Answer:
The angle between the two vectors is 84.813°.
Step-by-step explanation:
Statement is incomplete. Complete form is presented below:
Let be (6,-3, 1) and (8, 9, -11) vector with same origin. Find the angle between the two vectors.
Let [tex]\vec u = \langle 6, -3, 1 \rangle[/tex] and [tex]\vec v = \langle 8,9,-11 \rangle[/tex], the angle between the two vectors is determined from definition of dot product:
[tex]\theta = \cos^{-1} \left(\frac{\vec u \,\bullet \,\vec v}{\|\vec u\|\cdot \|\vec v\|} \right)[/tex] (1)
Where:
[tex]\vec u[/tex], [tex]\vec v[/tex] - Vectors.
[tex]\|\vec u\|[/tex], [tex]\|\vec v\|[/tex] - Norms of each vector.
Note: The norm of a vector in rectangular form can be determined by either the Pythagorean Theorem or definition of Dot Product.
If we know that [tex]\vec u = \langle 6,-3,1 \rangle[/tex] and [tex]\vec v = \langle 8, 9,-11 \rangle[/tex], then the angle between the two vectors is:
[tex]\theta = \cos^{-1}\left[\frac{(6)\cdot (8) + (-3)\cdot (9) + (1)\cdot (-11)}{\sqrt{6^{2}+(-3)^{2}+1^{2}}\cdot \sqrt{8^{2}+9^{2}+(-11)^{2}}} \right][/tex]
[tex]\theta \approx 84.813^{\circ}[/tex]
The angle between the two vectors is 84.813°.
