Question

Find two unit vectors perpendicular to both A = ˆi − 2ˆj + 3kˆ
r
and B = −2ˆi + 4ˆj
r
.

Answer 1

Answer on Question #46798 – Math – Analytic Geometry

Question:

Find two unit vectors perpendicular to both $\overline{A} = \overline{i} - 2\overline{j} + 3\overline{k}$ and $\overline{B} = -2\overline{i} + 4\overline{j}$.

Solution.

There are two ways to construct vectors perpendicular to the pair of given vectors: through scalar product and through cross (or vector) product.

1. Scalar (dot, inner) product.

Let $\overline{C} = x\overline{i} - y\overline{j} + z\overline{k}$ be the vector perpendicular to both given vectors.

Then, by properties of scalar product $\overline{A} \cdot \overline{C} = 0$ and $\overline{B} \cdot \overline{C} = 0$.

It means that each solution of system $\left\{ \begin{array}{l} x - 2y + 3z = 0 \\ -2x + 4y = 0 \end{array} \right.$ gives the required vector. To solve this system, set, for example, $y = 1$ and find other coordinates from system

$$\left\{ \begin{array}{l} x - 2 + 3z = 0 \\ -2x + 4 = 0 \end{array} \right.$$

From second equation we have $x = 2$, and then obtain $z = 0$ from the first equation. Hence $\overline{C} = 2\overline{i} + \overline{j}$ is perpendicular to both $\overline{A}$ and $\overline{B}$.

To obtain unit vectors we have to divide the vector by its magnitude:

$$\overline{n}_{1,2} = \pm \frac{\overline{C}}{|\overline{C}|} = \pm \frac{2\overline{i} + \overline{j}}{\sqrt{4 + 1}} = \pm \left(\frac{2}{\sqrt{5}}\overline{i} + \frac{1}{\sqrt{5}}\overline{j}\right)$$

2. Cross product

By properties of cross product $\overline{C} = \overline{A} \times \overline{B}$ is perpendicular to both $\overline{A}$ and $\overline{B}$.

Calculate

\overline{C} = \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ 1 & -2 & 3 \\ -2 & 4 & 0 \end{array} \right| = \left| \begin{array}{cc} -2 & 3 & \overline{i} - \left| \begin{array}{cc} 1 & 3 \\ -2 & 0 \end{array} \right| \overline{j} + \left| \begin{array}{cc} 1 & -2 \\ -2 & 4 \end{array} \right| \overline{k} = -12\overline{i} - 6\overline{j} \right|

After division by magnitude we obtain the same result

$$\overline{n}_{1,2} = \pm \frac{\overline{C}}{|\overline{C}|} = \pm \frac{-12\overline{i} - 6\overline{j}}{\sqrt{144 + 36}} = \pm \frac{-12\overline{i} - 6\overline{j}}{6\sqrt{5}} = \mp \left(\frac{2}{\sqrt{5}}\overline{i} + \frac{1}{\sqrt{5}}\overline{j}\right)$$

Answer: Two unit vectors are $\overline{n}_1 = \frac{2}{\sqrt{5}}\overline{i} + \frac{1}{\sqrt{5}}\overline{j}$ and $\overline{n}_2 = -\frac{2}{\sqrt{5}}\overline{i} - \frac{1}{\sqrt{5}}\overline{j}$

www.AssignmentExpert.com