Question

Find the angle between U =4i - 2j + 4k and V = 3i - 6j - 2k.

a. 68 degrees
b. 67 degrees
c. 58 degrees
d. 69 degrees

Answer 1

Find the angle between $U = 4i - 2j + 4k$ and $V = 3i - 6j - 2k$.

Solution

In our case we have such vectors: $\vec{U}(4; -2; 4)$ and $\vec{V}(3; -6; -2)$.

To find the angle between them we must find the cosine of that angle at first.

This cosine is equal to scalar multiplication of those vectors, divided by multiplication of sizes of those vectors.

The scalar multiplication of two vectors is equal to sum of multiplications of corresponding coordinates. $\vec{U} \cdot \vec{V} = u_1 v_1 + u_2 v_2 + u_3 v_3$, where $u_1, u_2, u_3$ are the coordinates of the first vector and $v_1, v_2, v_3$ are the coordinates of the second one.

The size of a vector is equal to square root from sum of squares of its coordinates.

$$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$$ $$\cos(\widehat{U; V}) = \frac{\vec{U} \cdot \vec{V}}{|\vec{U}| \cdot |\vec{V}|} = \frac{u_1 v_1 + u_2 v_2 + u_3 v_3}{\sqrt{u_1^2 + u_2^2 + u_3^2} \cdot \sqrt{v_1^2 + v_2^2 + v_3^2}}$$ $$\cos(\widehat{U; V}) = \frac{4 \cdot 3 + 2 \cdot 6 - 4 \cdot 2}{\sqrt{16 + 4 + 16} \cdot \sqrt{9 + 36 + 4}} = \frac{16}{42} = \frac{8}{21}.$$

Now when we have the cosine we can find the corresponding angle (using calculator or tables)

$$\arccos\left(\frac{8}{21}\right) \approx \arccos(0.38095) \approx 67.6{}^\circ = 67{}^\circ 36'.$$

$67{}^\circ 36'$ is closer to $68{}^\circ$ than to $67{}^\circ$.