Let u = i - 2 j + 2 k and v = i + 2 j + k , where -1 . The vector v is expressed as v = v ₁ + v ₂ , where v ₁ is parallel to u and v ₂ is perpendicular to u . If | v ₂| = 5 | v ₁| , then the value of | v ₂|^2 is equal to

Step-by-step solution

Given v = v ₁ + v ₂ with v ₁ u and v ₂ u , the vectors v ₁ and v ₂ are orthogonal components of v . By the Pythagorean theorem, | v |^2 = | v ₁|^2 + | v ₂|^2 . Since | v ₂| = 5 | v ₁| , we have | v ₂|^2 = 5| v ₁|^2 . Substituting this into the magnitude equation gives: | v |^2 = | v ₁|^2 + 5| v ₁|^2 = 6| v ₁|^2 The squared magnitude of v is: | v |^2 = ^2 + 2^2 + 1^2 = ^2 + 5 The magnitude of the parallel component v ₁ is the absolute value of the projection of v on u : | v ₁| = | u v | | u | Calculating the dot pro