Problem 55 - Algebra and Trigonometry

Question

In Exercises 53–56, let. $$\mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{v}=6 \mathbf{i}-\mathbf{j}, \mathbf{w}=-3 \mathbf{i}$$ Find each specified vector or scalar. $$|\mathbf{u}+\mathbf{v}|^{2}-\| \mathbf{u}-\left.\mathbf{v}\right|^{2}$$

Step-by-Step Solution

Answer

The result of the operation \(\| \mathbf{u} + \mathbf{v} \|^2 - \|\mathbf{u} - \mathbf{v}\|^2\) is -60.

Step 1: Add Vectors \(\mathbf{u}\) and \(\mathbf{v}\)

Vector addition is done component-wise. Thus the sum \(\mathbf{u}+\mathbf{v}\) is found by adding the \(\mathbf{i}\) components and the \(\mathbf{j}\) components of \(\mathbf{u}\) and \(\mathbf{v}\): \(( -2 \mathbf{i} + 3 \mathbf{j} ) + ( 6 \mathbf{i} - \mathbf{j} ) = 4 \mathbf{i} + 2 \mathbf{j}\)

Step 2: Subtract Vectors \(\mathbf{u}\) and \(\mathbf{v}\)

Subtracting vector \(\mathbf{v}\) from \(\mathbf{u}\) again involves subtracting the \(\mathbf{i}\) and \(\mathbf{j}\) components: \( -2 \mathbf{i} + 3 \mathbf{j} - ( 6 \mathbf{i} - \mathbf{j} ) = -8 \mathbf{i} + 4 \mathbf{j}\)

Step 3: Calculate the square of the magnitude of \(\mathbf{u}+\mathbf{v}\)

The square of the magnitude of a vector \(\mathbf{a}\) with components \(a_1, a_2\), is calculated using the formula \(\| \mathbf{a} \|^2 = a_1^2 + a_2^2\). So, for \(\| \mathbf{u}+\mathbf{v} \|^2\), we calculate \(4^2 + 2^2 = 20\).

Step 4: Calculate the square of the magnitude of \(\mathbf{u}-\mathbf{v}\)

Likewise, for \(\| \mathbf{u}-\mathbf{v} \|^2\), we calculate \(-8^2 + 4^2 = 80\).

Step 5: Subtract the magnitudes

Now we simply subtract the second result from the first: \(20 - 80 = -60\).

Key Concepts

Vector Addition

Vector addition is a foundational concept in vector algebra and involves adding two or more vectors to find their resultant. When adding vectors, we consider their direction and magnitude. In a coordinate system, vectors are broken down into components along the axes, typically denoted as \( \mathbf{i} \) and \( \mathbf{j} \) in two-dimensional space.
To add vectors \( \mathbf{u} \) and \( \mathbf{v} \), you simply add their respective components:

• \( \mathbf{i} \) Component: Add the \( \mathbf{i} \) components. For example, if \( \mathbf{u} = -2\mathbf{i} + 3\mathbf{j} \) and \( \mathbf{v} = 6\mathbf{i} - \mathbf{j} \), then the resulting \( \mathbf{i} \) component is \(-2 + 6 = 4 \).
• \( \mathbf{j} \) Component: Similarly, add the \( \mathbf{j} \) components: \( 3 - 1 = 2 \).

Thus, the sum of vectors \( \mathbf{u} \) and \( \mathbf{v} \) becomes \( 4\mathbf{i} + 2\mathbf{j} \). This vector represents the combined effect of both \( \mathbf{u} \) and \( \mathbf{v} \), both in terms of direction and magnitude.

Vector Subtraction

Vector subtraction is much like vector addition, except that it involves taking the difference between the corresponding components of two vectors, essentially adding a negative vector. Subtracting vectors allows us to find the difference in direction and magnitude between them.
For vectors \( \mathbf{u} \) and \( \mathbf{v} \), subtraction is done as follows:

• \( \mathbf{i} \) Component: Subtract the \( \mathbf{i} \) component of \( \mathbf{v} \) from that of \( \mathbf{u} \): \(-2 - 6 = -8 \).
• \( \mathbf{j} \) Component: Subtract the \( \mathbf{j} \) component of \( \mathbf{v} \) from that of \( \mathbf{u} \): \(3 - (-1) = 4 \).

Therefore, the vector resulting from \( \mathbf{u} - \mathbf{v} \) is \(-8\mathbf{i} + 4\mathbf{j} \). This operation effectively shows how much \( \mathbf{u} \) differs from \( \mathbf{v} \), pointing in the direction of the resultant vector.

Magnitude of a Vector

The magnitude or length of a vector is a measure of its size, considering both \( \mathbf{i} \) and \( \mathbf{j} \) components. It's crucial for understanding the vector's physical significance, such as its displacement or force.
To compute the magnitude, we use the Pythagorean theorem for the vector's components:

• Formula: If a vector has components \( a_1 \) and \( a_2 \), its magnitude is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).
• Squared Magnitude: For squared magnitude calculation, simply compute \( a_1^2 + a_2^2 \); beneficial in physics to avoid square roots.

Applying this concept to \( \mathbf{u} + \mathbf{v} \) (i.e., \( 4\mathbf{i} + 2\mathbf{j} \)), we find its squared magnitude: \[ 4^2 + 2^2 = 16 + 4 = 20 \]. For \( \mathbf{u} - \mathbf{v} \) (i.e., \(-8\mathbf{i} + 4\mathbf{j} \)), it computes to \[ (-8)^2 + 4^2 = 64 + 16 = 80 \]. These results illustrate the vector’s magnitude and how it changes in addition or subtraction processes.