Question

The angle between vectors `vecA = 3hati + 4hatj + 5hatk` and `vecB = 6hati + 8hatj + 10hatk` is

पर्याय

• zero

• 45°

• 90°

• 180°

Answer 1

zero

Explanation:

`vecA * vecB  = AB cos theta`   ......(i)

`vecA * vecB = (3hati + 4hatj + 5hatk)*(6hati + 8hatj + 10hatk)`

= `3 xx 6 + 4 xx 8 + 5 xx 10` = 100  .......(ii)

The magnitudes of A and B are

A = `sqrt((3)^2 + (4)^2 + (5)^2) = sqrt(50)` ......(iii)

B = `sqrt((6)^2 + (8)^2 + (10)^2) = sqrt(200)`  ......(iv)

Substituting the values from (ii), (iii) and (iv) in (i), we have

100 = `sqrt(50) xx sqrt(200) cos theta` = 100 cos θ

∴ cos θ = 1 or θ = zero.