Let S be the plane passing through the point (7,1,6) and that contains the line
r(t) = (1,0,2) + t(4,-1,3) where t ∈ ℝ
Q1) Find a cartesian equation of S
My attempt: I am not sure if I am doing this correctly and would like further help if I have missed something
The direction of r(t) is (4,-1,3) so (x-7)/4 = -(y-1) = (z-6)/3
or (4,-1,3)(x-7,y-1,z-6) = 0 which gives 4x – y + 3z = 45
Q2) Find the vector equation of the plane R which is perpendicular to S and contains the line r(t)
I am unsure how to attempt this question
Best Answer
Your answer to the first question is not correct, since $(1,0,2)$ does not belong to the plane $4x-y+3z=45$.
Let $A=(1,0,2)$, $B=(1,0,2)+(4,-1,3)=(5,-1,5)$, and let $C=(7,1,6)$. Then$$\vec{AB}\times\vec{AC}=(-7,2,10)$$and therefore a Cartesian equation of the plane $p_1$ that you're after is$$-7(x-1)+2y+10(z-2)=0(\iff-7x+2y+10z=13).$$
Now, if $D=A+\vec{AB}\times\vec{AC}=(-6,2,12)$, then the plane $p_2$ defined by $A$, $B$ and $D$ contains the line $\mathbf r$ and it is orthogonal to the plane $p_1$. The same method as above shows that a Cartesian equation of $p_2$ is $16x+61y-z=14$. See the picture below.