Let u , v , and w be three vectors such that | u | = 1 , | v | = 2 , and | w | = 3 . If | u + t v | 1 and | v + s w | 2 for all real numbers t and s , then the maximum possible value of | u + v + w |^2 is

Step-by-step solution

Given | u + t v | 1 for all t R . Squaring both sides, we get | u |^2 + t^2| v |^2 + 2t( u v ) 1 . Since | u | = 1 , this simplifies to 1 + 4t^2 + 2t( u v ) 1 4t^2 + 2t( u v ) 0 . For this quadratic in t to be non-negative for all real t , its discriminant must be less than or equal to zero: = 4( u v )^2 - 4(4)(0) 0 u v = 0 . Similarly, | v + s w | 2 for all s R . Squaring gives | v |^2 + s^2| w |^2 + 2s( v w ) 4 . Since | v | = 2 , we have 4 + 9s^2 + 2s( v w ) 4 9s^2 + 2s( v w ) 0 . Again, the discriminant must be