Problem 52 - Algebra and Trigonometry

Question

Find the unit vector in the same direction as \(\mathbf{V}\). \(\mathbf{v}=-5 \mathbf{i}+12 \mathbf{j}\)

Step-by-Step Solution

Answer

\(\mathbf{u} = -\frac{5}{13}\mathbf{i} + \frac{12}{13}\mathbf{j}\)

Step 1: Calculate the Magnitude

First, find the magnitude of the vector \(\mathbf{v} = -5\mathbf{i} + 12\mathbf{j}\). The magnitude is calculated using the formula \(\|\mathbf{v}\| = \sqrt{(-5)^2 + (12)^2}\).

Step 2: Simplify the Magnitude Calculation

Simplify the expression inside the square root: \(\|\mathbf{v}\| = \sqrt{25 + 144} = \sqrt{169}\).

Step 3: Determine the Magnitude

Determine the square root: \(\|\mathbf{v}\| = 13\).

Step 4: Apply the Unit Vector Formula

The unit vector \(\mathbf{u}\) in the same direction as \(\mathbf{v}\) is given by \(\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}\).

Step 5: Divide Each Component by the Magnitude

Divide each component of the vector \(\mathbf{v}\) by 13: \(\mathbf{u} = \frac{-5}{13}\mathbf{i} + \frac{12}{13}\mathbf{j}\).

Key Concepts

magnitude of a vector

To understand vectors, it's crucial to know how to find their magnitude. The magnitude of a vector symbolizes its length or size.
For the vector \(\text{{\mathbf{{v}}}} = -5\text{{\mathbf{{i}}}} + 12\text{{\mathbf{{j}}}}\), the magnitude \(\text{{\|\mathbf{{v}}\|}}\) is calculated using the Pythagorean theorem:
\[\text{{\|\mathbf{{v}}\|}} = \sqrt{(-5)^2 + (12)^2}\text{{.}}\]
This simplifies to:
\[\text{{\|\mathbf{{v}}\|}} = \sqrt{25 + 144} = \sqrt{169} = 13\text{{.}}\]
The magnitude helps us understand how 'long' the vector is.

direction vector

Knowing the direction of a vector is about understanding where the vector is pointing. The components tell us this. For the vector \(\text{{\mathbf{{v}}}} = -5\text{{\mathbf{{i}}}} + 12\text{{\mathbf{{j}}}}\), we see that:

• \(\text{{-5i}}\) tells us to move -5 units along the x-axis
• \(\text{{12j}}\) tells us to move 12 units along the y-axis

This gives a sense of the direction, showing how the vector moves in the 2-dimensional plane.

vector normalization

Vector normalization is the process of converting any vector into a unit vector. A unit vector has a magnitude of 1 and points in the same direction.

To normalize a vector \(\text{{\mathbf{{v}}}}\), you divide each component by the vector's magnitude. For \(\text{{\mathbf{{v}}}} = -5\text{{\mathbf{{i}}}} + 12\text{{\mathbf{{j}}}}\) with \(\text{{\|\mathbf{{v}}\|}} = 13\), we get the unit vector \(\text{{\mathbf{{u}}}}\) using:
\(\text{{\mathbf{{u}}}} = \frac{\text{{\mathbf{{v}}}}}{\text{{\|\mathbf{{v}}\|}}}\)
This becomes:

• \(\frac{-5}{13}\text{{\mathbf{{i}}}} + \frac{12}{13}\text{{\mathbf{{j}}}}\)

This vector now has a magnitude of 1 and points in the same direction as the original vector.

components of a vector

Vectors are often broken down into their components. These components show how much the vector moves in each dimension.

For the vector \(\text{{\mathbf{{v}}}} = -5\text{{\mathbf{{i}}}} + 12\text{{\mathbf{{j}}}}\), it has:

• A horizontal component of \(\text{{-5i}}\)
• A vertical component of \(\text{{12j}}\)

This is helpful because we can analyze each dimension separately, making complex calculations simpler. Each component contributes to the vector’s overall direction and magnitude.