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Linear Algebra Examples

y = - 5 i + 2 j

$y = - 5 i + 2 j$

Step 1

Apply the direction angle formula

θ = arctan (\frac{b}{a})

$θ = \arctan (\frac{b}{a})$ where

a = - 5

$a = - 5$ and

b = 2

$b = 2$ .

θ = arctan (\frac{2}{- 5})

$θ = \arctan (\frac{2}{- 5})$

Step 2

Solve for

θ

$θ$ .

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Step 2.1

Remove parentheses.

θ = arctan (\frac{2}{- 5})

$θ = \arctan (\frac{2}{- 5})$

Step 2.2

Simplify

arctan (\frac{2}{- 5})

$\arctan (\frac{2}{- 5})$ .

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Step 2.2.1

Move the negative in front of the fraction.

θ = arctan (- \frac{2}{5})

$θ = \arctan (- \frac{2}{5})$

Step 2.2.2

Evaluate

arctan (- \frac{2}{5})

$\arctan (- \frac{2}{5})$ .

θ = - 0.38050637

$θ = - 0.38050637$

θ = - 0.38050637

$θ = - 0.38050637$

θ = - 0.38050637

$θ = - 0.38050637$

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⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$