The shortest distance between the lines frac{ | vedclass

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

Question

(C) The given lines are $L_1: \frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $L_2: \frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For $L_1$, a p

Vedclass · Accepted Answer

$3\sqrt{5}$

Vedclass · Answer

(C) The given lines are $L_1: \frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $L_2: \frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For $L_1$, a point is $A(4, 3, 2)$ and the direction vector is $\vec{v}_1 = (1, 2, -3)$.For $L_2$, a point is $B(-2, 6, 5)$ and the direction vector is $\vec{v}_2 = (2, 4, -5)$.The vector $\vec{AB} = (-2-4, 6-3, 5-2) = (-6, 3, 3)$.The shortest distance $d$ between two skew lines is given by $d = \frac{|\vec{AB} \cdot (\vec{v}_1 	imes \vec{v}_2)|}{|\vec{v}_1 	imes \vec{v}_2|}$.First, calculate the cross product $\vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$.The magnitude is $|\vec{v}_1 	imes \vec{v}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4+1} = \sqrt{5}$.The dot product is $\vec{AB} \cdot (\vec{v}_1 	imes \vec{v}_2) = (-6, 3, 3) \cdot (2, -1, 0) = (-6)(2) + (3)(-1) + (3)(0) = -12 - 3 + 0 = -15$.Thus, $d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.

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:

Similar Questions

The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=\frac{z}{1}$ and $\frac{7-x}{2}=y-2=z-6$ is

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ Write its vector form.

The equation of the $x$-axis is:

The equation of the line that passes through the origin and is parallel to the $X$-axis is . . . . . . .

The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\bar{r} = \lambda(-3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module