Question

Find the angle between vector u= 3i + sqrt 3j and vector v=-2i-5j to the nearest degree a) 82 degrees b) 38 degrees c) 142 degrees d) 98 degrees

Answer 1

c) 142 degrees I think

Answer 2

The angle between u and v is 142°

C is correct

Explanation:

Given:
`\vec{u}=3i+√(3)j`

`\vec{v}=-2i-5j`

We need to find the angle between u and v vector.

Using dot product.

`\vec{u}\cdot \vec{v}=|u||v|\cos\theta`

where,

`\theta\text{ is angle between u and v}`

`(3i+√(3)j)\cdot (-2i-5j)=√(9+3)\cdot √(4+25)\cdot \cos\theta`

`-6-5√(3)=√(348)\cos\theta`

`\cos\theta=(-6-5√(3))/(√(348))`

`\theta=\cos^(-1)(-0.7859)`

`\theta=141.8`

Nearest degree

`\theta=142^\circ`

Hence, The angle between u and v is 142°