Q.(a) Find the vector equation (any form) or Cartesian equation of a plane passing through the points $(2, 2, 1)$, $(9, 3, 6)$ and perpendicular to the plane $2x+6y+6z=9$. OR

(a) finds a plane's normal as the cross product of the chord joining the two given points and the normal of the given plane, then writes the point-normal equation; (b) computes the angle between two curves at their intersection point from the tangent slopes there.

(a) Plane through two points, perpendicular to a given plane

1. Let $A(2,2,1)$, $B(9,3,6)$. Then $\vec{AB}=B-A=(7,1,5)$.
2. The normal of the given plane $2x+6y+6z=9$ is $\vec n_2=(2,6,6)$.
3. The required plane contains $\vec{AB}$ (since it passes through $A$ and $B$) and is perpendicular to the given plane, so its normal is $\vec n=\vec{AB}\times\vec n_2$.
4. $\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\7&1&5\\2&6&6\end{vmatrix}=\hat i(1\cdot6-5\cdot6)-\hat j(7\cdot6-5\cdot2)+\hat k(7\cdot6-1\cdot2)=(-24,-32,40)$.
5. Divide by $-8$ to simplify: $\vec n=(3,4,-5)$.
6. Plane through $A(2,2,1)$ with normal $(3,4,-5)$: $3(x-2)+4(y-2)-5(z-1)=0 \Rightarrow 3x+4y-5z-9=0$, i.e. $3x+4y-5z=9$.
7. Check: $B(9,3,6)$ gives $3(9)+4(3)-5(6)=27+12-30=9$ ✓. Perpendicularity: $(3,4,-5)\cdot(2,6,6)=6+24-30=0$ ✓.

(b) Angle between y = x^2 and x = y^2 at (1,1) …