Problem 51 - Precalculus with Limits

Question

Finding a Vector In Exercises \(49-52,\) find the vector \(v\) with the given magnitude and the same direction as u. \(\begin{array}{ll}{\text { Magnitude }} & {\text { Direction }} \\\ {\|\mathbf{y}\|=9} & {\mathbf{y}=\langle 2,5\rangle}\end{array}\)

Step-by-Step Solution

Answer

The vector \(v\) with magnitude \(9\) and the same direction as \(u=\langle2,5\rangle\) is \( y = \langle \frac{18}{\sqrt{29}}, \frac{45}{\sqrt{29}} \rangle \).

Step 1 - Find the magnitude of the given vector.

The given vector has components \(2\) and \(5\). The magnitude of any vector \(v = \langle a, b\rangle\) is calculated as \(\sqrt{a^{2} + b^{2}}\), hence here, \(\|v\| = \sqrt{2^{2} + 5^{2}} = \sqrt{29}\).

Step 2 - Calculate the unit vector

A unit vector is a vector of length 1 and it is obtained by dividing a vector by its magnitude. Therefore, the unit vector \(u\) of \(v\) is given by \(u = \frac{v}{\|v\|} = \frac{\langle2,5\rangle}{\sqrt{29}} = \langle \frac{2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \rangle \).

Step 3 - Calculate the vector with the desired magnitude

A vector with the same direction and a desired magnitude is given by multiplying the unit vector by this magnitude. Therefore, the vector \(y\) with magnitude \(9\) and same direction as \(v\) is \( y = 9 \cdot u = 9 \cdot \langle \frac{2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \rangle = \langle \frac{18}{\sqrt{29}}, \frac{45}{\sqrt{29}} \rangle \).

Key Concepts

Understanding Magnitude of a Vector

In vector calculus, the magnitude of a vector represents its length or size. Think of it as how long the vector is, which can be visualized as the distance from the origin to the point defined by the vector. Calculating the magnitude of a vector is straightforward. For a vector \(v = \langle a, b \rangle\), the magnitude is \(|v| = \sqrt{a^{2} + b^{2}}\). This formula stems from the Pythagorean theorem.

• Imagine a right triangle formed by the vector components \(a\) and \(b\).
• The vector acts as the hypotenuse of this triangle.
• Using the Pythagorean theorem gives you the length of this hypotenuse.

Understanding this helps in grasping that magnitudes are always non-negative, reflecting a scalar quantity like length in physical space.

Exploring the Unit Vector

A unit vector is a vector with a magnitude of exactly 1. It is primarily used to indicate direction. Any vector can be turned into a unit vector, which will then represent the same direction, but on a standardized scale.

• To find the unit vector of a given vector \(v\), divide each of its components by the magnitude of \(v\).
• For instance, the unit vector of our example \(\langle 2, 5 \rangle\) is \(<\frac{2}{\sqrt{29}}, \frac{5}{\sqrt{29}}>\).

The unit vector is useful because it retains the direction of the original vector but reduces the magnitude to 1, making it ideal for calculations where direction needs to be maintained without altering vector length.

Deciphering Vector Direction

The direction of a vector is an essential aspect as it tells you where the vector points. It's often represented by a unit vector to focus solely on the orientation without considering its size. Simply put, direction can be expressed through the angles a vector makes with the axes, or by its unit vector form.

• A vector like \(\langle 2, 5 \rangle\) points in both x and y directions.
• Its specific path can be fully described by its unit vector \(<\frac{2}{\sqrt{29}}, \frac{5}{\sqrt{29}}>\).

Understanding vector direction is key in mechanical systems, physics simulations, and navigation applications, where precise alignment of physical quantities is crucial.

Mastering Vector Multiplication

Vector multiplication can take various forms, but in this context, we’re discussing scalar multiplication.

• Scalar multiplication involves scaling a vector by a certain value, altering its magnitude but not its direction.
• In the exercise, multiplying the unit vector \(u\) by the magnitude 9 results in a new vector that is aligned with \(u\) but longer. The new vector is \(\langle \frac{18}{\sqrt{29}}, \frac{45}{\sqrt{29}} \rangle\).

This technique is useful in creating vectors of specific lengths while maintaining their original trajectories. It plays a crucial role in tasks like optimizing path lengths or converting between different units of measure in physics and engineering.