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Lines and Planes help please!

Plane Π\Pi has equation 5x3yz=15x-3y-z=1
a. Show that point P (2, 1, 6) lies in Π\Pi
b. Point Q has coordinates (7, -1, 2). Find the exact value of the sine of the angle between (PQ) and Π\Pi.
c. Find the exact distance PQ.
d. Hence find the exact distance of Q from Π\Pi.

I've done a, but I'm stuck on b.
What I've done is take the vector PQ and got (5, -2, -4) in a column as the direction. I've turned it into a line, (2, 1, 6) + (5, -2, -4). I've checked the answer, and it's
73\frac{\sqrt7}{3}
I'm guessing the opposite length is root 7 and the hypotenuse is 3, but I can't find 3 numbers with a magnitude of root 7..?
I'm definitely doing something wrong.

All help will be appreciated :smile:
Reply 1
You need an angle between two things. One of them is PQ, and you only need the vector PQ, not the equation of the line. The other thing is the plane. But there isn't a single direction vector for the plane, because it's two-dimensional. However, there is a single vector normal to the plane - do you know how to get this from the equation of the plane?

Once you have vector PQ and the normal vector to the plane, you can use the usual method for the angle between two vectors. But you need to consider how the angle between PQ and the normal is related to the angle between PQ and the plane itself - this may explain why you are asked for the sine of the angle and not the cosine.

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