What is the angle between # # and ##?

Question

What is the angle between #<9,-5,1 > # and #< -7,4,2 >#?

Oliver Atkinson · Accepted Answer

The angle is #=160.5^@#

The angle between #2# vectors #vecA# and #vecB# is given by the dot product definition.

#vecA.vecB=∥vecA∥*∥vecB∥costheta#

Where #theta# is the angle between #vecA# and #vecB#

The dot product is

#vecA.vecB=〈9,-5,1〉.〈-7,4,2〉=-63-20+2=-81#

The modulus of #vecA#= #∥〈9,-5,1〉∥=sqrt(81+25+1)=sqrt107#

The modulus of #vecB#= #∥〈-7,4,2〉∥=sqrt(49+16+4)=sqrt69#

So,

#costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-81/(sqrt107*sqrt69)=-0.943#

#theta=arccos(-0.943)=160.5^@#

Aaliyah Abdullah · Answer

undefined To find the angle between two vectors $ \mathbf{u} $ and $ \mathbf{v} $, you can use the formula:

$$ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} $$

Where:
- $ \theta $ is the angle between the two vectors.
- $ \mathbf{u} \cdot \mathbf{v} $ is the dot product of the two vectors.
- $ \|\mathbf{u}\| $ and $ \|\mathbf{v}\| $ are the magnitudes (or lengths) of the vectors.

Given the vectors $ \mathbf{u} = \langle 9, -5, 1 \rangle $ and $ \mathbf{v} = \langle -7, 4, 2 \rangle $, you can calculate the dot product and the magnitudes:

$$ \mathbf{u} \cdot \mathbf{v} = (9 \times -7) + (-5 \times 4) + (1 \times 2) $$
$$ \|\mathbf{u}\| = \sqrt{9^2 + (-5)^2 + 1^2} $$
$$ \|\mathbf{v}\| = \sqrt{(-7)^2 + 4^2 + 2^2} $$

After computing these values, substitute them into the formula to find $ \theta $, the angle between the two vectors.

What is the angle between #&lt;9,-5,1 &gt; # and #&lt; -7,4,2 &gt;#?

What is the angle between #<9,-5,1 > # and #< -7,4,2 >#?