Cfrgtkky

Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$?

My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help.

We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.

A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.

Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.

We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.

We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.

Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.

begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}

Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.

Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.