Question

Obtain a unit vector perpendicular to the plane of the vectors A^vector = 3i^cap - 4j^cap +2k^cap and B ^vector = i^cap + j ^ cap +3 k^ cap.

Answer 1

Answer on Question #70003 – Math – Linear Algebra

Question

Obtain a unit vector perpendicular to the plane of the vectors

A^vector = 3i^cap - 4j^cap + 2k^cap and B^vector = i^cap + j^cap + 3k^cap.

Solution

$$\begin{array}{l} \vec{a} = 3\vec{i} - 4\vec{j} + 2\vec{k}; \vec{b} = \vec{i} + \vec{j} + 3\vec{k}; \\ \vec{a} \ (3; -4; 2); \\ \vec{b} \ (1; 1; 3). \end{array}$$

Find the vector product (the cross product) of vectors $\vec{a}$ and $\vec{b}$ :

$$\begin{array}{l} \vec{c} = \left[ \vec{a} \times \vec{b} \right] = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 3 & -4 & 2 \\ 1 & 1 & 3 \end{array} \right| = \left| \begin{array}{cc} -4 & 2 \\ 1 & 3 \end{array} \right| \vec{i} - \left| \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} \right| \vec{j} + \left| \begin{array}{cc} 3 & -4 \\ 1 & 1 \end{array} \right| \vec{k} = \\ = -14\vec{i} - 7\vec{j} + 7\vec{k}, \quad \vec{c} \ (-14; -7; 7); \ |\vec{c}| = \sqrt{(-14)^2 + (-7)^2 + 7^2} = 7\sqrt{6}. \end{array}$$

It is known that $\vec{c} \perp \vec{a}, \vec{c} \perp \vec{b}$.

Find $\vec{d} \parallel \vec{c}; \ |\vec{d}| = 1$:

$$\vec{d} = \frac{1}{7\sqrt{6}} \vec{c};$$

$\vec{d}\left(-\frac{\sqrt{6}}{3}; -\frac{\sqrt{6}}{6}; \frac{\sqrt{6}}{6}\right)$ is one of the required unit vectors.

Answer: $\vec{d}\left(-\frac{\sqrt{6}}{3}; -\frac{\sqrt{6}}{6}; \frac{\sqrt{6}}{6}\right)$.

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