91Ó°ÊÓ

To find the magnitude of a vector, you use the formula:

• For a vector \( \mathbf{v} = \langle v_x, v_y \rangle \), the magnitude is \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \).

In the exercise, the vector \( \mathbf{v} \) was given as \( \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}} \right\rangle \). To find the magnitude, you substitute the components into the formula:\[|\mathbf{v}| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 } = \sqrt{\frac{1}{2} + \frac{1}{3} }\]This simplifies to \( \sqrt{\frac{5}{6}} \). The magnitude gives you an idea of the length of the vector in space. Similarly, the vector \( \mathbf{u} \) was also \( \sqrt{\frac{5}{6}} \) in magnitude.

The cosine of the angle between two vectors helps in understanding how parallel or perpendicular the vectors are. You can calculate it using the dot product formula:

• \( \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \)

If the cosine is close to 1, the vectors are nearly parallel. If it's close to 0, they are nearly perpendicular. In our exercise, the dot product \( \mathbf{v} \cdot \mathbf{u} \) was calculated as \( \frac{1}{6} \). The magnitude product \(|\mathbf{v}| |\mathbf{u}| \) gave us \( \frac{5}{6} \). Thus, the cosine of the angle \( \theta \) is:\[\cos(\theta) = \frac{\frac{1}{6}}{\frac{5}{6}} = \frac{1}{5}\]This indicates that the angle is not zero, thus \( \mathbf{v} \) and \( \mathbf{u} \) are neither fully parallel nor perpendicular but somewhere in between.

Projection of one vector onto another is like casting a shadow of that vector in the direction of the other. It's computed using the formula:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \)

This formula essentially projects vector \( \mathbf{u} \) onto \( \mathbf{v} \), showing the portion of \( \mathbf{u} \) that moves in the same direction as \( \mathbf{v} \). For the given exercise, the projection is given by:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\frac{1}{6}}{\frac{5}{6}} \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}} \right\rangle = \frac{1}{5} \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}} \right\rangle\]The resulting vector tells you how much of \( \mathbf{u} \) points in the direction of \( \mathbf{v} \). This concept is useful in various applications like physics and engineering.

Scalar component measures the "amount" of one vector in the direction of another. It provides a numerical length rather than a vector result.

• For vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \), it's given by: \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \)

This value helps you understand how much one vector "covers" of another along their route. In our specific problem:\[\text{Scalar component} = \frac{\frac{1}{6}}{ \sqrt{\frac{5}{6}} }\]It simplifies the vector \( \mathbf{u} \) down to a single value along the direction of \( \mathbf{v} \). It tells us that regardless of the larger picture, there's a measurable impact of \( \mathbf{u} \) towards \( \mathbf{v} \). This is often helpful in practical applications like determining force directions and motion along axes.