The line $L_1$ passes through point $A$ whose position vector is $3i – 5j + 4k$, and is parallel to the vector $3i + 4j + 2k$. The line $L_2$ passes through the point $B$ whose position vector is $2i + 3j + 5k$, and is parallel to the vector $i – j – 4k$. The point $P$ on $L_1$ and $Q$ on $L_2$ are such that $PQ$ is perpendicular to both $L_1$ and $L_2$. The plane $\pi_1$ contains $PQ$ and $L_1$, and the plane $\pi_2$ contains $PQ$ and $L_2$.
(i) Find the length of PQ.
Since $PQ$ is perpendicular to both lines, the cross product of both lines' directions is the direction vector of $PQ$
$\Rightarrow (3i + 4j + 2k) \times (i – j – 4k) = -7(2i – 2j + k).$
Hence $|PQ| = \sqrt{(4+4+1)} = \sqrt{9} = 3.$
(ii) Find a vector perpendicular to π1.
Since $\pi_1$ contains $PQ$ and $L_1$, the cross product of both direction vectors will be the normal vector of $\pi_1$
$\Rightarrow (2i – 2j + k) \times (3i + 4j + 2k) = -(8i + j – 14k).$
(iii) Find the perpendicular distance from $B$ to $\pi_1$.
$B$ was given as the position vector $(2i + 3j + 5k)$, which lies on the line $L_2$.
I can't figure out how to get the equation of $\pi_1$ because it involves getting a point on the plane (which would be $P$), and I can't figure out how to get point $P$ either.
If I had point $P$, I would have found the equation of $\pi_1$, and then, using the distance formula, found the shortest distance between $B$ and $\pi_1.$
How can I find this distance? Please explain
Best Answer
So I figured answer:
(iii) Using the distance formula, the shortest distance D from B to π1 is given by:
D = |ax + by + cz + d| / |n| where n is the normal to plane; a, b, c, d are the coefficients of the equation of the plane; x, y, z are the coordinates of the point from the plane.
π1: 8i + j - 14k = 75
B: 2i + 3j + 5k
D = |(8)(2) + (1)(3) + (-14)(5) + (-75)| / sqrt(64+1+196)
D = |-126| / sqrt(261)
D = 126/16.155 = 7.799 ~= 7.80