Find the angle between the two planes in degrees.Round to a tenth of a degree.
5x - 6y - 4z = -5
X + y + 5z = -1
A)114.5°
B)27.4°
C)117.4°
D)24.7°
E)2.0°

Question

Find the angle between the two planes in degrees.Round to a tenth of a degree. 
5x - 6y - 4z = -5
X + y + 5z = -1
A)114.5°
B)27.4°
C)117.4°
D)24.7°
E)2.0°

Quizplus · Accepted Answer

To find the angle between two planes, we need to find the cosine of the angle between their normal vectors. We can find the normal vectors by looking at the coefficients of x, y, and z in the equations of the planes. For the first plane, the normal vector is <5, -6, -4>, and for the second plane, the normal vector is <1, 1, 5>. We can find the cosine of the angle between these vectors using the dot product: <5, -6, -4> · <1, 1, 5> = 5(1) + (-6)(1) + (-4)(5) = -29 ||<5, -6, -4>|| ||<1, 1, 5>|| sqrt(5^2 + (-6)^2 + (-4)^2) sqrt(1^2 + 1^2 + 5^2) sqrt(77) sqrt(27) The cosine of the angle between the planes is therefore: cos θ = -29 / (sqrt(77) sqrt(27)) ≈ -0.957 Taking the inverse cosine, we get: θ ≈ 117.4° So the answer is (C).

Find the Angle Between the Two Planes in Degrees