Question

Find a vector of magnitude 5 that is parallel to \(-12 \mathbf{i}+9 \mathbf{j}\)

Short answer

-4\mathbf{i} + 3\mathbf{j}

Step-by-step solution

Find the magnitude of the given vector

To find a vector of magnitude 5 that is parallel to p{-12 \mathbf{i} + 9 \mathbf{j}}, start by determining the magnitude of the given vector using the formula for the magnitude of a vector \mathbf{v} = a\mathbf{i} + b\mathbf{j} which is \[ |\mathbf{v}| = \sqrt{a^2 + b^2} \]. So for the vector \mathbf{v} = -12\mathbf{i} + 9\mathbf{j}, the magnitude is \[ |\mathbf{v}| = \sqrt{(-12)^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \].

Determine the unit vector in the direction of the given vector

The unit vector \mathbf{u} in the direction of \mathbf{v} = -12\mathbf{i} + 9\mathbf{j} is found by dividing each component of the vector by its magnitude 15. This gives \[ \mathbf{u} = \frac{-12}{15} \mathbf{i} + \frac{9}{15} \mathbf{j} = -\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j} \].

Scale the unit vector to the desired magnitude

To find a vector with magnitude 5 that is parallel to \mathbf{v}, multiply the unit vector by 5.This results in \[ 5\mathbf{u} = 5 \left(-\frac{4}{5} \mathbf{i} + \frac{3}{5} \mathbf{j} \right) = -4 \mathbf{i} + 3 \mathbf{j} \].

Key concepts

vector calculation

When working with vectors, one of the fundamental tasks is calculating their magnitude. The magnitude of a vector is essentially its length and can be found by using the Pythagorean theorem generalized for vectors. For a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), the magnitude \( |\mathbf{v}| \) is calculated as \[ |\mathbf{v}| = \sqrt{a^2 + b^2} \].
In the given exercise, we start with the vector \( -12\mathbf{i} + 9\mathbf{j} \).
To find its magnitude, plug in the components into the formula: \[ |\mathbf{v}| = \sqrt{(-12)^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \].
This tells us that the original vector has a length of 15 units.

unit vector

A unit vector is a vector with a magnitude of 1. It points in the same direction as the original vector but has been normalized to a length of 1. To find the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} = -12\mathbf{i} + 9\mathbf{j} \), you divide each component of the vector by its magnitude. This results in:
\[ \mathbf{u} = \frac{-12}{15} \mathbf{i} + \frac{9}{15} \mathbf{j} = -\frac{4}{5} \mathbf{i} + \frac{3}{5} \mathbf{j} \].
The unit vector has the same direction as the original but a magnitude of 1. This is crucial for scaling vectors, as it allows us to easily adjust the vector's length while maintaining its direction.

scaling vectors

After finding the unit vector, the next step is to scale it to the desired magnitude. In our exercise, we need a vector of magnitude 5 that is parallel to \( -12\mathbf{i} + 9\mathbf{j} \). This is done by multiplying the unit vector by the desired magnitude:
\[ 5\mathbf{u} = 5 \left(-\frac{4}{5} \mathbf{i} + \frac{3}{5} \mathbf{j}\right) = -4 \mathbf{i} + 3 \mathbf{j} \].
Now, we have a vector \( -4\mathbf{i} + 3\mathbf{j} \) that is 5 units long and points in the same direction as the original vector. This process of scaling vectors is very useful when you need vectors of a specific length for applications in physics, engineering, and computer science.