Question

(6, - 3, 1) and (8,9,-11) find the angle between the vector​

Answer 1

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 $\vec u = \langle 6, -3, 1 \rangle$ and $\vec v = \langle 8,9,-11 \rangle$, the angle between the two vectors is determined from definition of dot product:

$\theta = \cos^{-1} \left(\frac{\vec u \,\bullet \,\vec v}{\|\vec u\|\cdot \|\vec v\|} \right)$ (1)

Where:

$\vec u$, $\vec v$ - Vectors.

$\|\vec u\|$, $\|\vec v\|$ - 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 $\vec u = \langle 6,-3,1 \rangle$ and $\vec v = \langle 8, 9,-11 \rangle$, then the angle between the two vectors is:

$\theta = \cos^{-1}\left[\frac{(6)\cdot (8) + (-3)\cdot (9) + (1)\cdot (-11)}{\sqrt{6^{2}+(-3)^{2}+1^{2}}\cdot \sqrt{8^{2}+9^{2}+(-11)^{2}}} \right]$

$\theta \approx 84.813^{\circ}$

The angle between the two vectors is 84.813°.