I have come across this question that I need a tip for.
Find the equation (general form) of the plane passing through the point $P(3,1,6)$ that is orthogonal to the vector $v=(1,7,-2)$.
I would be able to do this if it said "parallel to the vector"
I would set the equation up as
$(x,y,x) = (3,1,6) + t(1,7,-2)$
and go from there.
I don't get where I can get an orthogonal vector. Normally when I am finding an orthogonal vector I have two other vectors and do the cross product to find it.
I am thinking somehow I have to get three points on the plane, but I'm not sure how to go about doing that.
Any pointers?
thanks in advance.
Best Answer
For a plane in $\mathbb{R}^3$ with $\mathbf{r_0}$ a point that lies in the plane and $\mathbf{n}$ a vector normal to the plane, its equation can be given by: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad \mbox{where} \quad \mathbf{r}=(x,y,z) $$
To solve your question, let us first rearrange the above equation: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}-\mathbf{n}\cdot\mathbf{r_0}=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{r_0} $$
Substituting values, gives the following equation for the plane in Cartesian form: $$ (1,7,-2)\cdot(x,y,z) = (1,7,-2)\cdot(3,1,6) \quad \Rightarrow \quad x+7y-2z = -2 $$