Let $\mathbf{u} = \begin{bmatrix} -5 \\ 4 \\ 1 \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} -3 \\ -6 \\ 9 \end{bmatrix}$. Compute and compare $\mathbf{u} \cdot \mathbf{v}$, $\|\mathbf{u}\|^2$, $\|\mathbf{v}\|^2$, and $\|\mathbf{u} + \mathbf{v}\|^2$. Do not use the Pythagorean Theorem.

Neetesh Kumar | October 18, 2024

Linear Algebra Homework Help

This is the solution to Math2B Course: Linear Algebra
Assignment: Ch6 Section 1 Question Number 10
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Step-by-step solution:

Step 1: Compute $\mathbf{u} \cdot \mathbf{v}$

The dot product $\mathbf{u} \cdot \mathbf{v}$ is calculated as:

$\mathbf{u} \cdot \mathbf{v} = (-5)(-3) + (4)(-6) + (1)(9) = 15 - 24 + 9 = 0$

Thus, $\mathbf{u} \cdot \mathbf{v} = 0$.

Step 2: Compute $\|\mathbf{u}\|^2$

The square of the magnitude of $\mathbf{u}$ is given by:

$\|\mathbf{u}\|^2 = (-5)^2 + (4)^2 + (1)^2 = 25 + 16 + 1 = 42$

Thus, $\|\mathbf{u}\|^2 = 42$.

Step 3: Compute $\|\mathbf{v}\|^2$

The square of the magnitude of $\mathbf{v}$ is:

$\|\mathbf{v}\|^2 = (-3)^2 + (-6)^2 + (9)^2 = 9 + 36 + 81 = 126$

Thus, $\|\mathbf{v}\|^2 = 126$.

Step 4: Compute $\|\mathbf{u} + \mathbf{v}\|^2$

First, find $\mathbf{u} + \mathbf{v}$:

$\mathbf{u} + \mathbf{v} = \begin{bmatrix} -5 \\ 4 \\ 1 \end{bmatrix} + \begin{bmatrix} -3 \\ -6 \\ 9 \end{bmatrix} = \begin{bmatrix} -8 \\ -2 \\ 10 \end{bmatrix}$

Now, compute the square of the magnitude of $\mathbf{u} + \mathbf{v}$:

$\|\mathbf{u} + \mathbf{v}\|^2 = (-8)^2 + (-2)^2 + (10)^2 = 64 + 4 + 100 = 168$

Thus, $\|\mathbf{u} + \mathbf{v}\|^2 = 168$.

Final Results:

• $\mathbf{u} \cdot \mathbf{v} = 0$
• $\|\mathbf{u}\|^2 = 42$
• $\|\mathbf{v}\|^2 = 126$
• $\|\mathbf{u} + \mathbf{v}\|^2 = 168$

Since $\mathbf{u} \cdot \mathbf{v} = 0$ and $\|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 = \|\mathbf{u} + \mathbf{v}\|^2$, both statements imply that the vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.

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