I have been given 2 lines
L1 = < 3,- 1, 0> + x <2, 2 , -2>
L2 = < z, 0 , -2> + y <-1, B , 1>
I was asked to find z and B given that both lines intersect and they are perpendicular to eachother.
[Math] Vector lines that are perpendicular to each other
vectors
Related Solutions
It is important that you first normalize vectors $\vec{OA} \, ( = \vec{a})$ and $\vec{OB} \, ( = \vec{b})$. Please see the below diagram as an example. The directional vector of angle bisector is in direction $OE$ but if you do not normalize the vectors, you will get a vector in direction $OF$.
So, $\vec{OE} = \vec{OC} + \vec{OD}$
Unit vector $\vec{OC} = \displaystyle \frac{\vec{a}}{||a||}$
Unit vector $\vec{OD} = \displaystyle \frac{\vec{b}}{||b||}$
So, $\vec{OE} = \displaystyle \frac{\vec{a}}{||a||} + \frac{\vec{b}}{||b||}$
Directional vector for one of the angle bisectors = $\lambda_1 (||b|| \, \vec{a} + ||a|| \, \vec{b})$
Similarly the directional vector of the other angle bisector = $ \lambda_2 (||a|| \, \vec{b} - ||b|| \, \vec{a})$
Dot product of both angle bisectors = $\lambda_1 \lambda_2 \, (||b|| \, \vec{a} + ||a|| \, \vec{b}) \cdot (||a|| \, \vec{b} - ||b|| \, \vec{a}) = 0$
(as $\vec{a} \cdot \vec{a} = ||a||^2 \,$ and $ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$)
HINT
Here is a sketch of the solution, which you may be more interested in the future.
Take any vector $x\in\mathbb{R}^{3}\backslash\{0,w\}$ then remove its projection on the direction of $w$.
The result of such operation is going to be denoted by $u$, which is orthogonal to $w$.
Precisely, one has that \begin{align*} u = x - \frac{\langle x,w\rangle}{\|w\|^{2}}w \end{align*}
Once you have $u$ and $w$, you can take the cross product $v = u\times w$, and you are done.
Hopefully this helps !
Best Answer
Perpendicular, so the direction vectors should have a dot product equal to $0$: $$(2,2,-2) \cdot (-1,B,1) = 0 \iff \color{blue}{B =\ldots}$$ Then with that value of $B$, look for a point where the lines meet: $$(3,-1,0)+x(2,2,-2) = (z,0,-2)+y(-1,\color{blue}{B},1) \iff \left\{\begin{array}{rcl} 3+2x = z-y \\ -1+2x = \color{blue}{B}y \\ -2x = -2+y \end{array}\right. \iff \cdots$$ This is a system of three equations in the unknowns $x$, $y$ and $z$.