Question

Find the angle between v = 2i - j and w = 3i + 4jRound nearest tenth of a degree

Answer 1

The angle between two vectors

`\alpha = cos^(-1) ((2)/(5√(5) ) )`

∝ = 79.700°

Explanation:

Explanation

Given V = 2i - j and w = 3 i + 4 j

Let '∝' be the angle between the two vectors

`cos \alpha = (v^(-) .w^(-) )/(|v||w|)`

`cos \alpha = \frac{(2i-j).(3i+4j) }{\sqrt{2^(2)+1^(2) )\sqrt{3^(2) +4^(2) } } }`

`cos \alpha = \frac{(2(3)-4(1) }{\sqrt{5 )√(25 ) } } = (2)/(√(5))5 ) = (2)/(5√(5) )`

`cos\alpha = (2)/(5√(5) ) \\\alpha = cos^(-1) ((2)/(5√(5) ) )`

The angle between two vectors

∝ = 79.77°

Answer 2

`\theta=79.7^(\circ)`

Explanation:

Given that,

v = 2i - j and w = 3i + 4j

We need to find the angle between v and w.

Magnitude of |v|,
`|v|=√(2^2+(-1)^2) =\sqrt5`

Magnitude of |w|,
`|w|=√(3^2+4^2) =5`

The dot product of v and w,

`u{\cdot}w=2(3)+(-1)4\\\\=2`

The formula for the dot product is given by :

`u{\cdot}w=|u||w|\cos\theta\\\\\cos\theta=\frac{u{\cdot}w}\\\\=(2)/(\sqrt5* 5)\\\\\theta=\cos^(-1)((2)/(\sqrt5* 5))\\\\\theta=79.69^(\circ)\\\\=79.7^(\circ)`

So, the angle between u and v is
`79.7^(\circ)`.