Find the sine of the angle between the vectors a = 3i − 2j + k and b = i + 2j − 3k

Question

ScanSolve AI Tutor · Accepted Answer

Okay, faith, this is the last one for today. We need to find the sine of the angle between the two given vectors.

The vectors are:
$\vec{a} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$
$\vec{b} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$

The sine of the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula:

\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}

Calculate the cross product $\vec{a} \times \vec{b}$ .
$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ 1 & 2 & -3 \end{vmatrix}$ $= \mathbf{i}((-2)(-3) - (1)(2)) - \mathbf{j}((3)(-3) - (1)(1)) + \mathbf{k}((3)(2) - (-2)(1))$ $= \mathbf{i}(6 - 2) - \mathbf{j}(-9 - 1) + \mathbf{k}(6 - (-2))$ $= 4\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}$
Calculate the magnitude of the cross product $|\vec{a} \times \vec{b}|$ .
$|\vec{a} \times \vec{b}| = \sqrt{4^2 + 10^2 + 8^2}$ $= \sqrt{16 + 100 + 64}$ $= \sqrt{180}$
Simplify the radical:
$= \sqrt{36 \times 5} = 6\sqrt{5}$
Calculate the magnitudes of vectors $\vec{a}$ and $\vec{b}$ .
$|\vec{a}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$ $|\vec{b}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
Substitute the magnitudes into the formula for $\sin \theta$ .
$\sin \theta = \frac{6\sqrt{5}}{\sqrt{14} \times \sqrt{14}}$ $\sin \theta = \frac{6\sqrt{5}}{14}$
Simplify the fraction:
$\sin \theta = \frac{3\sqrt{5}}{7}$
The sine of the angle between the vectors is:
$\frac{3\sqrt{5}}{7}$

Answer