Finding the exact value of the sine of the angle between a line and a plane

Question

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.

Answer 1

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

Answer 2

Use the scalar product to find the cosine of the angle between $PQ$ and the normal. The angle between the line and the plane is $90 -$ this, so the sine of the required angle is the same as the cosine you obtain from the scalar product. You don't need to use cross product at all.

Answer 3

The angle angle between line $PQ$: $(5, -2, -4)$ & the normal to the plane:$(5, -3, -1)$ is given by using dot product as $$\cos^{-1}\left(\frac{(5, -2, -4)\cdot (5, -3, -1)}{|5, -2, -4||5, -3, -1|}\right)=\cos^{-1}\left(\frac{35}{\sqrt {45}\sqrt{35}}\right)=\cos^{-1}\left(\frac{\sqrt 7}{3}\right)$$ hence the angle between plane & the line $PQ$ $$\theta=\frac{\pi}{2}-\cos^{-1}\left(\frac{\sqrt 7}{3}\right)$$$$=\color{red}{\sin^{-1}\left(\frac{\sqrt 7}{3}\right)}$$ or $$\sin\theta=\frac{\sqrt7}{3}$$

Answer 4

You are using cross product instead of dot/scalar product. It should be:

$$ a . b = ||a||~||b|| sin \theta $$

$$ sin\theta = \frac{a.b}{||a||~||b||} = cos\alpha $$

$\theta$ is the angle between the line PQ and the plane.
$\alpha$ is the angle between the line PQ and the normal to the plane, (5,-3,-1).

$$ sin\theta = \frac{(5,-2,-4).(5,-3,-1)}{||(5,-2,-4)||.||(5,-3,-1)||} $$

$$ = \frac{35}{\sqrt{45} ~.~ \sqrt{35}} $$

$$ = \frac{7~.~5}{\sqrt{9}\sqrt{5} ~.~ \sqrt{7}\sqrt{5}} $$

$$ \implies sin\theta = \sqrt{7}/3 $$