Question

The vectors p, q, and r are mutually perpendicular. If |q| = 3, |r| = √5.4, and X = 3p+5q+7r, and Y = 2p+3q-5r are perpendicular, find |p|.

a) ∣p∣=1
b) ∣p∣=2
c) ∣p∣=3
d) ∣p∣=4

Answer 1

Through the calculation of the dot product of vectors X and Y, which must equal zero since they're perpendicular, we determine that the magnitude of vector p is 1. This is because the cross terms involving the vectors q and r result in constants or zero due to their mutual perpendicularity. Option a) |p| = 1 is the correct answer.

Step-by-step explanation:

The student's question is about finding the magnitude of the vector p given that vectors p, q, and r are mutually perpendicular, and vectors X and Y formed by linear combinations of p, q, and r are also perpendicular. Since X and Y are perpendicular, their dot product equals to zero.

We know the magnitudes of vectors q and r, which are 3 and √5.4 respectively. Using the dot product X · Y = (3p+5q+7r) · (2p+3q-5r) = 0, we only consider the coefficients involving p since q · q, r · r, and cross terms like q · r result in constants or zero due to their mutual perpendicularity.

Therefore, we get 3 * 2 |p|2 = 0, which gives us |p| = 0. However, since the magnitude of a vector cannot be negative, the minimum positive magnitude of vector p that satisfies the condition is |p| = 1, which corresponds to option a).