Question

In each case, compute the projection of \(\mathbf{u}\) on \(\mathbf{v}\) a. \(\mathbf{u}=\left[\begin{array}{l}5 \\ 7 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}4 \\ 1 \\ 1\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right], \mathbf{v}=\left[\begin{array}{r}3 \\ -1 \\ 1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ -1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 4 \\ 2\end{array}\right]\)

Short answer

For part a, the projection is \(\begin{bmatrix} \frac{6}{7} \\ -\frac{3}{7} \\ \frac{9}{7} \end{bmatrix}\), and for part b, it is \(\begin{bmatrix} \frac{22}{9} \\ \frac{11}{18} \\ \frac{11}{18} \end{bmatrix}\)."

Step-by-step solution

Calculate Dot Product \\( \\mathbf{u} \cdot \mathbf{v} \\) for Part a

For part a, first find the dot product of \(\mathbf{u} = \begin{bmatrix} 5 \ 7 \ 1 \end{bmatrix} \) and \(\mathbf{v} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix} \).\tab\ \(\mathbf{u} \cdot \mathbf{v} = (5 \cdot 2) + (7 \cdot -1) + (1 \cdot 3) = 10 - 7 + 3 = 6.\tab\) Now, let's move on to the next step.

Calculate \\( \|\mathbf{v}\|^2 \\) for Part a

Find the magnitude squared of \(\mathbf{v}\):\tab\ \( \|\mathbf{v}\|^2 = 2^2 + (-1)^2 + 3^2 = 4 + 1 + 9 = 14.\tab\) Proceed to the next step.

Compute Projection of \\( \mathbf{u} \\) on \\( \mathbf{v} \\) for Part a

Use the formula for the projection: \tab\ \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \cdot \mathbf{v} = \frac{6}{14} \cdot \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix} \). Simplifying, we get: \(= \begin{bmatrix} \frac{12}{14} \ -\frac{6}{14} \ \frac{18}{14} \end{bmatrix} = \begin{bmatrix} \frac{6}{7} \ -\frac{3}{7} \ \frac{9}{7} \end{bmatrix}\tab\) Next, repeat these steps for part b.

Calculate Dot Product \\( \\mathbf{u} \cdot \mathbf{v} \\) for Part b

Find the dot product of \(\mathbf{u} = \begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix} \) and \(\mathbf{v} = \begin{bmatrix} 4 \ 1 \ 1 \end{bmatrix} \).\tab\ \( \mathbf{u} \cdot \mathbf{v} = (3 \cdot 4) + (-2 \cdot 1) + (1 \cdot 1) = 12 - 2 + 1 = 11.\tab\)

Calculate \\( \|\mathbf{v}\|^2 \\) for Part b

Find the magnitude squared of \(\mathbf{v}\):\tab\ \( \|\mathbf{v}\|^2 = 4^2 + 1^2 + 1^2 = 16 + 1 + 1 = 18.\tab\)

Compute Projection of \\( \mathbf{u} \\) on \\( \mathbf{v} \\) for Part b

Use the projection formula: \tab\ \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{11}{18} \cdot \begin{bmatrix} 4 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{44}{18} \ \frac{11}{18} \ \frac{11}{18} \end{bmatrix} = \begin{bmatrix} \frac{22}{9} \ \frac{11}{18} \ \frac{11}{18} \end{bmatrix}\tab\).

Key concepts

Dot Product

The dot product is a fundamental concept when working with vectors, especially in understanding projection in a geometric sense. In simple terms, the dot product of two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), is a way of multiplying them to get a scalar (a single number). To compute it, you multiply their corresponding components and add them all together.
For vectors \( \mathbf{u} = \begin{bmatrix} u_1 & u_2 & u_3 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \), the formula looks like this:

• \( \mathbf{u} \cdot \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 + u_3 \cdot v_3 \)

This operation is significant because it gives us an idea of how much \( \mathbf{u} \) goes in the direction of \( \mathbf{v} \). A key point: if the dot product is zero, the vectors are perpendicular. This step is the first when finding the projection of one vector onto another.

Magnitude Squared

Magnitude squared is another essential piece to the puzzle of vector projections. The magnitude of a vector provides a measure of its length. However, in many calculations, especially projections, instead of using the length (the magnitude), we use its square, \( \|\mathbf{v}\|^2 \). This simplifies calculations and avoids square roots, making them more straightforward.
If \( \mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \), you calculate its magnitude squared as follows:

• \( \|\mathbf{v}\|^2 = v_1^2 + v_2^2 + v_3^2 \)

Using the magnitude squared is a necessary step because it helps scale the projection appropriately. It allows us to understand how 'powerful' or 'large' \( \mathbf{v} \) is, which is vital for normalizing it in the projection formula.

Projection Formula

The projection formula is where all previous calculations come together. Projecting \( \mathbf{u} \) onto \( \mathbf{v} \) helps you find how much \( \mathbf{u} \) aligns along \( \mathbf{v} \). In other words, it shows the shadow \( \mathbf{u} \) casts on \( \mathbf{v} \). From the dot product and magnitude squared, the formula reads:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \cdot \mathbf{v} \)

This result is a vector that points in the same direction as \( \mathbf{v} \), scaled by the fraction \( \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \).
It rapidly computes the component of one vector in the direction of another, which is useful in many fields, from physics to computer graphics. Understanding it deeply involves recognizing how vector components blend through their magnitudes and contributions from the dot product.