What is the angle between # # and # #?

Question

What is the angle between #<5 , 2 , -1 > # and # < 1, -3 , 1 > #?

Brooklyn Anderson · Accepted Answer

#costheta=(<5,2,-1>*<1,-3,1>)/(sqrt(5^2+2^2+(-1)^2)sqrt(1^2+(-3)^2+1^2))->costheta = (5(1)+2(-3)+(-1)1)/(sqrt(30) sqrt(11))->theta=cos^-1(-2/sqrt330)#
#theta=96.32^@# To find the angle between the vectors use the formula #cos theta=(a*b)/(|a||b|#. Note that a and b are vectors and #a*b#is the dot product of the vectors

Grace Ashcraft · Answer

undefined To find the angle between two vectors, you can use the dot product formula:

$$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \cos(\theta) $$

Where:
- $ \mathbf{a} $ and $ \mathbf{b} $ are the vectors,
- $ |\mathbf{a}| $ and $ |\mathbf{b}| $ are the magnitudes of the vectors, and
- $ \theta $ is the angle between the vectors.

Given vectors $ \mathbf{a} = \langle 5, 2, -1 \rangle $ and $ \mathbf{b} = \langle 1, -3, 1 \rangle $, we can calculate their dot product:

$$ \mathbf{a} \cdot \mathbf{b} = (5)(1) + (2)(-3) + (-1)(1) $$

$$ \mathbf{a} \cdot \mathbf{b} = 5 - 6 - 1 $$

$$ \mathbf{a} \cdot \mathbf{b} = -2 $$

Next, we need to find the magnitudes of the vectors:

$$ |\mathbf{a}| = \sqrt{5^2 + 2^2 + (-1)^2} = \sqrt{30} $$

$$ |\mathbf{b}| = \sqrt{1^2 + (-3)^2 + 1^2} = \sqrt{11} $$

Now, substitute the dot product and magnitudes into the formula:

$$ -2 = \sqrt{30} \cdot \sqrt{11} \cdot \cos(\theta) $$

$$ \cos(\theta) = \frac{-2}{\sqrt{30} \cdot \sqrt{11}} $$

$$ \cos(\theta) = \frac{-2}{\sqrt{330}} $$

Finally, find the angle $ \theta $ by taking the inverse cosine:

$$ \theta = \arccos\left(\frac{-2}{\sqrt{330}}\right) $$

Calculate the value of $ \theta $, which will give you the angle between the two vectors.

What is the angle between #&lt;5 , 2 , -1 &gt; # and # &lt; 1, -3 , 1 &gt; #?

What is the angle between #<5 , 2 , -1 > # and # < 1, -3 , 1 > #?