Let $\underline v = h\underline i+l\underline j-3\underline k=(h, l, -3)$, $\underline w=\underline i-\underline j-\underline k=(1, -1, -1)$, $\underline t=\underline i-\underline k=(1, 0, -1)$ and $\underline s=\underline i+\underline j=(1, 1, 0)$. For which values of $h$ and $l$ is $\underline v$ orthogonal to $\underline s$ and coplanar to $\underline w$ and $\underline t$?
- for all pairs of $h$ and $l$ satisfying $h=-l$;
- for only one pair $(h, l)$ in which $h$ satisfies $h^3-4h^2+9=0$;
- for infinite pairs $(h, l)$; in one of them $h=0$ and $l=2$;
- for only one pair $(h, l)$ in which $h$ satisfies $h^3+2h^2+9=0$;
- No one of previous answers.
My attempt. The condition that $\underline v$ is coplanar to $\underline w$ and $\underline t$ implies that there exist coefficients $a$ and $b$ such that $$ \underline v = a\underline w + b\underline t=a(1, -1, -1)+b(1, 0, -1)=(a+b, -a, -a-b). $$ So $(a+b, -a,-a-b)=(h, l, -3)$, from which we get $$ a = -l, \quad b = h+l,\quad h = 3. $$ The condition that $\underline v$ is orthogonal to $\underline s$ implies that $$ \underline v\cdot\underline s =h+l=0, $$ that is $h=-l$.
So, I think that the correct answer is $1$. Is this attempt correct?
Edit. Also probably answer $2$ can be the correct one. Because if $h=3$ necesserely $l=-3$.
You are correct, the condition that $\underline v$ is orthogonal to $\underline s$ implies that $$ \underline v\cdot\underline s =h+l=0, $$ that is $h=-l$.
For the condition to be coplanar to $\underline w$ and $\underline t$ we can use cross product that is
$$n=w\times t=(1,0,1)$$
and by dot product
$$v \cdot n = (h,-h,-3)\cdot (1,0,1)=h-3=0\implies h=3$$
Thus the correct one should be the option $2$.