Question

given that vector A + vector b + vector C is equal to zero out of three vectors two equal in magnitude and the magnitude of the third vector is root 2 times that of either of the two vectors having equal magnitude then the angle between vector are given by

Answer 1

Answer:

90° , 135° &  135°

Step-by-step explanation:

Let say magnitude of A & B = Equal  M  

C = M√2

ACosA + BcosB + C CosC  = 0

McosA + MCosB + M√2CosC = 0

McosA + MCosB = -M√2CosC

Squaring both sides

M²cos²A + M²Cos²B + 2M²cosACosB= 2M²Cos²C

AsinA + BSinB + C SinC = 0

MSinA + MSinB + M√2SinC = 0

M²Sin²A + M²Sin²B + 2M²SinASinB= 2M²Sin²C

Adding both

M² + M² + 2M²(cosACosB + SinASinB) = 2M²

=> cosACosB + SinASinB = 0

=> Cos(A-B) = 0

=> A - B = 90°

Angle between A & B = 90°

M²Sin²A + M²Sin²B + 2M²SinASinB= 2M²Sin²C

=> M²Sin²(90 + B) + M²Sin²B + 2M²Sin(90 + B)SinB= 2M²Sin²C

Sin(90 + B) = CosB

=> M²Cos²B + M²Sin²B + 2M²CosBSinB= 2M²Sin²C

=> M² + M²Sin2B = 2M²Sin²C

=> M²Sin2B = M²(2sin²C - 1)

=> M²Sin2B = - M²Cos2C

=> Sin2B  = -Cos2C

=>  -Sin2B = Cos(2C)

=> Sin(-2B) = Cos2C

=> Cos(90 + 2B) = Cos2C

=>2B  + 90 = 2C

=> 2(90 + A) + 90 = 2C

=> 270 = 2(C - A)

=> C - A = 135°

Angle between C & A = 135°