Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <2, -4>, v = <3, -8>

Answer 1

Hello,


`\vec{u}=2\vec{i}-4\vec{j}\\ \vec{v}=3\vec{i}-8\vec{j}\\ \vec{u}*\vec{v}=(2\vec{i}-4\vec{j})*(3\vec{i}-8\vec{j})=6+32=38\\ =|u|.|v|.cos(\alpha)=√(20).√(73).cos(\alpha)\\ cos(\alpha)=(38)/(√(20.73))\\\\ ==\ \textgreater \ \alpha=6.00900595`

≈6°

Answer 2

The angle between the given vectors to the nearest tenth of a degree is:

6 degree

Explanation:

Let α be the angle between two vectors.

The cosine of the angle between two vectors is given by:

`\cos \alpha=(u\cdot v)/(|u||v|)`

Here we have vector u as:

`u=<2,-4>`

and

`v=<3,-8>`

`u\cdot v=<2,-4>\cdot <3,-8>\\\\u\cdot v=2* 3+(-4)* (-8)\\\\u\cdot v=6+32\\\\u\cdot v=38`

Also,

`|u|=√(2^2+(-4)^2)\\\\|u|=√(4+16)\\\\|u|=√(20)\\\\|u|=2√(5)`

and

`|v|=√(3^2+(-8)^2)\\\\|v|=√(9+64)\\\\|v|=√(73)`

Hence, we get:

`\cos \alpha=(38)/(2√(5)\cdot √(73))\\\\\cos \alpha=0.9945\\\\i.e.\\\\\alpha=\arccos 0.9945\\\\\alpha=6.01198`

which to the nearest tenth of a degree is: 6°