0
$\begingroup$

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.

  • 0
    Are you familiar with vector calculus, or at least linear algebra? – 2017-02-02
  • 0
    I am. I just need to brush up the concepts, so if you could give me a clue how to proceed? – 2017-02-02
  • 0
    You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points. – 2017-02-02
  • 0
    Could you start with the solution please? – 2017-02-02

1 Answers 1

0

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

A vector equation for $\overleftrightarrow{AB}$ is $\vec{r}=\langle1,2,4\rangle+t\langle2,2,1\rangle$ for $t\in\mathbb{R}$.

Point $Q$ must lie on $\overleftrightarrow{AB}$, so the position vector for $Q$ is $\vec{Q}=\langle1,2,4\rangle+q\langle2,2,1\rangle=\langle1+2q,2+2q,4+q\rangle$ for some $q\in\mathbb{R}$.

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, $\vec{PQ}\perp\vec{AB}$.

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

Since $\vec{PQ}\perp\vec{AB}$, then $\vec{PQ}\cdot\vec{AB}=0$.

\begin{align} \vec{PQ}\cdot\vec{AB}&=0\\ \langle2q-1,2q-2,q+1\rangle\cdot\langle2,2,1\rangle&=0\\ 4q-2+4q-4+q+1&=0\\ 9q&=5\\ q&=\frac{5}{9} \end{align}

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

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