Question

Find a unit vector \((a)\) in the direction of \(\mathbf{u}\) and \((\mathbf{b})\) in the direction opposite \(\mathbf{u}\) \(\mathbf{u}=\langle 8,0,0\rangle\)

Short answer

The unit vector in the direction of \(\mathbf{u}\) is \(\mathbf{a}=\langle 1,0,0\rangle\), and the unit vector in the direction opposite to \(\mathbf{u}\) is \(\mathbf{b}=\langle -1,0,0\rangle\).

Step-by-step solution

Identify Given Vector

The given vector is \(\mathbf{u}=\langle 8,0,0\rangle\). The magnitude or length of this vector can be calculated with the general formula \(\|\mathbf{u}\|=\sqrt{u_{1}^2+u_{2}^2+u_{3}^2}\), where \(u_{1}\), \(u_{2}\), and \(u_{3}\) are the components of the vector.

Calculate Magnitude of Given Vector

Plugging the components of \(\mathbf{u}\) into the general formula yields \(\|\mathbf{u}\|=\sqrt{8^2+0^2+0^2} = 8\). The magnitude of \(\mathbf{u}\) is 8.

Compute Unit Vector in Direction of Given Vector

A unit vector in the direction of a given vector \(\mathbf{u}\) can be calculated with the formula \(\mathbf{a}=\mathbf{u}/\|\mathbf{u}\|\). Therefore, \(\mathbf{a}=\langle 8,0,0\rangle/8 = \langle 1,0,0\rangle\).

Compute Unit Vector in Direction Opposite to Given Vector

To find a unit vector in the direction opposite to \(\mathbf{u}\), simply compute the negative of the unit vector calculated in Step 3, to get \(\mathbf{b}= -\mathbf{a} = -\langle 1,0,0\rangle = \langle -1,0,0\rangle\). Thus, \(\mathbf{b}\) is the unit vector in the direction opposite to \(\mathbf{u}\).

Key concepts

Vector Magnitude

Understanding vector magnitude is fundamental to many fields of mathematics and physics. It's essentially a measure of how long the vector is, representing the distance from its starting point (usually the origin) to its endpoint. For any vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude is given by \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).

In our exercise example, where the given vector is \( \mathbf{u} = \langle 8,0,0 \rangle \), computing the magnitude is straightforward since it only has one non-zero component. The formula reduces to \( \| \mathbf{u} \| = \sqrt{8^2 + 0^2 + 0^2} = 8 \). Thus, the magnitude of \( \mathbf{u} \) is 8, which tells us the vector's size or length in the space it occupies.

Direction of a Vector

Once you have a grasp of vector magnitude, it is much easier to appreciate the concept of the direction of the vector. The direction tells us where the vector is headed in space and is described by the vector's components. If a vector \( \mathbf{v} \) has a non-zero magnitude, a unit vector \( \mathbf{a} \) in the direction of \( \mathbf{v} \) can be found by dividing every component by its magnitude, \( \mathbf{a} = \frac{\mathbf{v}}{\| \mathbf{v} \|} \).

In our exercise, the unit vector in the direction of \( \mathbf{u} \langle 8,0,0 \rangle \) is \( \mathbf{a} = \frac{\mathbf{u}}{\| \mathbf{u} \|} = \langle \frac{8}{8}, \frac{0}{8}, \frac{0}{8} \rangle = \langle 1,0,0 \rangle \), which indicates that the vector is pointing along the positive x-axis. This unit vector has a magnitude of 1 and retains the original direction of \( \mathbf{u} \).

Opposite Direction Vector

Sometimes, it's not enough to just know the direction in which a vector points; we may also need to know the direction that is exactly opposite to it. This concept is relatively simple: the opposite direction vector, often denoted \( \mathbf{b} \), has the same magnitude but points in the opposite way. Just as pushing a swing forwards and then backwards, you'll apply the same force in two opposite directions.

In vector terms, if \( \mathbf{a} \) is a unit vector, then its opposite is \( \mathbf{b} = -\mathbf{a} \). From our textbook exercise, we start with \( \mathbf{a} = \langle 1,0,0 \rangle \), so flipping its direction gives us \( \mathbf{b} = -\mathbf{a} = -\langle 1,0,0 \rangle = \langle -1,0,0 \rangle \). This unit vector \( \mathbf{b} \) points along the negative x-axis, exactly opposite to \( \mathbf{a} \) while still possessing a magnitude of 1.