I'm currently working on a linear algebra problem set and I'm struggling with the following questions:
a) Find a vector that is orthogonal to both $(1,1,2)$ and $(-3,4,1)$, and then find an equation for the line that goes through the point $P_0:(0,1,2)$ and has direction $v$.
b) Find the point $Q$ on the line from part a) that minimizes the distance between $Q$ and the point $P:(2,3,3)$.
c) Find the point $Q$ in the plane $x+2y+z+1=0$ that minimizes the distance between $Q$ and the point $P:(2,3,3)$.
For part a), I know that two vectors are orthogonal if their dot product is $0$, so I tried to find a vector that satisfies this condition. I found the cross product of the given vectors to get $v=(1,1,-1)$, and the line is $P_0(t)=(0,1,2)+t(1,1,-1)$.
For part b), I know that I need to use the shortest distance formula to find the point $Q$, but I'm having trouble setting up the equation to minimize the distance. I found that the point $Q$ on the line $P_0(t)$ that is closest to the point $P=(2,3,3)$ is $Q=(\frac{5}{3},\frac{8}{3},\frac{4}{3})$. Then, the distance between $P$ and the line is $|PQ|=\sqrt{\frac{13}{3}}$.
For part c), I calculated the equation of the line $P(t)=(2,3,3)+t(1,2,1)$ and substituted it into the equation of the plane $x-y+3z=6$ to get $t=\frac{5}{3}$. Then, the intersection point is $P(\frac{5}{3})=(\frac{8}{3},\frac{13}{3},\frac{8}{3})$.
Any help or hints would be greatly appreciated! Thank you in advance.
Note that
$$(-9,-7,7) \cdot (1,1,2)=-9-7+14=-2\neq0$$
is not a correct choice for the orthogonal vector. We can proceed as follows
Part a)
By cross product
$$v=(1,1,2)\wedge (-3,4,1)=(7,7,-7)$$
which is the same as $(1,1,-1)$ and the line is $$P_0 (t)=P_0+vt=(0,1,2)+t(1,1,-1)$$
Part b)
An effective way is consider and minimize $|P-P_0(t)|^2$ by derivative to find $t_Q$ such that $Q=P_0(t_Q)$.
Part c)
A vector normal to the plane in given by $v=(1,2,1)$ then consider the line
$$P(t)=P+vt=(2,3,3)+t(1,2,1)$$
and finally its intersection with the given plane.