I have the following problem on my Calc III homework:
Find the line through $(8, 1, −5)$ that intersects and is perpendicular to the line $x = −1 + t, y = −3 + t, z = −1 + t$. (HINT: If $(x_0, y_0, z_0)$ is the point of intersection, find its coordinates. Enter your answers as a comma-separated list of equations.)
I'm not quite sure where to start. I wanted to first use a system of equations, but that would just cancel $t$, and get me #=other#. I wanted to use the fact that perpendicular vectors should get a dot product of zero, but I'm not quite sure how to use the point here.
I found this post here, but don't get the explanation given with it.
Any help would be great! I am new to vectors, and my textbook hasn't been the greatest at explaining these concepts (at least to me).
Best Answer
Let you want the line $\ell$ so $(8, 1, −5)\in\ell$ and $(x_0, y_0, z_0)\in\ell$, then $\vec{v}=(x_0-8, y_0-1, z_0+5)$ lies on $\ell$.
given line is $$\frac{x+1}{\color{blue}{1}}=\frac{y+3}{\color{blue}{1}}=\frac{z+1}{\color{blue}{1}}$$ with $\vec{v}\perp(1,1,1)$ so $(x_0-8, y_0-1, z_0+5).(1,1,1)=0$ or $x_0+y_0+z_0=4$. In the other hand $(x_0, y_0, z_0)$ lies on given line so $$\frac{x_0+1}{\color{blue}{1}}=\frac{y_0+3}{\color{blue}{1}}=\frac{z_0+1}{\color{blue}{1}}=k$$ these shows $k=3$ and you have the intersect point.