If the shortest distance between the linesL_1 | vedclass

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

Question

(B) The lines are given by:$L_1: \overrightarrow{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$$L_2: \overrightarrow{r

Vedclass · Accepted Answer

$387$

Vedclass · Answer

(B) The lines are given by:$L_1: \overrightarrow{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$$L_2: \overrightarrow{r} = (2\hat{i} + 3\hat{j} + 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$Let $\overrightarrow{a_1} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\overrightarrow{a_2} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.Let $\overrightarrow{p} = \hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{q} = 2\hat{i} + 3\hat{j} + \hat{k}$.Then $\overrightarrow{a_2} - \overrightarrow{a_1} = (2-2)\hat{i} + (3-1)\hat{j} + (5-3)\hat{k} = 0\hat{i} + 2\hat{j} + 2\hat{k}$.The cross product $\overrightarrow{p} 	imes \overrightarrow{q}$ is:$\overrightarrow{p} 	imes \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 4 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3-12) - \hat{j}(1-8) + \hat{k}(3+6) = -15\hat{i} + 7\hat{j} + 9\hat{k}$.The magnitude $|\overrightarrow{p} 	imes \overrightarrow{q}| = \sqrt{(-15)^2 + 7^2 + 9^2} = \sqrt{225 + 49 + 81} = \sqrt{355}$.The shortest distance is given by $d = \left| \frac{(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{p} 	imes \overrightarrow{q})}{|\overrightarrow{p} 	imes \overrightarrow{q}|} ight|$.$d = \left| \frac{(0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{j} + 9\hat{k})}{\sqrt{355}} ight| = \left| \frac{0 + 14 + 18}{\sqrt{355}} ight| = \frac{32}{\sqrt{355}}$.Comparing with $\frac{m}{\sqrt{n}}$,we have $m = 32$ and $n = 355$.Since $\operatorname{gcd}(32, 355) = 1$,the value of $m+n = 32 + 355 = 387$.

Similar Questions

The orthocentre of the triangle formed by the points $(2,1,5)$,$(3,2,3)$,and $(4,0,4)$ is

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

The equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module