The shortest distance between the lines frac{ | vedclass

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

Question

(A) The given lines are $L_1: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $L_2: \frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$.Points on the lines are

Vedclass · Accepted Answer

$\frac{18}{\sqrt{5}}$

Vedclass · Answer

(A) The given lines are $L_1: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $L_2: \frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$.Points on the lines are $A(3, 2, 1)$ and $B(-3, 6, 5)$.The direction vectors are $\vec{b_1} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} + 3\hat{k}$.The vector $\vec{AB} = (-3-3)\hat{i} + (6-2)\hat{j} + (5-1)\hat{k} = -6\hat{i} + 4\hat{j} + 4\hat{k}$.The shortest distance $d$ is given by $d = \frac{|\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.First,calculate the cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(9 - (-1)) - \hat{j}(6 - (-2)) + \hat{k}(2 - 6) = 10\hat{i} - 8\hat{j} - 4\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{10^2 + (-8)^2 + (-4)^2} = \sqrt{100 + 64 + 16} = \sqrt{180} = 6\sqrt{5}$.The scalar triple product $|\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2})| = |(-6)(10) + (4)(-8) + (4)(-4)| = |-60 - 32 - 16| = |-108| = 108$.Thus,the shortest distance $d = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$.

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

Similar Questions

The vector equation of the line joining the points $i - 2j + k$ and $-2j + 3k$ is

If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

The angle between a line with direction ratios $2, 2, 1$ and the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

The Cartesian equation of the line passing through the points $A(2, 2, 1)$ and $B(1, 3, 0)$ is

Let $P$ be the point $(10, -2, -1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1, 7, 6)$ on the line passing through the points $(2, -5, 11)$ and $(-6, 7, -5)$. Then the length of the line segment $PQ$ is equal to ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module