Question

Three vectors A, B, C in three-dimensional space satisfy the following properties

Answer 1

Answer:

$\frac{7}{8}\pi$

Step-by-step explanation:

observe

||a–b+c|| = ||a+b+c||

(a-b+c)² = (a+b+c)²

(a+b+c)² – (a-b+c)² = 0

((a+b+c)+(a-b+c))((a+b+c)–(a-b+c)) = 0

(2a+2c)(2b) = 0

(a+c)b = 0

a•b + c•b = 0

||a||×||b||×cos(π/8) + ||c||×||b||×cos(θ) = 0

$\cos(\theta)=-\frac{||a||\times ||b|| \times \cos(\frac{\pi}{8})}{||c||\times ||b||}=-\cos(\frac{\pi}{8})\\ \theta=\frac{7}{8}\pi$