The shortest distance between the skew lines | vedclass

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

Question

(B) The shortest distance $d$ between two skew lines $r = a_1 + t b_1$ and $r = a_2 + s b_2$ is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 	imes b_2

Vedclass · Accepted Answer

$\frac{22}{\sqrt{17}}$

Vedclass · Answer

(B) The shortest distance $d$ between two skew lines $r = a_1 + t b_1$ and $r = a_2 + s b_2$ is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 	imes b_2)|}{|b_1 	imes b_2|}$.Here,$a_1 = -\hat{i} + 3\hat{k}$,$b_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$,$a_2 = 3\hat{i} + \hat{j} - \hat{k}$,and $b_2 = 2\hat{i} - \hat{j} + 2\hat{k}$.First,calculate $a_2 - a_1 = (3 - (-1))\hat{i} + (1 - 0)\hat{j} + (-1 - 3)\hat{k} = 4\hat{i} + \hat{j} - 4\hat{k}$.Next,calculate the cross product $b_1 	imes b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 6 \ 2 & -1 & 2 \end{vmatrix} = \hat{i}(6 - (-6)) - \hat{j}(4 - 12) + \hat{k}(-2 - 6) = 12\hat{i} + 8\hat{j} - 8\hat{k}$.The magnitude $|b_1 	imes b_2| = \sqrt{12^2 + 8^2 + (-8)^2} = \sqrt{144 + 64 + 64} = \sqrt{272} = \sqrt{16 	imes 17} = 4\sqrt{17}$.The dot product $(a_2 - a_1) \cdot (b_1 	imes b_2) = (4\hat{i} + \hat{j} - 4\hat{k}) \cdot (12\hat{i} + 8\hat{j} - 8\hat{k}) = (4 	imes 12) + (1 	imes 8) + (-4 	imes -8) = 48 + 8 + 32 = 88$.Therefore,$d = \frac{|88|}{4\sqrt{17}} = \frac{22}{\sqrt{17}}$.

The shortest distance between the skew lines $r = (-\hat{i} + 3\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $r = (3\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} - \hat{j} + 2\hat{k})$ is

Similar Questions

Prove that the line through $A(0,-1,-1)$ and $B(4,5,1)$ intersects the line through $C(3,9,4)$ and $D(-4,4,4)$.

The shortest distance between the lines $L_1: \bar{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + \hat{j} - \hat{k})$ and $L_2: \bar{r} = \hat{j} + \hat{k} + \mu(\hat{j} + 2\hat{k} - \hat{i})$ is equal to:

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

The shortest distance between the lines ${r_1} = 4i - 3j - k + \lambda (i - 4j + 7k)$ and ${r_2} = i - j - 10k + \mu (2i - 3j + 8k)$ is

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module