Problem 18 - Essential Calculus Early Transcendentals

Question

Find a vector that has the same direction as \(\langle- 2,4,2\rangle\) but has length \( 6 .\)

Step-by-Step Solution

Answer

The vector with the same direction as \( \langle -2, 4, 2 \rangle \) and length 6 is \( \langle -\sqrt{6}, 2\sqrt{6}, \sqrt{6} \rangle \).

Step 1: Determine the Magnitude of the Original Vector

First, we find the magnitude of the vector \( \langle -2, 4, 2 \rangle \). The magnitude \( \| \mathbf{v} \| \) is calculated using the formula \( \sqrt{x^2 + y^2 + z^2} \). Substituting the values from the vector, we get:\[\| \mathbf{v} \| = \sqrt{(-2)^2 + 4^2 + 2^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}.\]

Step 2: Normalize the Vector

To find a unit vector in the same direction, we divide each component of the vector \( \langle -2, 4, 2 \rangle \) by its magnitude \( 2\sqrt{6} \). This gives us the unit vector:\[\mathbf{u} = \left\langle \frac{-2}{2\sqrt{6}}, \frac{4}{2\sqrt{6}}, \frac{2}{2\sqrt{6}} \right\rangle = \left\langle \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle.\]

Step 3: Scale the Unit Vector to the Desired Length

To achieve the desired length of 6, multiply the unit vector by 6. This scales it to the required magnitude while maintaining direction:\[\mathbf{w} = 6 \times \left\langle \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle = \left\langle \frac{-6}{\sqrt{6}}, \frac{12}{\sqrt{6}}, \frac{6}{\sqrt{6}} \right\rangle.\]

Step 4: Simplify the Resulting Vector

Simplify each component of the vector \( \left\langle \frac{-6}{\sqrt{6}}, \frac{12}{\sqrt{6}}, \frac{6}{\sqrt{6}} \right\rangle \) by rationalizing the denominator:\[\mathbf{w} = \left\langle -\sqrt{6}, 2\sqrt{6}, \sqrt{6} \right\rangle.\]

Key Concepts

Understanding the Magnitude of a Vector

The magnitude of a vector is akin to its length or size in space. To visualize this, imagine a straight line drawn from the origin of a coordinate system to the point defined by the vector's components. The magnitude tells us how long this line is.

It is essential for tasks that involve scaling or normalizing vectors, like turning a vector into a unit vector.

To calculate the magnitude of a vector with components \(x, y, z\), we use the formula:

• \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \)

For the vector \( \langle -2, 4, 2 \rangle \), the magnitude calculation is:

• \( \| \mathbf{v} \| = \sqrt{(-2)^2 + 4^2 + 2^2} = \sqrt{24} = 2\sqrt{6} \)

This value indicates the vector’s "distance" from the origin, providing a basis for further transformations like normalization.

Creating a Unit Vector

A unit vector is a vector that points in the same direction as the original vector, but has a length of exactly 1. Creating a unit vector is an essential step in normalizing a vector, and it helps when determining direction without regard to magnitude.

To find a unit vector, you simply divide each component of the vector by its magnitude. Given our vector \( \langle -2, 4, 2 \rangle \), and its magnitude \(2\sqrt{6}\), the unit vector becomes:

• \( \mathbf{u} = \left\langle \frac{-2}{2\sqrt{6}}, \frac{4}{2\sqrt{6}}, \frac{2}{2\sqrt{6}} \right\rangle = \left\langle \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle \)

This unit vector maintains the direction of the original vector but is scaled to have a length of 1.

Scaling Vectors to Desired Lengths

Scaling a vector essentially means changing its magnitude while ensuring it continues to point in the same direction. This is particularly useful when you need a vector of a specific length.

After finding a unit vector, the next step is scaling it by the desired magnitude. If we need our vector \( \mathbf{u} \) to have a magnitude of 6, we multiply each component of the unit vector by 6:

• \( \mathbf{w} = 6 \left\langle \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle = \left\langle \frac{-6}{\sqrt{6}}, \frac{12}{\sqrt{6}}, \frac{6}{\sqrt{6}} \right\rangle \)

Finally, simplify each component by rationalizing the denominator to convert back into a standard form. The resulting scaled vector \( \mathbf{w} = \langle -\sqrt{6}, 2\sqrt{6}, \sqrt{6} \rangle \) maintains the direction but now has the desired magnitude of 6.

This process illustrates how we can control both the direction and length of vectors in multiple contexts, from physics to computer graphics.