91Ó°ÊÓ

The dot product is a way to multiply two vectors, resulting in a scalar (a single number). This operation is incredibly useful as it helps to measure the angle between two vectors, and plays a key role in many vector calculations. To find the dot product of two vectors, \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), you multiply the corresponding components and add them together: \[\mathbf{u} \cdot \mathbf{v} = (u_1 \cdot v_1) + (u_2 \cdot v_2)\]
In our example, the vectors were \( \mathbf{u} = \langle -4, 7 \rangle \) and \( \mathbf{v} = \langle 3, 5 \rangle \). So the dot product was calculated as:

• Multiply \( -4 \) by \( 3 \), getting \( -12 \).
• Multiply \( 7 \) by \( 5 \), getting \( 35 \).
• Add these two results: \( -12 + 35 = 23 \).

This gives a dot product of 23, which tells us about how aligned the two vectors are.
The value being positive also indicates they are pointing, roughly, in the same direction.

The magnitude of a vector measures its length or size. It is contrived by utilizing the components of the vector. For a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\), its magnitude, represented by \(||\mathbf{u}||\), can be calculated using the following formula:\[||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2}\]
The magnitude provides insights into how long the vector is, irrespective of its direction. For vector \( \mathbf{u} = \langle -4, 7 \rangle \,\) we proceed to calculate:

• Square \( -4 \,\) resulting in \( 16 \).
• Square \( 7 \,\) yielding \( 49 \).
• Add these values: \( 16 + 49 = 65 \).
• Finally, take the square root: \( \sqrt{65} \).

So, the magnitude of vector \( \mathbf{u} \) is \( \sqrt{65} \,\) indicating its length.

The scalar projection of a vector onto another gives a measure of how much one vector 'extends' onto or along the direction of another vector. It essentially provides a shadow or footprint of one vector onto another when the light (so to speak) is cast perpendicularly.
To compute the scalar projection of \( \mathbf{v} \) onto \( \mathbf{u} \,\) employ the formula:\[ \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||} \]
For our vectors:

• We have already found that \( \mathbf{u} \cdot \mathbf{v} = 23 \).
• The magnitude \( ||\mathbf{u}|| = \sqrt{65} \).
• Thus, the scalar projection is \( \frac{23}{\sqrt{65}} \).

This result tells us the effective length of \( \mathbf{v} \) in the direction of vector \( \mathbf{u} \), simplified without direction indication.

Expressing vectors in component form essentially breaks them down into their critical parts or elements that can be used to easily identify direction and magnitude. After arriving at the vector projection \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} \,\) you'll want to convert it into component form, so it becomes clear and straightforward.
You find the projection using the formula:\[\operatorname{proj}_{\mathbf{u}} \mathbf{v} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||^2} \right) \mathbf{u}\]In the context of our particular vectors:

• Calculate \( \frac{23}{65} \) which effectively multiplies each component of \( \mathbf{u} \).
• Multiply \( \frac{23}{65} \) by \( -4 \) resulting in \( -\frac{92}{65} \).
• Then multiply \( \frac{23}{65} \) by \( 7 \) which gives \( \frac{161}{65} \).
• Thus, the vector in component form is \( \langle -\frac{92}{65}, \frac{161}{65} \rangle\).

In this form, the vector projection is clearly broken down into numerical values that correspond to each dimension's contribution, giving you great insight into its real geometric configuration.