Question

Show that there is a 96.7°angle between the vectors 5a +37 and 3ā – 5 if ā= -i+j – k and 7 = 3i + 2j – 3k .

Answer 1

Solution:- Let the vectoes are vi = 5a + 3 b V2 = 34 b a - ā=-i+jok & b = 31 toj 1 3K Н VI = 5(-itj - k) + 3( 31 +23-3k ) = -51 +5j-sk + gi +6; -gk vi 41 +11j-14k V2 = 31-i+j-k) - (31 + 2] -3k ) -31 + 3) -3% -31 - 2; +3K V2 こ -61 + j-ok To find formula is cos d = ă. b Täib cos d T - ViV2 IVIT 1V21 (41 +11-14k) (-6i +j-ok) 142 +112+-14j?. /F632+12+02 -24 +11 to 16+ 121 +196 36 +1 +0 = 13 333 37

il -13 (:1333 = 37x9 = 3√37 537 3837 ) -13 + cos 2 13 c d = car-l (-13 T d ñ 96.72575224° ㅋ e 96.7° T Angle between two vectors is 96.7°