What is the angle between #<-4 , 0 , -1 > # and # < 8 , 6 , 3 > #?

Let the vectors be called AC and AB. Then the vector BC = AB - AC = <12, 6, 4>. The sides a, b and c of the #triangle#ABC are 14, #sqrt#17 and #sqrt#109. The angle at A is arccos (( b.b + c.c - a.a ) / (2bc).= arccos( -0.8131) =144.4 deg.

To find the angle between two vectors \( \mathbf{u} = <-4, 0, -1> \) and \( \mathbf{v} = <8, 6, 3> \), you can use the dot product formula: \[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \cdot |\mathbf{v}| \cdot \cos(\theta) \] Where: - \( \mathbf{u} \cdot \mathbf{v} \) is the dot product of the two vectors, - \( |\mathbf{u}| \) and \( |\mathbf{v}| \) are the magnitudes of the vectors, - \( \theta \) is the angle between the vectors. First, calculate the dot product of the vectors: \[ \mathbf{u} \cdot \mathbf{v} = (-4)(8) + (0)(6) + (-1)(3) = -32 + 0 - 3 = -35 \] Then, find the magnitudes of the vectors: \[ |\mathbf{u}| = \sqrt{(-4)^2 + (0)^2 + (-1)^2} = \sqrt{16 + 0 + 1} = \sqrt{17} \] \[ |\mathbf{v}| = \sqrt{(8)^2 + (6)^2 + (3)^2} = \sqrt{64 + 36 + 9} = \sqrt{109} \] Now, plug these values into the formula: \[ -35 = \sqrt{17} \cdot \sqrt{109} \cdot \cos(\theta) \] \[ \cos(\theta) = \frac{-35}{\sqrt{17} \cdot \sqrt{109}} \] \[ \theta = \arccos \left( \frac{-35}{\sqrt{17} \cdot \sqrt{109}} \right) \] Calculate the arccosine of the value to find the angle \( \theta \). This angle will be in radians. If you want the result in degrees, convert it using the appropriate conversion factor.