What is the angle between #<8,-1,2> # and #<-4,-1,3>#?

The angle is given by the dot product definition

Here, we have

Therefore,

To find the angle between two vectors ( \mathbf{a} ) and ( \mathbf{b} ), you can use the dot product formula:

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

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

Given vectors ( \mathbf{a} = \langle 8, -1, 2 \rangle ) and ( \mathbf{b} = \langle -4, -1, 3 \rangle ), you can calculate the dot product:

[ \mathbf{a} \cdot \mathbf{b} = (8 \cdot -4) + (-1 \cdot -1) + (2 \cdot 3) ]

[ \mathbf{a} \cdot \mathbf{b} = -32 + 1 + 6 ]

[ \mathbf{a} \cdot \mathbf{b} = -25 + 6 ]

[ \mathbf{a} \cdot \mathbf{b} = -19 ]

Next, calculate the magnitudes of the vectors:

[ |\mathbf{a}| = \sqrt{8^2 + (-1)^2 + 2^2} = \sqrt{64 + 1 + 4} = \sqrt{69} ] [ |\mathbf{b}| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26} ]

Now, substitute the values into the formula:

[ -19 = \sqrt{69} \cdot \sqrt{26} \cdot \cos(\theta) ]

[ \cos(\theta) = \frac{-19}{\sqrt{69} \cdot \sqrt{26}} ]

[ \theta = \arccos\left(\frac{-19}{\sqrt{69} \cdot \sqrt{26}}\right) ]

[ \theta \approx 106.56^\circ ]

So, the angle between the vectors ( \langle 8, -1, 2 \rangle ) and ( \langle -4, -1, 3 \rangle ) is approximately ( 106.56^\circ ).