Problem 39 - Algebra and Trigonometry

Question

\(37-40\) Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=\langle 10,-1\rangle, \quad \mathbf{v}=\langle- 2,-2\rangle $$

Step-by-Step Solution

Answer

Magnitudes are \(|\mathbf{u}| = \sqrt{101}, |\mathbf{v}| = 2\sqrt{2}, |2\mathbf{u}| = 2\sqrt{101}, |\frac{1}{2}\mathbf{v}| = \sqrt{2}, |\mathbf{u} + \mathbf{v}| = \sqrt{73}, |\mathbf{u} - \mathbf{v}| = \sqrt{145}, |\mathbf{u}| - |\mathbf{v}| = \sqrt{101} - 2\sqrt{2}.\)

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

To find the magnitude of \( \mathbf{u} = \langle 10, -1 \rangle \), use the formula \( |\mathbf{u}| = \sqrt{x^2 + y^2} \), where \( x = 10 \) and \( y = -1 \). Therefore, \( |\mathbf{u}| = \sqrt{10^2 + (-1)^2} = \sqrt{100 + 1} = \sqrt{101} \).

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

For \( \mathbf{v} = \langle -2, -2 \rangle \), use the same approach. \( |\mathbf{v}| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

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

First, find \( 2\mathbf{u} = 2\langle 10, -1 \rangle = \langle 20, -2 \rangle \). The magnitude is \( |2\mathbf{u}| = \sqrt{20^2 + (-2)^2} = \sqrt{400 + 4} = \sqrt{404} = 2\sqrt{101} \).

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

Calculate \( \frac{1}{2}\mathbf{v} = \frac{1}{2}\langle -2, -2 \rangle = \langle -1, -1 \rangle \). The magnitude is \( \left| \frac{1}{2}\mathbf{v} \right| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \).

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

Add \( \mathbf{u} + \mathbf{v} = \langle 10, -1 \rangle + \langle -2, -2 \rangle = \langle 8, -3 \rangle \). Calculate the magnitude: \( |\mathbf{u} + \mathbf{v}| = \sqrt{8^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \).

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

Subtract \( \mathbf{u} - \mathbf{v} = \langle 10, -1 \rangle - \langle -2, -2 \rangle = \langle 12, 1 \rangle \). Calculate the magnitude: \( |\mathbf{u} - \mathbf{v}| = \sqrt{12^2 + 1^2} = \sqrt{144 + 1} = \sqrt{145} \).

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

Subtract the magnitudes: \( |\mathbf{u}| - |\mathbf{v}| = \sqrt{101} - 2\sqrt{2} \).

Key Concepts

Vector Addition

Vector addition is like combining directions and distances. If you have two vectors, say \( \mathbf{u} = \langle x_1, y_1 \rangle \) and \( \mathbf{v} = \langle x_2, y_2 \rangle \), adding them together results in a new vector where each component is the sum of the corresponding components of the vectors. This means:

• For the x-components: \( x_1 + x_2 \)
• For the y-components: \( y_1 + y_2 \)

The resulting vector is \( \mathbf{u} + \mathbf{v} = \langle x_1 + x_2, y_1 + y_2 \rangle \). In our original exercise, adding vectors \( \mathbf{u} = \langle 10, -1 \rangle \) and \( \mathbf{v} = \langle -2, -2 \rangle \) gives us \( \langle 8, -3 \rangle \). This concept highlights how vector addition can help determine the combined effect of two separate influences.

Vector Subtraction

Vector subtraction is very similar to vector addition but with a key difference. Instead of combining, you are finding the difference between vectors. Think of it as reversing the direction of the second vector before combining it with the first. For vectors \( \mathbf{u} = \langle x_1, y_1 \rangle \) and \( \mathbf{v} = \langle x_2, y_2 \rangle \), subtracting \( \mathbf{v} \) from \( \mathbf{u} \) looks like this:

• For the x-components: \( x_1 - x_2 \)
• For the y-components: \( y_1 - y_2 \)

The result is \( \mathbf{u} - \mathbf{v} = \langle x_1 - x_2, y_1 - y_2 \rangle \). In our exercise, subtracting \( \mathbf{v} = \langle -2, -2 \rangle \) from \( \mathbf{u} = \langle 10, -1 \rangle \) results in \( \langle 12, 1 \rangle \). This operation is useful for finding the net change between two vectors.

Scalar Multiplication

Scalar multiplication is when you stretch or shrink a vector by multiplying it by a number called a scalar. This operation changes the magnitude of the vector but not its direction if the scalar is positive. For a vector \( \mathbf{u} = \langle x, y \rangle \) and scalar \( k \), scalar multiplication is done component-wise:

• Multiply each component by the scalar: \( \langle kx, ky \rangle \)

In the example provided, if you multiply \( \mathbf{u} = \langle 10, -1 \rangle \) by 2, you get \( \langle 20, -2 \rangle \). Similarly, multiplying \( \mathbf{v} = \langle -2, -2 \rangle \) by \( \frac{1}{2} \) yields \( \langle -1, -1 \rangle \). Scalar multiplication is handy for adjusting vectors to different scales.

Magnitude of a Vector

Magnitude of a vector, often referred to as the length or norm of the vector, is a measurement of its size. It tells us how long the vector is in a Euclidean space, regardless of its direction. You determine this by using the Pythagorean theorem for vectors in the form \( \mathbf{u} = \langle x, y \rangle \). The formula is:

• Magnitude: \( |\mathbf{u}| = \sqrt{x^2 + y^2} \)

In the exercise, for \( \mathbf{u} = \langle 10, -1 \rangle \), the magnitude is \( \sqrt{101} \). For \( \mathbf{v} = \langle -2, -2 \rangle \), it is \( 2\sqrt{2} \). Knowing how to calculate magnitudes is crucial for understanding the size difference between vectors and is essential in physics and engineering applications.