Question

Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular to the plane 4x+5y+6z=14. Write the equation in the form indicated.

Answer 1

The plane we want to find has general equation

$a(x-2)+b(y-5)+c(z-7)=0$

with $a,b,c$ not equal to 0, and has normal vector

$\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k$

$\vec n$ is perpendicular to both the normal vector of the other plane, which is $4\,\vec\imath+5\,\vec\jmath+6\,\vec k$, as well as the tangent vector to the line $\vec r(t)$, which is $\vec\imath+2\,\vec\jmath+9\,\vec k$.

This means the dot product of $\vec n$ with either vector is 0, giving us

$\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}$

Suppose we fix $c=1$. Then the system reduces to

$\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}$

and we get

$(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10$

$a+2(-10)=-9\implies a=11$

Then one equation for the plane could be

$\boxed{11(x-2)-10(y-5)+(z-7)=0}$

or in standard form,

$\boxed{11x-10y+z=-21}$

The solution is unique up to non-zero scalar multiplication, which is to say that any equation $(11x-10y+z)k=-21k$ would be a valid answer. For example, suppose we instead let $c=2$; then we would have found $a=22$ and $b=-20$, but clearly dividing both sides of the equation

$22(x-2)-20(y-5)+2(z-7)=0$

by 2 gives the same equation as before.