I have the following parametric equation for line g: $$ x=3t\land y=-7+5t\land z=2+2t $$ I have to find the equation of a line perpendicular to $g$ and going through point $Q(3,-2,4)$ which lies on $g$. I know I have to use the dot product but can somebody tell me the whole procedure.
Thank you in advance :]
The equation of line can be written as: $$L:\frac{x-0}{3}=\frac{y+7}{5}=\frac{z-2}{2}=t$$ So the leading vector of the line is indeed, $\vec{u}(3,5,2)$. Since you are supposed to find the equation of perpendicular line to $L$ so, you have to find an appropriate vector $\vec{v}(a,b,c)$ such that $$\vec{u}\cdot\vec{v}=0$$ You may use some other ways to find such this vector, but I suggest you in this problem to find $a$,$b$ and $c$ but examine some values for them such that: $$3a+5b+2c=0$$ For example, $$a=+1,~b=-1,~c=+1$$ Now write the equation of your line.