Problem 15 - Algebra and Trigonometry

Question

(a) find the dot product v \(\cdot \mathbf{w} ;\) (b) find the angle between \(\mathbf{v}\) and \(\mathbf{w} ;\) (c) state whether the vectors are parallel, orthogonal, or neither. $$ \mathbf{v}=3 \mathbf{i}+4 \mathbf{j}, \quad \mathbf{w}=-6 \mathbf{i}-8 \mathbf{j} $$

Step-by-Step Solution

Answer

The dot product is -50. The angle between the vectors is \( \pi \) radians (180 degrees). The vectors are parallel.

Step 1: Find the dot product \( \mathbf{v} \cdot \mathbf{w} \)

The dot product of two vectors \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \) and \( \mathbf{w} = c \mathbf{i} + d \mathbf{j} \) is given by the formula: \( \mathbf{v} \cdot \mathbf{w} = ac + bd \). For the given vectors, \( \mathbf{v} = 3 \mathbf{i} + 4 \mathbf{j} \) and \( \mathbf{w} = -6 \mathbf{i} - 8 \mathbf{j} \), substitute the values: \( 3 \times (-6) + 4 \times (-8) \). Compute the arithmetic: \( -18 + (-32) = -50 \). So, \( \mathbf{v} \cdot \mathbf{w} = -50 \).

Step 2: Compute the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \)

The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \) is given by \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \). For \( \mathbf{v} = 3 \mathbf{i} + 4 \mathbf{j} \), compute the magnitude: \( ||\mathbf{v}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). For \( \mathbf{w} = -6 \mathbf{i} - 8 \mathbf{j} \), compute the magnitude: \( ||\mathbf{w}|| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).

Step 3: Find the cosine of the angle between \( \mathbf{v} \) and \( \mathbf{w} \)

The cosine of the angle \( \theta \) between two vectors is given by: \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||} \). Substitute the known values: \( \cos \theta = \frac{-50}{5 \cdot 10} = \frac{-50}{50} = -1 \).

Step 4: Determine the angle between \( \mathbf{v} \) and \( \mathbf{w} \)

Since \( \cos \theta = -1 \), the angle \( \theta \) is \( \pi \) radians (180 degrees), which means the vectors are pointing in exactly opposite directions.

Step 5: Classify the relationship between the vectors

Vectors are parallel if the angle between them is 0 or \( \pi \) radians. Since the angle is \( \pi \) radians, the vectors \( \mathbf{v} \) and \( \mathbf{w} \) are parallel.

Key Concepts

Vector Magnitude

Understanding vector magnitude is essential. The magnitude of a vector is like its length.
To find it, we use the Pythagorean theorem.
For a vector in 2D like \( \mathbf{v} = 3\mathbf{i} + 4\mathbf{j} \), calculate the magnitude as \( ||\mathbf{v}|| = \sqrt{3^2 + 4^2} = 5 \).
Similarly, for \( \mathbf{w} = -6\mathbf{i} - 8\mathbf{j} \), the magnitude is \( ||\mathbf{w}|| = \sqrt{(-6)^2 + (-8)^2} = 10 \).
Think of the magnitude as the distance from the vector’s tail to its tip.

Angle Between Vectors

The angle between vectors tells us how they point relative to each other.
To find this angle, we use the dot product formula: \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||} \).
If \ \( \mathbf{v}\) and \( \mathbf{w}\) \ are given as above, compute \( \cos \theta \): \( \cos \theta = \frac{-50}{5 \cdot 10} = -1 \).
The value \( -1 \) indicates they have an angle of \( \pi \) radians (180 degrees).
This means they point in exactly opposite directions.

Parallel Vectors

Vectors align in the same or exactly opposite directions are parallel.
If the angle between them is 0 or \( \pi \) radians, the vectors are parallel.
When \( \cos \theta \) is \(1\) (same direction) or \(-1\) (opposite direction), the vectors are parallel.
In our problem, the angle between \( \mathbf{v}\) and \( \mathbf{w}\) is \( \pi \) radians.
So, \( \mathbf{v}\) and \( \mathbf{w}\) are parallel but point in opposite directions.

Orthogonal Vectors

Orthogonal vectors meet at a right angle. This means the dot product is zero.
For vectors \( \mathbf{v}\) and \( \mathbf{w}\), use \( \mathbf{v} \cdot \mathbf{w} \).
If the dot product is 0, the vectors are orthogonal.
In our exercise, \( \mathbf{v} \cdot \mathbf{w} \) equals \(-50\) not zero.
Hence, \( \mathbf{v}\) and \( \mathbf{w}\) are not orthogonal.
Orthogonality implies perpendicular vectors, useful in many applications.