Question

if the dot product of two non - zero vectors \\(\mathbf{v}_1\\) and \\(\mathbf{v}_2\\) is zero, what does this tell us?\
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a. \\(\mathbf{v}_1\\) is a component of \\(\mathbf{v}_2\\).\
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b. \\(\mathbf{v}_1 = \mathbf{v}_2\\)\
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c. \\(\mathbf{v}_1\\) is perpendicular to \\(\mathbf{v}_2\\).\
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d. \\(\mathbf{v}_1\\) is parallel to \\(\mathbf{v}_2\\).

Explanation:

Step1: Recall dot product formula

The dot product of two vectors $\mathbf{v_1}$ and $\mathbf{v_2}$ is defined as $\mathbf{v_1} \cdot \mathbf{v_2} = \|\mathbf{v_1}\| \|\mathbf{v_2}\| \cos\theta$, where $\theta$ is the angle between the vectors, and $\|\mathbf{v_1}\|, \|\mathbf{v_2}\|$ are the magnitudes of the vectors.

Step2: Apply given condition

We know $\mathbf{v_1} \cdot \mathbf{v_2} = 0$, and $\|\mathbf{v_1}\|
eq 0$, $\|\mathbf{v_2}\|
eq 0$ (non-zero vectors). Substitute into the formula:
$$0 = \|\mathbf{v_1}\| \|\mathbf{v_2}\| \cos\theta$$

Step3: Solve for $\theta$

Since $\|\mathbf{v_1}\| \|\mathbf{v_2}\|
eq 0$, we divide both sides by this product:
$$\cos\theta = 0$$
The angle $\theta$ that satisfies this is $\theta = 90^\circ$ (or $\frac{\pi}{2}$ radians), meaning the vectors are perpendicular.

Answer:

C. $\mathbf{v_1}$ is perpendicular to $\mathbf{v_2}$.