QuestionGiven $\mathbf{u}_{1}=(6,-1)$ and $\mathbf{u}_{2}=(3,2)$ , if we let $\mathbf{v}_{1}=\mathbf{u}_{1}$ , use the Gram-Schmidt process to find $\mathbf{v}_{2}$ If needed, enter your answers as fractions, not decimals.
This question accepts'answers that are in a form like " $(-1,3)$ " or " $(3,7,3 z)$ ". The entries can be numbers or formulas. Help | Preview

Studdy Solution

STEP 1

1. We are using the Gram-Schmidt process to orthogonalize the given vectors.
2. The vectors $\mathbf{u}_1$ and $\mathbf{u}_2$ are in $\mathbb{R}^2$ .
3. $\mathbf{v}_1$ is already given as $\mathbf{u}_1$ .
4. We need to find $\mathbf{v}_2$ such that it is orthogonal to $\mathbf{v}_1$ .

STEP 2

1. Assign $\mathbf{v}_1$ to $\mathbf{u}_1$ .
2. Compute the projection of $\mathbf{u}_2$ onto $\mathbf{v}_1$ .
3. Subtract the projection from $\mathbf{u}_2$ to find $\mathbf{v}_2$ .

STEP 3

Assign $\mathbf{v}_1$ to $\mathbf{u}_1$ :
$\mathbf{v}_1 = \mathbf{u}_1 = (6, -1)$

STEP 4

Compute the projection of $\mathbf{u}_2$ onto $\mathbf{v}_1$ . The formula for the projection of a vector $\mathbf{a}$ onto a vector $\mathbf{b}$ is given by:
$\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}$
First, calculate the dot product $\mathbf{u}_2 \cdot \mathbf{v}_1$ :
$\mathbf{u}_2 \cdot \mathbf{v}_1 = (3, 2) \cdot (6, -1) = 3 \cdot 6 + 2 \cdot (-1) = 18 - 2 = 16$
Next, calculate $\mathbf{v}_1 \cdot \mathbf{v}_1$ :
$\mathbf{v}_1 \cdot \mathbf{v}_1 = (6, -1) \cdot (6, -1) = 6 \cdot 6 + (-1) \cdot (-1) = 36 + 1 = 37$
Now, compute the projection:
$\text{proj}_{\mathbf{v}_1} \mathbf{u}_2 = \frac{16}{37} \mathbf{v}_1 = \frac{16}{37} (6, -1) = \left(\frac{96}{37}, -\frac{16}{37}\right)$

STEP 5

Subtract the projection from $\mathbf{u}_2$ to find $\mathbf{v}_2$ :
$\mathbf{v}_2 = \mathbf{u}_2 - \text{proj}_{\mathbf{v}_1} \mathbf{u}_2 = (3, 2) - \left(\frac{96}{37}, -\frac{16}{37}\right)$
Calculate each component:
$\mathbf{v}_2 = \left(3 - \frac{96}{37}, 2 + \frac{16}{37}\right)$
Simplify the components:
$\mathbf{v}_2 = \left(\frac{111}{37} - \frac{96}{37}, \frac{74}{37} + \frac{16}{37}\right) = \left(\frac{15}{37}, \frac{90}{37}\right)$
The orthogonal vector $\mathbf{v}_2$ is:
$\boxed{\left(\frac{15}{37}, \frac{90}{37}\right)}$

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