Two planes have eqns r.(3i – 4j + 2k) = 5 and r = λ (2i + j + 5k) + μ(i – j – 2k), where λ and μ are scalar parameters. Find the acute angle between the planes, giving your answer to the nearest degree.

Summary of solution: To find the angle between the planes, we must find the normal vector to each plane and then use the scalar product to find the angle between these two normal vectors. Breaking this down into simple steps: 1) Find the normal to the first plane: As the first plane is in the form r.n = a.n, which means the plane is perpendicular to the vector n and passes through the point with position vector a, it is simple enough to deduce that the normal vector is 3i - 4j + 2k. 2) Find the normal to the second plane: The normal to the plane must be perpendicular to both vectors in the vector equation. To find the vector that satisfies this, we cross product the two vectors (this can be demonstrated more easily using the whiteboard). The calculated normal vector is 3i + 9j - 3k. This can be simplified to i + 3j - k to give a vector of the same direction. 3) Use scalar product on the two vectors: To summarise so far, we have calculated that the normal vectors to the two planes are 3i - 4j + 2k and i + 3j - k. The scalar product equation is a.b =|a||b|cosθ. So, to start, a.b = (3)(1) + (-4)(3) + (2)(-1) = -11. The magnitude of the two vectors are |a| = rt(9+16+4) = rt(29) and |b| = rt (1+9+1) = rt(11). Therefore cosθ = (-11)/[rt(29)*rt(11)] = -0.61588.... Finally, θ = 128.01.... except that the question asks for the acute angle, therefore θ = 180 - 128.01... = 52 degrees to the nearest degree.