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 first line is $L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$.$A$ point on $L_{1}$ is $\vec{a}_{1} = (-7, 6, 0)$ and its direction vector

Vedclass · Accepted Answer

$2 \sqrt{29}$

Vedclass · Answer

(A) The first line is $L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$.$A$ point on $L_{1}$ is $\vec{a}_{1} = (-7, 6, 0)$ and its direction vector is $\vec{b}_{1} = (-6, 7, 1)$.The second line is $L_{2}: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}$.$A$ point on $L_{2}$ is $\vec{a}_{2} = (7, 2, 6)$ and its direction vector is $\vec{b}_{2} = (-2, 1, 1)$.The vector $\vec{a}_{2} - \vec{a}_{1} = (7 - (-7), 2 - 6, 6 - 0) = (14, -4, 6)$.The cross product $\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -6 & 7 & 1 \ -2 & 1 & 1 \end{vmatrix} = \hat{i}(7-1) - \hat{j}(-6+2) + \hat{k}(-6+14) = 6\hat{i} + 4\hat{j} + 8\hat{k} = (6, 4, 8)$.The magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{6^2 + 4^2 + 8^2} = \sqrt{36 + 16 + 64} = \sqrt{116} = 2\sqrt{29}$.The shortest distance $d = \left| \frac{(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} 	imes \vec{b}_{2})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight| = \left| \frac{(14, -4, 6) \cdot (6, 4, 8)}{2\sqrt{29}} ight| = \left| \frac{84 - 16 + 48}{2\sqrt{29}} ight| = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.

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

Similar Questions

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

The angle between the lines $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-5}{1}$ and $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-5}{2}$ is $ . . . . . . $. (in $^{\circ}$)

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.

If the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{-2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular, then the value of $p$ is . . . . . . .

The coordinates of the point at which the $XZ$-plane divides the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module