Question

A and B are two vectors. If |Al =
3, IBI = 4 and A+BI = 1, then |A-
Blis​

Answer 1

Explanation:

Given ∣

a

∣=1,

∣

∣

∣

∣

b

∣

∣

∣

∣

=4

a

.

b

=2 and

c

=(2

a

×

b

)−3

b

a

.

b

=2⇒∣a∣∣b∣.cos(a,b)=2

1×4×cosθ=2

cosθ=

2

1

⇒θ=60

∘

(Angle b/w

a

,

b

)

a

×

b

=∣a∣∣b∣.sinθ=1×4×sin60=4×

2

2

=2

3

Given

c

=2

a

×

b

−3

b

squaring on both sides

∣

c

∣

2

=

∣

∣

∣

2

a

ˉ

×

b

ˉ

∣

∣

∣

2

+9

∣

∣

∣

∣

b

∣

∣

∣

∣

2

−2×(3

a

(2

a

×

b

))

∣

c

∣

2

=4

∣

∣

∣

∣

a

×

b

∣

∣

∣

∣

2

+9

∣

∣

∣

∣

b

∣

∣

∣

∣

2

−0

(∵ b ,2 a × b are perpendicular to each other) ∣ c ∣ 2 =4×(2 3 ) 2 +9(4) 2

∣c∣ 2 =4×12+9×16=48+144=192 ∣c∣

2=192⇒ ∣c∣=8 3c =(2 a × b )−3 b

dot product on both sides with bc . b =(2 a × b ). b −3 b . b c . b=0−3∣b∣ 2

(∵ b 12 a × b and cos90=0)

∴∣c∣.∣b∣.cos(α)=−3∣b∣

cos(α)= 83−3×4 = 2− 3−α= 65π

solution