JEE Main 2024 (Online) 1st February Morning ShiftQuestion 1
Let the line of the shortest distance between the lines $L_1:\vec{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda\left(\hat{i}-\hat{j}+\hat{k}\right)$ and $L_2:\vec{r}=\left(4\hat{i}+5\hat{j}+6\hat{k}\right)+\mu\left(\hat{i}+\hat{j}-\hat{k}\right)$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha,\beta,\gamma)$ is the mid point of the line segment $PQ$, then $2(\alpha+\beta+\gamma)$ is equal to $\_\_\_\_\_\_$ .
3D Geometry
Answer$21$
Solution
Common perpendicular conditions with $d_1=(1,-1,1)$, $d_2=(1,1,-1)$ give $t=-s$ and $3-2s=0$: $s=\frac32$, $t=-\frac32$. $P=(\frac52,\frac12,\frac92)$, $Q=(\frac52,\frac72,\frac{15}2)$, midpoint $(\frac52,2,6)$, so $2(\alpha+\beta+\gamma)=21$.