Problem 26 - Algebra and Trigonometry

Question

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle- 2,-1\rangle $$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = 6\sqrt{2}\), \(|\mathbf{v}| = \sqrt{5}\), \(|2\mathbf{u}| = 12\sqrt{2}\), \(\left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2}\sqrt{5}\), \(|\mathbf{u} + \mathbf{v}| = \sqrt{89}\), \(|\mathbf{u} - \mathbf{v}| = \sqrt{65}\), \(|\mathbf{u}| - |\mathbf{v}| = 6\sqrt{2} - \sqrt{5}\)."

Step 1: Calculate Magnitude of Vector \(\mathbf{u}\)

The magnitude of a vector \(\mathbf{u} = \langle a, b \rangle\) is given by \(|\mathbf{u}| = \sqrt{a^2 + b^2}\). For \(\mathbf{u} = \langle -6, 6 \rangle\), calculate: \[|\mathbf{u}| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\]

Step 2: Calculate Magnitude of Vector \(\mathbf{v}\)

The magnitude of vector \(\mathbf{v} = \langle -2, -1 \rangle\) is calculated similarly: \[|\mathbf{v}| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

Step 3: Calculate Magnitude of Scaled Vector \(2\mathbf{u}\)

Scaling a vector by a constant \(c\) multiplies its magnitude by \(|c|\). Given \(\mathbf{u} = \langle -6, 6 \rangle\), the magnitude of \(2\mathbf{u}\) is:\[|2\mathbf{u}| = 2|\mathbf{u}| = 2 \times 6\sqrt{2} = 12\sqrt{2}\]

Step 4: Calculate Magnitude of Scaled Vector \(\frac{1}{2}\mathbf{v}\)

Scaling vector \(\mathbf{v}\) by \(\frac{1}{2}\), the magnitude becomes: \[\left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2}|\mathbf{v}| = \frac{1}{2}\sqrt{5}\]

Step 5: Calculate Magnitude of Sum of Vectors \(\mathbf{u} + \mathbf{v}\)

First, determine \(\mathbf{u} + \mathbf{v} = \langle -6, 6 \rangle + \langle -2, -1 \rangle = \langle -8, 5 \rangle\).Then, compute its magnitude:\[|\mathbf{u} + \mathbf{v}| = \sqrt{(-8)^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}\]

Step 6: Calculate Magnitude of Difference of Vectors \(\mathbf{u} - \mathbf{v}\)

Determine the vector \(\mathbf{u} - \mathbf{v} = \langle -6, 6 \rangle - \langle -2, -1 \rangle = \langle -4, 7 \rangle\).Then, calculate its magnitude:\[|\mathbf{u} - \mathbf{v}| = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}\]

Step 7: Calculate Difference of Magnitudes \(|\mathbf{u}| - |\mathbf{v}|\)

The difference of magnitudes is simply:\[|\mathbf{u}| - |\mathbf{v}| = 6\sqrt{2} - \sqrt{5}\]

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is crucial as it represents the vector's length in the Cartesian plane. Imagine you have a vector \( \mathbf{u} = \langle a, b \rangle \). The magnitude is essentially the distance from the origin to the point \( (a, b) \). This is calculated using the Pythagorean theorem.

To find the magnitude \( |\mathbf{u}| \), we apply the formula \( |\mathbf{u}| = \sqrt{a^2 + b^2} \). For example, if \( \mathbf{u} = \langle -6, 6 \rangle \), the magnitude would be:

\[|\mathbf{u}| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\]

This calculation provides us with a scalar value, which is the length of vector \( \mathbf{u} \). It’s important to remember that the magnitude is always a non-negative number, regardless of the direction of the vector's components.

Vector Addition

Vector addition involves combining two or more vectors to obtain a resultant vector. For two vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), their sum \( \mathbf{u} + \mathbf{v} \) is calculated by adding corresponding components.

The general rule for addition is:

\[\mathbf{u} + \mathbf{v} = \langle a + c, b + d \rangle\]

Let's consider vectors \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle -2, -1 \rangle \). Their sum would be:

\[\mathbf{u} + \mathbf{v} = \langle -6 + (-2), 6 + (-1) \rangle = \langle -8, 5 \rangle\]

The magnitude of this resultant vector represents the "net" effect of combining these directions and magnitudes. Vector addition is visualized by placing the tail of one vector at the head of the other, and the resultant vector spans from the tail of the first to the head of the last.

Vector Scaling

When scaling a vector, you multiply it by a scalar value, which stretches or shrinks the vector. Let's say we have a vector \( \mathbf{u} \) and we scale it by a constant \( c \). The new vector, denoted as \( c\mathbf{u} \), is just the original vector stretched/shrunk by \( c \).

The magnitude of this new vector is calculated by multiplying the magnitude of the original vector \( |c||\mathbf{u}| \). For instance, consider scaling \( \mathbf{u} = \langle -6, 6 \rangle \) by 2:

\[|2\mathbf{u}| = 2|\mathbf{u}| = 2 \times 6\sqrt{2} = 12\sqrt{2}\]

Scaling by half, for vector \( \mathbf{v} \):

\[\left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2}|\mathbf{v}| = \frac{1}{2}\sqrt{5}\]

Scaling changes the vector’s length but not its direction unless the scalar is negative, which reverses it. It's crucial for applications needing proportional adjustments, such as physics and engineering problems.

Difference of Vectors

Vector subtraction is finding the difference between two vectors, which results in another vector. Given vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), their difference \( \mathbf{u} - \mathbf{v} \) is computed by subtracting corresponding components:

\[\mathbf{u} - \mathbf{v} = \langle a - c, b - d \rangle\]

This operation gives us the vector pointing from the tip of \( \mathbf{v} \) to the tip of \( \mathbf{u} \).Consider, if \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle -2, -1 \rangle \), the result is:

\[\mathbf{u} - \mathbf{v} = \langle -6 - (-2), 6 - (-1) \rangle = \langle -4, 7 \rangle\]

Finally, the magnitude of this vector gives the "absolute" difference in distance given their directions, calculated similarly to previous magnitudes. This concept is vital for navigation, relative movement, and adjusting quantities in real-life scenarios.