Equations of Planes — Question 3

Question 3 – Solution

Strategy Plane orientation is controlled by the normals $\vec n_k=\left\langle k,2,-1\right\rangle$ and $\vec m=\left\langle 2,-1,2\right\rangle$.

Step 1: Perpendicularity Calculate $$\vec n_k\cdot\vec m=k(2)+2(-1)+(-1)(2)=2k-4.$$ Perpendicular normals require $2k-4=0$, hence $\boxed{k=2}$.

Step 2: Parallelism Parallelism would require $\left\langle k,2,-1\right\rangle=\lambda\left\langle 2,-1,2\right\rangle$. From $2=-\lambda$, $\lambda=-2$; but the third coordinate would require $-1=2(-2)=-4$, a contradiction. Hence .

Step 3: The $60^\circ$ condition The angle formula gives $$\frac{|2k-4|}{3\sqrt{k^2+5}}=\frac 12.$$ Cross-multiply and square: $$2|2k-4|=3\sqrt{k^2+5}\Longrightarrow 16(k-2)^2=9(k^2+5).$$ Expanding gives $16k^2-64k+64=9k^2+45$, so $$7k^2-64k+19=0.$$ The quadratic formula yields $$k=\frac{64\pm\sqrt{64^2-4(7)(19)}}{14} =\frac{64\pm 18\sqrt{11}}{14},$$ and therefore $$\boxed{k=\frac{32\pm 9\sqrt{11}}7}.$$

Verification Neither root makes the original denominator zero, and substitution into the unsquared equation gives the positive value $1/2$, so no extraneous root remains.