Find line passing through a point, perpendicular to another line in $\mathbb{R}^3$

I am given a point $A(3,1,-2)$ and I am asked to find the line passing through it, being orthogonal to line $x = t-1, y=t-2, z=t-1$.

My process was:

The line is $ε(t) = (-1,-2,-1) + t(1,1,1)$ and we know the line is parallel to the vector $(1,1,1)$. So I tried to find an orthogonal vector to that vector, let $w$, we should have $$ w \cdot (1,1,1) = 0 \Rightarrow z = -x-y$$ so $$\vec{w} = x(1,0,-1) + y(0,1,-1)$$ So let $(1,0,-1)$ be the orthogonal vector . Now using the rule I find that the line is $$ x = 3 + (1-3)t,\, y = 1 + ( 0 -1)t, \,z = -2 + (-1 - (-2))t$$ or $$l(t) = (3,1,-2) + (-2,-1,1)t$$ which is wrong because the one I have to find is : $$l(t) = (3,1,-2) + (-2,-1,3)t$$ I might be close, but I am wrong. What am I doing wrong?

Solution 1:

Finally the line you write is,
$l(t) = (3,1,-2) + (-2,-1,1)t$ but vector $(-2, -1, 1)$ is not orthogonal to the vector $(1, 1, 1)$.

You choose $(1, 0, -1)$ as orthogonal vector and then subtract the given point from it, which is incorrect and does not give you anything meaningful.

Here is one of the correct approaches: Instead of choosing an arbitrary vector, first vector between the given point $A$ and a point $P$ on the line. When this vector is orthogonal to the given line, its dot product with vector $(1, 1, 1)$ will be zero.

As you mentioned, a point $P$ on the given line is $x = t-1, y=t-2, z=t-1$.

So, $\vec {AP} = ((t-1-3), (t-2 - 1), (t - 1 + 2))$
$= (t-4, t - 3, t + 1)$

Now for the vector to be orthogonal to the given line,
$(t - 4, t - 3, t + 1) \cdot (1, 1, 1) = 0 \implies t = 2$

So the desired line is $(3, 1, -2) + ( - 2, - 1, 3) s$