find the distance from (4,-2,6) to the xy plane and yz plane

Question

I am doing cal 3 h.w the text book only show area from two points..."the distance formula in three dimension".. i do know how to do the two points, but this one point question is confusing. Please explain how to find between xy and yz plane. Thank You.

Answer 1

In the $xy$-plane, we have $z=0$ and in the $yz$-plane, we have $x=0$.

The point nearest to $(4/-2/6)$ in the $xy$-plane is therefore $(4/-2/0)$ with distance $6$ and the point nearest to $(4/-2/6)$ in the $yz$-plane is $(0/-2/6)$ with distance $4$.

If you prefer to use the distance formula : The $xy$-plane has equation $z=0$ , in vector-form $$\pmatrix{0\\0\\1}x=0$$ , and the $yz$-plane has equation $x=0$ , in vector form $$\pmatrix{1\\0\\0}x=0$$