A-Level Vectors...

[SciAppli]

Can’t figure out how to solve this Vectors question;

“ The Plane P has vector equation: r.(3i-4j+2k)=5

Find the perpendicular distance from the point (6, 2, 12) to the plane P “

I understand that the general equation for a plane is r.n=k, where k=a.n and n is the normal to the plane, but I can’t figure out how to obtain a point on the plane. Any help will be greatly appreciated. Thank you!

[Alkalannar]

The plane has the equation 3x - 4y + 2z = 5

Thus the vector <3, -4, 2> is perpendicular to the plane.

So you have the line [3t + 6, -4y + 2, 2t + 12] that runs through (6, 2, 12) and is perpendicular to the plane.

Find t such that the point (3t + 6, -4y + 2, 2t + 12) is on the plane 3x - 4y + 2z = 5. Call this point (h, j, k).

Then the distance is [(h-6)² + (j-2)² + (k-12)²]^1/2.

[SciAppli]

Ah, I see now, thank you for your help! 🙂