Line L_1 passes through the point (1,2,3) and is parallel to the z-axis.

Question

Line $L_1$ passes through the point $(1,2,3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda,5,6)$ and is parallel to the $y$-axis. Let, for $\lambda=\lambda_1,\lambda_2\ (\lambda_2<\lambda_1)$, the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1,\lambda_2,7)$ from the line $L_1$ is:

Accepted Answer

$$3=\frac{|[(\lambda-1)\hat i+3\hat j+3\hat k]\cdot(\hat k\times\hat j)|}{|\hat k\times\hat j|}=|\lambda-1| \implies \lambda=4\text{ or }-2.$$
Since $\lambda_2<\lambda_1$: $\lambda_2=-2,\ \lambda_1=4$.

Point: $(4,-2,7)$. Since $L_1$ is parallel to the $z$-axis through $(1,2,3)$ (so $x=1,y=2$ for all points), the square of the distance is
$(4-1)^2+(-2-2)^2=9+16=25.$

Solutions · 1