec{a}_1 = 2 \hat{i}-\hat{k}$ and is parallel to $\vec{b}_1 = 3 \hat{i}-\hat{j}+2 \hat{k}$.Line $L_2$ passes through $\vec{a}_2 = 2 \hat{i}+\hat{j}-3 \hat{k}$ and is parallel to $\vec{b}_2 = \hat{i}-2 \hat{j}+\hat{k}$.The shortest distance $d$ between two skew lines is given by the formula: $d = \frac{|(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 	imes \vec{b}_2|}$.First,calculate the cross product $\vec{b}_1 	imes \vec{b}_2$:$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 2 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-1 + 4) - \hat{j}(3 - 2) + \hat{k}(-6 + 1) = 3 \hat{i} - \hat{j} - 5 \hat{k}$.The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.Next,calculate $\vec{a}_2 - \vec{a}_1 = (2 \hat{i} + \hat{j} - 3 \hat{k}) - (2 \hat{i} - \hat{k}) = \hat{j} - 2 \hat{k}$.Now,calculate the dot product: $(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (3 \hat{i} - \hat{j} - 5 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 2 \hat{k}) = (3)(0) + (-1)(1) + (-5)(-2) = 0 - 1 + 10 = 9$.Therefore,the shortest distance is $d = \frac{|9|}{\sqrt{35}} = \frac{9}{\sqrt{35}}$.

Let $L_1$ (respectively $L_2$) be the line passing through $2 \hat{i}-\hat{k}$ (respectively $2 \hat{i}+\hat{j}-3 \hat{k}$) and parallel to $3 \hat{i}-\hat{j}+2 \hat{k}$ (respectively $\hat{i}-2 \hat{j}+\hat{k}$). Then the shortest distance between the lines $L_1$ and $L_2$ is equal to

Similar Questions

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

If the position vectors of two points $P$ and $Q$ are respectively $9\hat{i} - \hat{j} + 5\hat{k}$ and $\hat{i} + 3\hat{j} + 5\hat{k}$,and the line segment $PQ$ intersects the $YOZ$ plane at a point $R$,then the ratio $PR : RQ$ is equal to

If the distance of point $B(2, 1, -3)$ from the line passing through the point $A(4, -2, 2)$ and parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is $x$, then $x^4 + x^2 + 541 =$

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module