Problem 13 - Calculus: Early Transcendentals

Question

Find the scalar and vector projections of \langle 1,1,1\rangle onto \(\langle 3,2,1\rangle .\)

Step-by-Step Solution

Answer

Scalar projection is \(\frac{3}{7}\), and vector projection is \(\langle \frac{9}{7}, \frac{6}{7}, \frac{3}{7} \rangle\)."

Step 1: Dot Product Calculation

First, calculate the dot product of vectors \( \langle 1,1,1 \rangle \) and \( \langle 3,2,1 \rangle \). The dot product is given by \( \langle a_1, a_2, a_3 \rangle \cdot \langle b_1, b_2, b_3 \rangle = a_1b_1 + a_2b_2 + a_3b_3 \). Compute:\[ 1 \times 3 + 1 \times 2 + 1 \times 1 = 3 + 2 + 1 = 6 \].

Step 2: Magnitude of the Projection Vector

Next, calculate the magnitude of the vector \( \langle 3,2,1 \rangle \) squared. This is given by \( \| \langle 3,2,1 \rangle \|^2 = 3^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14 \).

Step 3: Scalar Projection

Calculate the scalar projection using the formula:\[ \text{scalar projection} = \frac{\text{dot product}}{\text{magnitude squared}} = \frac{6}{14} = \frac{3}{7} \].

Step 4: Vector Projection Formula

The vector projection is computed using the formula:\[ \text{vector projection} = \frac{\text{dot product}}{\| \text{magnitude} \|^2} \times \text{vector} = \frac{6}{14} \times \langle 3,2,1 \rangle \].

Step 5: Calculating the Vector Projection

Multiply each component of \( \langle 3,2,1 \rangle \) by \( \frac{3}{7} \) to find the vector projection:\[ \langle 3,2,1 \rangle \times \frac{3}{7} = \langle 3 \times \frac{3}{7}, 2 \times \frac{3}{7}, 1 \times \frac{3}{7} \rangle = \langle \frac{9}{7}, \frac{6}{7}, \frac{3}{7} \rangle \].

Key Concepts

Dot Product

The dot product is a fundamental operation when dealing with vectors. It allows you to extract important geometric information, such as the angle between two vectors. For our vectors \( \langle 1,1,1 \rangle \) and \( \langle 3,2,1 \rangle \), the dot product is calculated by multiplying corresponding components and then summing these products. This results in \( 1 \times 3 + 1 \times 2 + 1 \times 1 = 6 \).
The dot product is often used because its result can indicate the extent to which one vector goes in the direction of the other.
A crucial point to remember is that a dot product of zero signifies that the vectors are perpendicular to one another.

Scalar Projection

The scalar projection can be thought of as how much of one vector lies in the direction of another vector. This value gives us a measure of the length of the shadow cast by one vector onto another, assuming the light source is perpendicular to the second vector.
To find this, you take the dot product of the two vectors and then divide by the magnitude of the vector onto which you are projecting. In our exercise, the scalar projection of \( \langle 1,1,1 \rangle \) onto \( \langle 3,2,1 \rangle \) is \( \frac{6}{14} = \frac{3}{7} \). This tells us that along the direction of \( \langle 3,2,1 \rangle \), the component of \( \langle 1,1,1 \rangle \) is \( \frac{3}{7} \) of the full magnitude along that line.

Magnitude

The magnitude of a vector in space is essentially its length. To calculate it, you use the square root of the sum of the squares of its components. This formula stems from the Pythagorean theorem.
For the vector \( \langle 3,2,1 \rangle \), the magnitude is \( \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \).
Magnitudes are important because they provide scale. Without magnitude, understanding the real-world impact or length of a vector would be difficult. The squared value of the magnitude, found before calculating projections, eliminates the square root to make calculations simpler in some cases, as seen here with \( 14 \) in our solution.

Vectors in 3D Space

Vectors in 3D space are a way to represent quantities that have both magnitude and direction in three-dimensions—space that we are familiar with in everyday life. Each vector is defined by three components (x, y, z), which specify directions along each of the three axes.
The vector \( \langle 1,1,1 \rangle \) and \( \langle 3,2,1 \rangle \) tell us that both move away from the origin in three dimensional space, but in different directions.
When working in 3D space, both the direction and the magnitude provide a full picture of the vector's influence or impact. Understanding vectors in this context allows for applications in physics, engineering, and computer graphics, among other fields. They help us model real-world scenarios accurately with both physical and abstract concepts.