Compute the distance between the point R(1, 1, 1) and the line (0, -1, 1) + t(1, -2, 2)
Can someone please show me how to answer this in full? It was in a test recently and is still bugging me as the answer has not been posted yet. Thanks a lot in advance
When people say "distance", they mostly mean "perpendicular distance" / "shortest distance".
In this case, there are a couple of ways to go about it. The way that I will choose to solve it is as follows:
Let $L$ be the given line, $R$ be the given point to find the shortest distance form
Find the general equation of a point $P$ on the line $L$
Find the exact instance of the point $P$ (called $Q$) such that the vector from $R$ to $P$ is perpendicular to $L$ (that is, we find the point on $L$ such that it gives the perpendicular).
Find the distance from $Q$ to $R$ which is the distance we want.
Since we know the parametric equation of the line $L$, any point $P = (0 + 1t, -1 -2t, 1 + 2t)$
The vector from $R$ to $P$ is $P - R = (t - 1, -2t -2, 2t)$.
For the vector to be perpendicular, we need the dot product of the vector with the direction of line to be $0$.
Hence,
$$ (t - 1, -2t -2, 2t) \cdot (1, -2, 2) = 0 \\ 1 \cdot (t - 1) -2(-2t - 2) +2 \cdot 2t = 0\\ 9t + 3 = 0 \\ t = -\frac{1}{3} $$ Hence, the point that gives us perpendicular distance is $Q = P(\frac{-1}{3})$
To finish off the problem, just find the distance from $R$ to $Q$ using the distance formula.