Question

What are the values of x for which the angle between the vectors? `2x^2hati + 3xhatj + hatk` and `hati - 2hatj + x^2hatk` are obtuse?

Answer 1

Given, `veca = 2x^2hati + 3xhatj + hatk` and `vecb = hati - 2hatj + x^2hatk`

Angle between `veca` and `vecb` is obtuse.

⇒ cos θ < 0

cos θ = `(veca.vecb)/(|veca||vecb|)`

`|veca||vecb| > 0`

∴ `veca . vecb < 0`

⇒ `veca.vecb`

= (2x²) (1) + (3x) (−2) + (1) (x²)

= 2x² – 6x + x² 

= 3x² − 6x

= 3x (x – 2)

3x (x − 2) < 0

Divide by 3 (positive, inequality unchanged):

x (x − 2) < 0

Now x (x − 2) is negative between its roots x = 0 and x = 2.

0 < x < 2

x ∈ (0, 2)