Problem 36 - Precalculus

Question

Find a unit vector in the direction of the given vector. $$v=(3,4)$$

Step-by-Step Solution

Answer

The unit vector in the direction of \( v = (3,4) \) is \( \left( \frac{3}{5}, \frac{4}{5} \right) \).

Step 1: Determine the Magnitude of the Vector

To find a unit vector, we first need the magnitude (or length) of the vector. The magnitude is given by the formula: \( \| v \| = \sqrt{x^2 + y^2} \). For the vector \( v = (3, 4) \), substitute the values into the formula: \( \| v \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Step 2: Divide Each Component by the Magnitude

Now that we know the magnitude of the vector is 5, we can find the unit vector by dividing each component of the vector by its magnitude. The formula for the unit vector \( u \) is: \( u = \left( \frac{x}{\| v \|}, \frac{y}{\| v \|} \right) \). Substitute \( x = 3 \) and \( y = 4 \) into the equation: \( u = \left( \frac{3}{5}, \frac{4}{5} \right) \).

Step 3: Validate the Unit Vector

Finally, ensure the resulting unit vector has a magnitude of 1. Calculate the magnitude of \( u = \left( \frac{3}{5}, \frac{4}{5} \right) \) as follows: \( \| u \| = \sqrt{\left( \frac{3}{5} \right)^2 + \left( \frac{4}{5} \right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \). This confirms that \( u \) is indeed a unit vector.

Key Concepts

Vector Magnitude

Understanding the magnitude of a vector is essential in vector mathematics.
The magnitude determines the length of a vector. It’s calculated using the formula:
\( \| v \| = \sqrt{x^2 + y^2} \). This comes from the Pythagorean theorem, where each component is squared and summed, and then the square root is taken.
For example, consider the vector \( v = (3, 4) \).

• Square each component: \( 3^2 = 9 \) and \( 4^2 = 16 \).
• Add these squares: \( 9 + 16 = 25 \).
• Finally, take the square root: \( \sqrt{25} = 5 \).

The magnitude of vector \( v \) is 5. This shows the distance from the origin to the point (3, 4) in the coordinate plane.

Unit Vector Calculation

A unit vector is a vector with a magnitude of 1, but it maintains the same direction as the original vector.
The calculation involves dividing each component of the vector by its magnitude.
So, if the original vector is \( v = (3, 4) \), we already know its magnitude is 5.
The formula for the unit vector \( u \) is:

• For the x-component: \( \frac{3}{5} \).
• For the y-component: \( \frac{4}{5} \).

Thus, the unit vector \( u \) is \( \left( \frac{3}{5}, \frac{4}{5} \right) \).
This process normalizes the vector, providing a dimensionless representation that retains direction.

Magnitude of a Vector

It's crucial to verify the magnitude of your calculated unit vector to ensure accuracy.
A unit vector must have a magnitude of exactly 1. Let's check the unit vector \( u = \left( \frac{3}{5}, \frac{4}{5} \right) \):

• Square both components: \( \left( \frac{3}{5} \right)^2 = \frac{9}{25} \) and \( \left( \frac{4}{5} \right)^2 = \frac{16}{25} \).
• Add these squared components: \( \frac{9}{25} + \frac{16}{25} = \frac{25}{25} \).
• The result is \( \sqrt{1} = 1 \).

This confirms the unit vector has the correct magnitude. This validation step ensures that the unit vector \( u \) is indeed correctly scaled to a magnitude of 1, exemplifying proper unit vector calculation.