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The dot product is a fundamental operation for vectors that helps determine the relationship between them. It combines two vectors and results in a scalar—essentially a single number. The formula for the dot product between two vectors \( \vec{v} = \langle v_x, v_y \rangle \) and \( \vec{w} = \langle w_x, w_y \rangle \) is:

• \( \vec{v} \cdot \vec{w} = v_x \times w_x + v_y \times w_y \)

In simple terms, you multiply the corresponding components (x with x, y with y) and then add those products together.
This operation can tell us how "aligned" two vectors are. If their dot product is zero, the vectors are perpendicular to each other.
In our example, the dot product of \( \vec{v} = \langle -8, 3 \rangle \) and \( \vec{w} = \langle 2, 6 \rangle \) is calculated as follows:

• \( (-8 \times 2) + (3 \times 6) = -16 + 18 = 2 \)

Thus, the dot product is 2, indicating that they are neither perfectly aligned nor orthogonal.

Vector projection is a tool that helps project one vector onto another. It determines the shadow of one vector onto the other if light were shining from the side.
To compute the projection of vector \( \vec{v} \) onto vector \( \vec{w} \), we use the formula:

• \( \operatorname{proj}_{\vec{w}}(\vec{v}) = \left( \frac{\vec{v} \cdot \vec{w}}{\vec{w} \cdot \vec{w}} \right) \vec{w} \)

This formula scales the vector \( \vec{w} \) by the fraction of \( \vec{v} \) that aligns with \( \vec{w} \).
Given \( \vec{v} = \langle -8, 3 \rangle \) and \( \vec{w} = \langle 2, 6 \rangle \), we first find \( \vec{w} \cdot \vec{w} = 40 \), and then calculate:

• \( \operatorname{proj}_{\vec{w}}(\vec{v}) = \frac{2}{40} \times \langle 2, 6 \rangle = \langle \frac{1}{10}, \frac{3}{10} \rangle \)

This shows that \( \vec{v} \)'s shadow on \( \vec{w} \) is quite small, given by the vector \( \langle \frac{1}{10}, \frac{3}{10} \rangle \).

Finding the angle between vectors is a useful way to determine the orientation of one vector relative to another. To find this angle \( \theta \), use the dot product in conjunction with magnitudes:

• \( \cos \theta = \frac{\vec{v} \cdot \vec{w}}{||\vec{v}|| \cdot ||\vec{w}||} \)

Here, \( ||\vec{v}|| \) and \( ||\vec{w}|| \) represent the magnitudes of the vectors \( \vec{v} \) and \( \vec{w} \), calculated as the lengths of the vectors.
For \( \vec{v} = \langle -8, 3 \rangle \) and \( \vec{w} = \langle 2, 6 \rangle \):

• \( ||\vec{v}|| = \sqrt{73}, \quad ||\vec{w}|| = \sqrt{40} \)
• Thus, \( \cos \theta = \frac{2}{\sqrt{73} \cdot \sqrt{40}} \)
• Solving for \( \theta \approx 86.99^\circ \)

This means that the vectors are nearly perpendicular to each other, with a wide angle of approximately 87 degrees.

Orthogonal vectors are two vectors at right angles (90 degrees) to each other. Their dot product is zero because their components effectively "cancel out" any alignment.
In the context of linear algebra, being orthogonal indicates independence; they don't rely on one another.
Consider \( \vec{q} = \vec{v} - \operatorname{proj}_{\vec{w}}(\vec{v}) \), which subtracts the projection of \( \vec{v} \) onto \( \vec{w} \) from \( \vec{v} \). This new vector \( \vec{q} \) is guaranteed to be orthogonal to \( \vec{w} \), as demonstrated by:

• \( \vec{q} = \langle -8.1, 2.7 \rangle \)
• \( \vec{q} \cdot \vec{w} = (-8.1 \times 2) + (2.7 \times 6) = -16.2 + 16.2 = 0 \)

Since the dot product is zero, vector \( \vec{q} \) is indeed orthogonal to vector \( \vec{w} \), supporting their independence and orthogonal nature.