I have done part (a). For part (b), I know the principle of how to do it, I tried to use the cross product to find the exact value of the sine of the angle.
So I found PQ which is $$\begin{pmatrix}7-2\\-1-1\\2-6\end{pmatrix}$$
$$=\begin{pmatrix}5\\-2\\-4\end{pmatrix}$$
Then I do the cross product between $$\begin{pmatrix}5\\-2\\-4\end{pmatrix}$$ and $$\begin{pmatrix}5\\-3\\-1\end{pmatrix}$$ which the normal vector for the plane equation given
and using $$||{\bf a} \times {\bf b}|| = ||{\bf a}|| \, ||{\bf b}|| \sin \theta,$$ I got the answer to be $\sqrt (490)/\sqrt (490)$ which makes $sin\theta =1$ but the answer states that $sin\theta = \sqrt 7/3$. Where did I go wrong?
Thank you and sorry in advance for any wrong tags and title labelling.
Best Answer
See the determinant is $-10i+15j-5k$ so its $\sqrt{350}$ and rhs is $\sqrt{45\times 35}.sin\theta=\sqrt{1575}.sin\theta$ so $$\sqrt{\frac{350}{1575}}=sin\theta$$ solving you get $sin(\theta)=\sqrt{7}/3$. Thats all