Problem 41 - Precalculus

Question

Find a unit vector in the direction of the given vector. $$v=(-9,-12)$$

Step-by-Step Solution

Answer

The unit vector in the direction of \( v = (-9, -12) \) is \( \left( -\frac{3}{5}, -\frac{4}{5} \right) \).

Step 1: Find the Magnitude of the Vector

To find a unit vector in the direction of a given vector, we first need to find the magnitude (or length) of the vector. The magnitude of a vector \( v=(x, y) \) is calculated using the formula:\[||v|| = \sqrt{x^2 + y^2}\]For the vector \( v = (-9, -12) \), its magnitude is:\[||v|| = \sqrt{(-9)^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\]

Step 2: Divide Each Component by the Magnitude

Once the magnitude of the vector has been calculated, a unit vector in the direction of the original vector is found by dividing each component of the vector by its magnitude. For the vector \( v = (-9, -12) \) with a magnitude of 15, divide both components by 15:\[v_{unit} = \left( \frac{-9}{15}, \frac{-12}{15} \right) = \left( -\frac{3}{5}, -\frac{4}{5} \right)\]

Step 3: Verify the Unit Vector

To ensure that \( v_{unit} = \left( -\frac{3}{5}, -\frac{4}{5} \right) \) is indeed a unit vector, calculate its magnitude. The magnitude of a unit vector should be 1. Calculate the magnitude:\[||v_{unit}|| = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1\]Since the magnitude is 1, \( v_{unit} \) is indeed a unit vector.

Key Concepts

Magnitude of a vector

The magnitude of a vector is like its length or size. Think of it like measuring how long it is, just like you'd measure the length of a stick. To find it, we use a special formula that combines all the parts (or components) of the vector together. For a vector written as \( v = (x, y) \), the magnitude is calculated using the formula:

• \(||v|| = \sqrt{x^2 + y^2}\)

This formula uses the Pythagorean Theorem from geometry. You take each part of the vector, square it, add those squares together, and then find the square root of that sum.

For example, if you have a vector \( v = (-9, -12) \), you will:

• Square each part: \((-9)^2 = 81\) and \((-12)^2 = 144\).
• Add them together: \(81 + 144 = 225\).
• Then take the square root: \(\sqrt{225} = 15\).

Now, you know the magnitude or the length of the vector is 15. This step is vital because it's the first part of finding a unit vector that moves in the same direction as our original vector.

Direction of a vector

The direction of a vector shows you where the vector is pointing in space. If you think about vectors as arrows, the direction is simply the way the arrow is pointing. Direction is important because it tells you *where* the vector is headed, not just how far.

To find a unit vector in the same direction as an original vector, you must use both the magnitude and the vector components. Once you have the magnitude, you can then find the unit vector by carefully adjusting the vector's size to "1" without changing its direction.

This process involves dividing each component of the vector by the magnitude, as seen with our vector example, \( (-9, -12) \). The unit vector direction is found by calculating:

• \(v_{unit} = \left( \frac{-9}{15}, \frac{-12}{15} \right) = \left( -\frac{3}{5}, -\frac{4}{5} \right)\)

Even though the numbers are smaller now, the arrow would still point in the same direction. This is because a unit vector scales the original vector down to a length of exactly one but keeps pointing the same way.

Vector components

Vectors have components that tell you how the vector stretches across dimensions. For simple vectors, these components are often given in two parts, such as \( x \) and \( y \), which represent the horizontal and vertical directions respectively.

The components are like coordinates that help you navigate through space. In the vector \( v = (-9, -12) \), the component \(-9\) tells you how far the vector moves horizontally (along the x-axis), while \(-12\) indicates the movement vertically (along the y-axis).

Each component plays a crucial role in computing both the magnitude and direction. By changing the size or value of a component, you directly affect the vector itself. When calculating a unit vector, you modify these components by dividing each one by the vector's magnitude. This scaling process allows the vector to maintain its direction but reduces its length to one unit, ensuring it acts as a standard "unit vector." The final outcome provides a simplified way to understand both the orientation and basic structure of the vector itself.