Find the position vector g of the point G, where O is the origin, so that OG is perpendicular to the line (7,1,2) + t(5,0,-1).

Question

I have:

OG = g - o = (x, y, z) - (0, 0, 0) = (x, y, z)

So (7,1,2) + t(5, 0, -1) is the normal to (x,y,z).

g. (5, 0, -1) = 0

That is 5x + 0y -1z = 0

I'm so stuck. How do I solve?

How do I get from that (the plane?) back to the vector OG? I'm bamboozled.

I can't use g. n = o . n.

I am barking up the complete wrong tree here it seems. Am I able to use OG = g - o because o is the zero vector?

This is my first foray into this area of maths so any explanation behind what you write would be appreciated greatly. The only information I can find is how to find the equation of the plane. But I don't need the plane, I just need the point G and its position vector.

Answer 1

Hint...I assume that $G$ is the foot of the perpendicular from the origin to the line. In which case, since $G$ lies on the line, let$$\overrightarrow{OG}=\left(\begin{matrix}7+5t\\1\\2-t\end{matrix}\right)$$

Since $\overrightarrow{OG}$ is perpendicular to the line itself, then use $$\overrightarrow{OG}\cdot\left(\begin{matrix}5\\0\\-1\end{matrix}\right)=0$$ and find the value of $t$ and hence the position of $G$