Question

Show that vectors `vec a = 2 hat i + 5 hat j - 6 hat k` and `vec b = hat i + 5/2 hat j - 3 hat k` are parallel.

Answer 1

Let the angle between the two vectors be θ.

∴ cos θ = `(vec a * vec b)/(|vec a| |vec b|)`

= `((2 hat i + 5 hat j - 6 hat k) * (hat i + 5/2 hat j - 3 hat k))/(sqrt (2^2 + 5^2 + (-6)^2) xx sqrt(1^2 + (5/2)^2 + (-3)^2))`

= `(2 + 25/2 + 18)/(sqrt 65 xx sqrt (65//4))`

= `(65//2)/(65//2)`

= 1

⇒ θ = cos⁻¹ (1) = 0°

⇒ Two vectors are parallel.

Alternate method:

`vec a = 2 hat i + 5 hat j - 6 hat k = 2 (hat i + 5/2 hat j - 3hat k) = 2 vec b`

Since `vec a` is a scalar multiple of `vec b`, the vectors are parallel.