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The dot product, also known as the scalar product, is a way of multiplying two vectors to yield a scalar (a single number). This operation can tell us how much two vectors align with each other. To find the dot product of vectors \(\mathbf{v}\) and \(\mathbf{u}\), we use the formula:\[\mathbf{v} \cdot \mathbf{u} = v_x \cdot u_x + v_y \cdot u_y + v_z \cdot u_z.\]Consider \(\mathbf{v} = -\mathbf{i} + \mathbf{j}\) and \(\mathbf{u} = \sqrt{2}\mathbf{i} + \sqrt{3}\mathbf{j} + 2\mathbf{k}\). To compute their dot product, substitute each component into the formula:

• \(-1 \times \sqrt{2}\) for \(\mathbf{i}\)
• \(1 \times \sqrt{3}\) for \(\mathbf{j}\)
• \(0 \times 2\) for \(\mathbf{k}\) since \(\mathbf{v}\) has no \(\mathbf{k}\) component

Adding these together gives us:\[-\sqrt{2} + \sqrt{3}.\]Hence, the dot product is \(-\sqrt{2} + \sqrt{3}.\)

The magnitude, or length, of a vector is a measure of how long the vector is in Euclidean space. It is computed using the Pythagorean theorem, extended into three dimensions. The magnitude of a vector \(\mathbf{a}\) is calculated with the formula:\[|\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}.\]For the vector \(\mathbf{v} = -\mathbf{i} + \mathbf{j}\):

• Its magnitude is \(\sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}.\)

For the vector \(\mathbf{u} = \sqrt{2}\mathbf{i} + \sqrt{3}\mathbf{j} + 2\mathbf{k}\):

• Its magnitude is \(\sqrt{(\sqrt{2})^2 + (\sqrt{3})^2 + 2^2} = \sqrt{9} = 3.\)

These computations reflect how far each vector extends from the origin in a three-dimensional space.

To determine the angle between two vectors, we use the dot product. Specifically, the cosine of the angle \(\theta\) between vectors \(\mathbf{v}\) and \(\mathbf{u}\) can be found by rearranging the dot product formula:\[\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|}.\]This formula comes in handy when comparing the direction of vectors, since the cosine value will be close to 1 if they point in a similar direction, and close to -1 if they point opposite to each other. Using our values:

• \(\mathbf{v} \cdot \mathbf{u} = -\sqrt{2} + \sqrt{3}\)
• \(|\mathbf{v}| = \sqrt{2}\)
• \(|\mathbf{u}| = 3\)

The calculated cosine of their angle is:\[\frac{-\sqrt{2} + \sqrt{3}}{3\sqrt{2}}.\]This gives us a sense of their alignment in space.

Vector projection calculates how much of one vector lies in the direction of another. It's a way of "casting" one vector onto another. The formula to find the projection of vector \(\mathbf{u}\) onto \(\mathbf{v}\) is:\[\text{proj}_{\mathbf{v}}\mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v}.\]For vectors \(\mathbf{v} = -\mathbf{i} + \mathbf{j}\) and \(\mathbf{u} = \sqrt{2}\mathbf{i} + \sqrt{3}\mathbf{j} + 2\mathbf{k}\), first find the dot product \(\mathbf{v} \cdot \mathbf{u}\) and the magnitude squared of \(\mathbf{v}\), which is \(|\mathbf{v}|^2 = 2\).Substitute into the formula:

• \(\text{proj}_{\mathbf{v}}\mathbf{u} = \left( \frac{-\sqrt{2} + \sqrt{3}}{2} \right)(-\mathbf{i} + \mathbf{j})\)
• This results in: \(\left( \frac{\sqrt{2} - \sqrt{3}}{2} \right) \mathbf{i} + \left( \frac{-\sqrt{2} + \sqrt{3}}{2} \right) \mathbf{j}\)

This projection shows the part of \(\mathbf{u}\) that is aligned with \(\mathbf{v}\).