Q.(a) Find the cartesian equation of the plane containing the line $\dfrac{x-2}{2} = \dfrac{y-2}{3} = \dfrac{z-1}{-2}$ and passing through the point $(-1, 1, -1)$. OR

(a) finds the plane containing a given line and passing through an extra point using a normal built from the cross product; (b) factorises a degree-11 polynomial by grouping into $(x^5-1)(x^6+1)$ and solves each factor by De Moivre's theorem.

(a) Plane containing the line, through $(-1,1,-1)$

1. The line $\dfrac{x-2}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-2}$ passes through $A(2,2,1)$ with direction vector $\vec d=(2,3,-2)$.
2. The required plane contains this line (so it contains $A$ and is parallel to $\vec d$) and also passes through $B(-1,1,-1)$. Compute $\vec{AB}=B-A=(-3,-1,-2)$.
3. A normal to the plane is $\vec n=\vec d\times\vec{AB}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-2\\-3&-1&-2\end{vmatrix}$
4. $=\hat i\big(3(-2)-(-2)(-1)\big)-\hat j\big(2(-2)-(-2)(-3)\big)+\hat k\big(2(-1)-3(-3)\big)=\hat i(-8)-\hat j(-10)+\hat k(7)=(-8,10,7)$.
5. Plane through $A(2,2,1)$ with normal $(-8,10,7)$: $-8(x-2)+10(y-2)+7(z-1)=0$.
6. Simplify: $-8x+16+10y-20+7z-7=0\Rightarrow -8x+10y+7z-11=0$, i.e. $8x-10y-7z+11=0$.
7. Check: $B(-1,1,-1)$: $8(-1)-10(1)-7(-1)+11=-8-10+7+11=0$ ✓; $A(2,2,1)$: $16-20-7+11=0$ ✓; and $\vec n\cdot\vec d=-8(2)+10(3)+7(-2)=-16+30-14=0$ ✓ (normal is perpendicular to the line).

(b) Solve $x^{11}-x^6+x^5-1=0$

1. Group: $x^{11}-x^6+x^5-1 = x^6(x^5-1)+1\cdot(x^5-1) = (x^5-1)(x^6+1)$.
2. So either $x^5-1=0$ or $x^6+1=0$.
3. $x^5=1$: the five 5th roots of unity, $x=\cos\dfrac{2k\pi}{5}+i\sin\dfrac{2k\pi}{5}$, $k=0,1,2,3,4$ — i.e. $x=1$ and four complex roots at $72^\circ,144^\circ,216^\circ,288^\circ$. …