If the shortest distance between the lines ov | vedclass

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

Question

(B) The lines are given by $\overline{r}_1 = \overline{a}_1 + \lambda \overline{b}_1$ and $\overline{r}_2 = \overline{a}_2 + \mu \overline{b}_2$,where

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The lines are given by $\overline{r}_1 = \overline{a}_1 + \lambda \overline{b}_1$ and $\overline{r}_2 = \overline{a}_2 + \mu \overline{b}_2$,where $\overline{a}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k}$,$\overline{b}_1 = \hat{i} - 2 \hat{j} + 2 \hat{k}$,$\overline{a}_2 = -4 \hat{i} - \hat{k}$,and $\overline{b}_2 = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$.Shortest distance $d = \frac{|(\overline{a}_2 - \overline{a}_1) \cdot (\overline{b}_1 	imes \overline{b}_2)|}{|\overline{b}_1 	imes \overline{b}_2|}$.First,calculate $\overline{b}_1 	imes \overline{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8\hat{i} + 8\hat{j} + 4\hat{k}$.Magnitude $|\overline{b}_1 	imes \overline{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.Now,$\overline{a}_2 - \overline{a}_1 = (-4 - \alpha)\hat{i} - 2\hat{j} - 3\hat{k}$.$(\overline{a}_2 - \overline{a}_1) \cdot (\overline{b}_1 	imes \overline{b}_2) = (-4 - \alpha)(8) + (-2)(8) + (-3)(4) = -32 - 8\alpha - 16 - 12 = -60 - 8\alpha$.Given $d = 9$,so $\frac{|-60 - 8\alpha|}{12} = 9 \implies |-60 - 8\alpha| = 108$.Since $\alpha > 0$,$-60 - 8\alpha$ is negative,so $60 + 8\alpha = 108 \implies 8\alpha = 48 \implies \alpha = 6$.

Similar Questions

The equation of the line passing through the point $Q(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

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

If $\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}$ is the equation of the line passing through the points $(1, 2, -1)$ and $(-1, 0, 1)$,then the direction ratios $(l, m, n)$ are:

The point $P$ lies on the line $AB$,where $A \equiv (2, 4, 5)$ and $B \equiv (1, 2, 3)$. If the $z$-coordinate of point $P$ is $3$,then its $y$-coordinate is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module