$$π1:( 1+k , -3 , 6 )$$
$$π2: ( 1 , 5+k , 3k)$$
Does anyone know the values of k that would make these parallel and perpendicular, ive been trying for hours but nothing seems to work
The original equation is
$$π1:(1+k)x−3y+6z−4=0$$
$$π2:x+(5+k)y+3kz+1=0$$
$\endgroup$ 21 Answer
$\begingroup$Hints:
- If $v_1$ and $v_2$ are perpendicular, $v_1 \cdot v_2 = 0$.
- If $v_1$ and $v_2$ are parallel, $v_2 = av_2$ for some constant $a$.
