Question

Find a unit vector in the direction of: a. \(\left[\begin{array}{r}7 \\ -1 \\ 5\end{array}\right]\) b. \(\left[\begin{array}{r}-2 \\ -1 \\ 2\end{array}\right]\)

Short answer

a. \(\left[\frac{7\sqrt{3}}{15}, \frac{-\sqrt{3}}{15}, \frac{\sqrt{3}}{3}\right]\); b. \(\left[\frac{-2}{3}, \frac{-1}{3}, \frac{2}{3}\right]\).\)

Step-by-step solution

Understanding Unit Vectors

A unit vector is a vector that has a magnitude of 1 and points in the same direction as a given vector. To find the unit vector in the direction of a vector, divide each component of the vector by the vector's magnitude.

Step 2a: Calculate Magnitude for Vector a

Calculate the magnitude of the vector \(\mathbf{a} = \left[\begin{array}{r}7 \ -1 \ 5\end{array}\right]\) using the formula for magnitude \(\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\). So, \(\|\mathbf{a}\| = \sqrt{7^2 + (-1)^2 + 5^2} = \sqrt{49 + 1 + 25} = \sqrt{75}\).

Step 3a: Determine Unit Vector for Vector a

Divide each component of the vector \(\mathbf{a}\) by its magnitude: \[\mathbf{u}_a = \frac{1}{\sqrt{75}} \left[\begin{array}{r}7 \ -1 \ 5\end{array}\right] = \left[\begin{array}{r}\frac{7}{\sqrt{75}} \ \frac{-1}{\sqrt{75}} \ \frac{5}{\sqrt{75}}\end{array}\right].\] Simplifying further, you get \[\mathbf{u}_a = \left[\begin{array}{r}\frac{7}{5\sqrt{3}} \ \frac{-1}{5\sqrt{3}} \ \frac{5}{5\sqrt{3}}\end{array}\right] = \left[\begin{array}{r}\frac{7\sqrt{3}}{15} \ \frac{-\sqrt{3}}{15} \ \frac{\sqrt{3}}{3}\end{array}\right].\]

Step 4b: Calculate Magnitude for Vector b

Calculate the magnitude of the vector \(\mathbf{b} = \left[\begin{array}{r}-2 \ -1 \ 2\end{array}\right]\) using the magnitude formula: \(\|\mathbf{b}\| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\).

Step 5b: Determine Unit Vector for Vector b

Divide each component of the vector \(\mathbf{b}\) by its magnitude: \[\mathbf{u}_b = \frac{1}{3} \left[\begin{array}{r}-2 \ -1 \ 2\end{array}\right] = \left[\begin{array}{r}\frac{-2}{3} \ \frac{-1}{3} \ \frac{2}{3}\end{array}\right].\] This is the desired unit vector.

Key concepts

Magnitude

When dealing with vectors, understanding the concept of magnitude is crucial. The magnitude of a vector is a measure of its length or how "big" the vector is.
For a vector in three-dimensional space, the magnitude is found using the formula \[\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\]where \(x\), \(y\), and \(z\) are the components of the vector.
This formula is derived from the Pythagorean theorem, extending the concept of distance to three dimensions.

• A vector like \(\mathbf{a} = \left[\begin{array}{r}7 \ -1 \ 5\end{array}\right]\) has a magnitude of \(\sqrt{75}\).
• Another vector, \(\mathbf{b} = \left[\begin{array}{r}-2 \ -1 \ 2\end{array}\right]\), has a magnitude of 3.

Understanding magnitude helps in various applications, especially when scaling vectors or converting them to unit vectors.

Vector Direction

The direction of a vector is essentially what orientation it has in space. Although the magnitude gives the size, the direction tells us where the vector points.
Vectors are represented by components that indicate their influence along the axis they belong.

• For \(\mathbf{a} = \left[\begin{array}{r}7 \ -1 \ 5\end{array}\right]\), the direction points inwardly based on the vector’s components 7 towards the x-axis, -1 towards the y-axis, and 5 towards the z-axis.
• The vector \(\mathbf{b} = \left[\begin{array}{r}-2 \ -1 \ 2\end{array}\right]\) points in the opposite way toward components on negative x and y axes and positive z axis.

Determining the vector direction is pivotal when analyzing forces, movements, or other directional measurements in physics and engineering.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector. A unit vector retains the original vector's direction but changes its magnitude to precisely 1.
To normalize a vector, you divide each component of the vector by its magnitude.

• For the vector \(\mathbf{a}\), with magnitude \(\sqrt{75}\), normalization is achieved by performing: \[\mathbf{u}_a = \frac{1}{\sqrt{75}} \left[\begin{array}{r}7 \ -1 \ 5\end{array}\right] = \left[\begin{array}{r}\frac{7}{5\sqrt{3}} \ \frac{-1}{5\sqrt{3}} \ \frac{5}{5\sqrt{3}}\end{array}\right]\].
• For \(\mathbf{b}\), dividing each component by 3 results in: \[\mathbf{u}_b = \frac{1}{3} \left[\begin{array}{r}-2 \ -1 \ 2\end{array}\right] = \left[\begin{array}{r}\frac{-2}{3} \ \frac{-1}{3} \ \frac{2}{3}\end{array}\right]\].

Normalization is essential in various fields including computer graphics, where directions often need to be consistent.