Answer to Question #84921 in Mechanics | Relativity for Onuoha

Question

Find the angle between the two vectors
A= 2i +3j +k
B = I - 6j +k

Answer 1

We can find the angle between the two vectors from the formula:

$$cos\alpha = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{\lvert \overrightarrow{A} \rvert \cdot \lvert \overrightarrow{B} \rvert},$$

here,

$$\overrightarrow{A} \cdot \overrightarrow{B}$$

is the dot product of two vectors

$$\overrightarrow{A}$$

and

$$\overrightarrow{B}$$

;

$$\lvert \overrightarrow{A} \rvert$$

,

$$\lvert \overrightarrow{B} \rvert$$

are the magnitudes of the vectors

$$\overrightarrow{A}$$

and

$$\overrightarrow{B}$$

, respectively.

Let's first find the dot product of two vectors

$$\overrightarrow{A}$$

and

$$\overrightarrow{B}$$

:

$$\overrightarrow{A} \cdot \overrightarrow{B} = A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z = 2 \cdot 1 + 3 \cdot (-6) + 1 \cdot 1 = -15.$$

Then, let's find the magnitudes of the vectors

$$\overrightarrow{A}$$

and

$$\overrightarrow{B}$$

:

$$\lvert \overrightarrow{A} \rvert =\sqrt{A_x^2 + A_y^2 + A_z^2} = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14},$$ $$\lvert \overrightarrow{B} \rvert =\sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{1^2 + (-6)^2 + 1^2} = \sqrt{38}.$$

Finally, we can calculate the angle between two vectors

$$\overrightarrow{A}$$

and

$$\overrightarrow{B}$$

:

$$\alpha = arccos(\frac{\overrightarrow{A} \cdot \overrightarrow{B}}{\lvert \overrightarrow{A} \rvert \cdot \lvert \overrightarrow{B} \rvert}),$$ $$\alpha = arccos(\frac{-15}{\sqrt{14} \cdot \sqrt{38}}) = 130^ \circ.$$

Answer:

$$\alpha = 130^ \circ$$

.