What is the angle between ## and ##?

Question

What is the angle between #<1,5,7 ># and #< 6,8,1 >#?

Emery Adkins · Accepted Answer

52.49 deg, nearly.

use the angle between two vectors is arc cos (cross product of the components) / (product of the lengths of the vectors). Here, it is arc cos (6 + 40 + 7) / (#sqrt#75 #sqrt#101))

Carter Anderson · 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 1, 5, 7 \rangle $ and $ \mathbf{b} = \langle 6, 8, 1 \rangle $, we first find their magnitudes:

$$ |\mathbf{a}| = \sqrt{1^2 + 5^2 + 7^2} = \sqrt{1 + 25 + 49} = \sqrt{75} $$
$$ |\mathbf{b}| = \sqrt{6^2 + 8^2 + 1^2} = \sqrt{36 + 64 + 1} = \sqrt{101} $$

Next, we calculate the dot product:

$$ \mathbf{a} \cdot \mathbf{b} = (1)(6) + (5)(8) + (7)(1) = 6 + 40 + 7 = 53 $$

Now, we can find the angle $ \theta $ using the dot product formula:

$$ 53 = \sqrt{75} \times \sqrt{101} \times \cos \theta $$

$$ \cos \theta = \frac{53}{\sqrt{75} \times \sqrt{101}} $$

$$ \theta = \arccos \left( \frac{53}{\sqrt{75} \times \sqrt{101}} \right) $$

$$ \theta \approx 28.38^\circ $$

So, the angle between the vectors $ \langle 1, 5, 7 \rangle $ and $ \langle 6, 8, 1 \rangle $ is approximately $ 28.38^\circ $.

What is the angle between #&lt;1,5,7 &gt;# and #&lt; 6,8,1 &gt;#?

What is the angle between #<1,5,7 ># and #< 6,8,1 >#?