Question goes:
Find the position vector of the point P on the line AB such that OP is perpendicular to AB.
A has a position vector 7i-8j+7k, B has 4i+7j+4k
and O is the origin.
I started by finding out the line BA, which is: r = <7,-8,7> + t<1,-5,1>
Why is it BA though, and not AB? I first tried to make it AB (which is b-a), but got the wrong position vector and direction vector. AB is what it's supposed to be isn't it?
I know what formula to probably have to use, OP * AB = 0 means the lines are perpendicular. I thought I'd have to denote P by (x,y,z) and find out OP by p – o, which would result in (x,y,z) as well.
After this I got lost, and couldn't get the right answer.
The final answer should be 5i+2j+5k
Best Answer
$\vec {OP}$ $\perp$ $\vec {AB}:$
$\small{((7,-8,7)+t(1,-5,1))\cdot (1,-5,1)=0;}$
$54+t(27)=0;$
$t=-2.$
$\vec {OP}= (7,-8,7)-2(1,-5,1)=(5,2,5)$.