Find the shortest distance between the lines | vedclass

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

Question

(A) The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$.The shortest distance $d$ betwee

Vedclass · Accepted Answer

$2 \sqrt{29}$ units

Vedclass · Answer

(A) The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$.The shortest distance $d$ between two lines $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ is given by:$d = \frac{|\det(A)|}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2} + (c_{1}a_{2}-c_{2}a_{1})^{2} + (a_{1}b_{2}-a_{2}b_{1})^{2}}}$ where $A = \begin{bmatrix} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \ a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \end{bmatrix}$.Comparing the equations,we get:$(x_{1}, y_{1}, z_{1}) = (-1, -1, -1)$ and $(a_{1}, b_{1}, c_{1}) = (7, -6, 1)$.$(x_{2}, y_{2}, z_{2}) = (3, 5, 7)$ and $(a_{2}, b_{2}, c_{2}) = (1, -2, 1)$.Calculating the determinant:$\det(A) = \begin{vmatrix} 3-(-1) & 5-(-1) & 7-(-1) \ 7 & -6 & 1 \ 1 & -2 & 1 \end{vmatrix} = \begin{vmatrix} 4 & 6 & 8 \ 7 & -6 & 1 \ 1 & -2 & 1 \end{vmatrix}$$= 4(-6+2) - 6(7-1) + 8(-14+6) = 4(-4) - 6(6) + 8(-8) = -16 - 36 - 64 = -116$.Calculating the denominator:$\sqrt{(-6+2)^{2} + (1+7)^{2} + (-14+6)^{2}} = \sqrt{(-4)^{2} + (8)^{2} + (-8)^{2}} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$.Wait,re-calculating the cross product vector $\vec{n} = \vec{v_{1}} 	imes \vec{v_{2}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & -6 & 1 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-6+2) - \hat{j}(7-1) + \hat{k}(-14+6) = -4\hat{i} - 6\hat{j} - 8\hat{k}$.Magnitude $|\vec{n}| = \sqrt{(-4)^{2} + (-6)^{2} + (-8)^{2}} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$.Shortest distance $d = \frac{|-116|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$ units.

Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$.

Similar Questions

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

$A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$ are the vertices of a triangle $ABC$. The direction cosines of the internal bisector of $\angle BAC$ are

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ is

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module