If we reject the null hypothesis when it is true, we might be making

Given $|\vec{a}| = 5$∣a∣=5, $|\vec{b}| = 2$∣b∣=2, and $\vec{a}\cdot\vec{b} = 6$a⋅b=6

Using the dot product formula $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$a⋅b=∣a∣∣b∣cosθ:

$$6 = 5 \times 2 \times \cos\theta$$6=5×2×cosθ

$$6 = 10\cos\theta$$6=10cosθ

$$\cos\theta = \frac{6}{10} = 0.6$$cosθ=106​=0.6

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Using the identity $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1:

$$\sin^2\theta = 1 - \cos^2\theta$$sin2θ=1−cos2θ

$$\sin^2\theta = 1 - (0.6)^2$$sin2θ=1−(0.6)2

$$\sin^2\theta = 1 - 0.36$$sin2θ=1−0.36

$$\sin^2\theta = 0.64$$sin2θ=0.64

$$\sin\theta = 0.8$$sinθ=0.8

Note: The angle between two vectors is always between $0°$0° and $180°$180°, so $\sin\theta$sinθ is positive.

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Using the cross product magnitude formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$∣a×b∣=∣a∣∣b∣sinθ:

$$|\vec{a} \times \vec{b}| = 5 \times 2 \times 0.8$$∣a×b∣=5×2×0.8

$$|\vec{a} \times \vec{b}| = 10 \times 0.8$$∣a×b∣=10×0.8

$$|\vec{a} \times \vec{b}| = 8$$∣a×b∣=8

Therefore, $|\vec{a} \times \vec{b}| = 8$∣a×b∣=8