Looking for tips with regard to this problem.
Let $L$ be the line with parametric equations
$$\left\{\begin{align}
x &= 1−2t\\
y &= 4−t\\
z &= 5+t
\end{align}\right.$$
Find the vector equation for a line that passes through the point $P(−5, −4, 1)$ and intersects $L$ at a point that is exactly $4$ units from the point $Q(1, 4, 5)$. Note that there are two possible correct answers.
Best Answer
The point of intersection is somewhere on the line $L$, where each point $(x,y,z)$ can be represented by the parametric equations $A(1-2t,4-t,5+t)$. We know that $|\vec{QA}|=4$, so we have \begin{align} 4&=\sqrt{(1-2t-1)^2+(4-t-4)^2+(5+t-5)^2}\\ 16&=(2t)^2+(-t)^2+(t)^2\\ 16&=6t^2\\ t&=\pm\frac4{\sqrt6} \end{align}
All we need now is to substitute $t$ into point $A$, then write a vector equation of a line that passes through $A$ and $P$. Can you take it from here?