Problem 79 - Precalculus with Calculus Previews

Question

Find a vector \(\mathbf{u}\) that is four times as long as \(\mathbf{v}=\mathbf{i}\) \(-\mathbf{j}+\mathbf{k}\) in the same direction as \(\mathbf{v}\).

Step-by-Step Solution

Answer

The vector \( \mathbf{u} \) is \( \left( \frac{4}{\sqrt{3}}, \frac{-4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \right) \).

Step 1: Understand the Vector Notation

The vector \( \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k} \) represents a vector in a 3-dimensional space. The vector components are (1, -1, 1). This means that the vector \( \mathbf{v} \) moves 1 unit in the x-direction, -1 unit in the y-direction, and 1 unit in the z-direction.

Step 2: Calculate the Magnitude of \( \mathbf{v} \)

The magnitude (or length) of a vector \( \mathbf{v} = (a, b, c) \) is given by the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). For \( \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k} \), the magnitude is \( \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \).

Step 3: Find the Unit Vector in the Direction of \( \mathbf{v} \)

The unit vector \( \mathbf{u_v} \) in the direction of \( \mathbf{v} \) is calculated by dividing each component of \( \mathbf{v} \) by its magnitude. Thus, \( \mathbf{u_v} = \left( \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \).

Step 4: Scale the Unit Vector

To find a vector \( \mathbf{u} \) that is four times as long as \( \mathbf{v} \) in the same direction, scale the unit vector \( \mathbf{u_v} \) by 4. This means \( \mathbf{u} = 4 \times \mathbf{u_v} = \left( \frac{4}{\sqrt{3}}, \frac{-4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \right) \).

Step 5: Conclusion

The vector \( \mathbf{u} = \left( \frac{4}{\sqrt{3}}, \frac{-4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \right) \) is four times as long as \( \mathbf{v} \) and points in the same direction.

Key Concepts

Vector Magnitude

The magnitude of a vector is a measure of its length, much like the length of a line segment in geometry. This is essential in understanding vectors as it provides a sense of how far the vector stretches in space. To find the magnitude of a vector \( \mathbf{v} = (a, b, c) \), you use the formula for the Euclidean norm:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \)

This formula calculates the distance from the origin \((0, 0, 0)\) to the point \((a, b, c)\).
For instance, in the exercise above, the vector \( \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k} \) translates to components (1, -1, 1).
Using the magnitude formula,

• \( \| \mathbf{v} \| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \)

This tells us that \( \mathbf{v} \) has a length of \( \sqrt{3} \) units in the 3-dimensional space.

Unit Vector

A unit vector is a vector with a magnitude of 1. It is important because it gives you a direction in space without any specific magnitude. To find the unit vector \( \mathbf{u_v} \) in the direction of a given vector \( \mathbf{v} \), you divide each of the vector's components by its magnitude:

• \( \mathbf{u_v} = \left( \frac{a}{\| \mathbf{v} \|}, \frac{b}{\| \mathbf{v} \|}, \frac{c}{\| \mathbf{v} \|} \right) \)

In our example, the vector \( \mathbf{v} \) has components (1, -1, 1), and its magnitude is \( \sqrt{3} \). Thus, the unit vector in the direction of \( \mathbf{v} \) is determined as follows:

• \( \mathbf{u_v} = \left( \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)

This result highlights that \( \mathbf{u_v} \) maintains the same direction as \( \mathbf{v} \) but adjusts the magnitude to 1.

Vector Scaling

Vector scaling is the operation of increasing or decreasing the length of a vector by multiplying it by a scalar, which is simply a real number. This process is useful when you need to adjust a vector's length without altering its direction.
To scale a vector \( \mathbf{v} \) by a scalar value \( k \), each component of the vector is multiplied by \( k \):

• \( \mathbf{v'} = (ka, kb, kc) \)

In the given problem, the task is to find a vector \( \mathbf{u} \) that is four times as long as the unit vector \( \mathbf{u_v} \). You achieve this by multiplying \( \mathbf{u_v} \) by 4, thus ensuring the direction remains unchanged while extending the vector:

• \( \mathbf{u} = 4 \times \mathbf{u_v} = \left( \frac{4}{\sqrt{3}}, \frac{-4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \right) \)

Thus, vector scaling allows you to elegantly transform a unit vector to any desired length.