Problem 38 - Precalculus Mathematics for Calculus

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 \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \quad \mathbf{v}=\mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = \sqrt{13}\), \(|\mathbf{v}| = \sqrt{5}\), \(|2 \mathbf{u}| = 2\sqrt{13}\), \(\left|\frac{1}{2} \mathbf{v}\right| = \frac{1}{2}\sqrt{5}\), \(|\mathbf{u}+\mathbf{v}| = \sqrt{2}\), \(|\mathbf{u}-\mathbf{v}| = \sqrt{34}\), \(|\mathbf{u}| - |\mathbf{v}| = \sqrt{13} - \sqrt{5}\).

Step 1: Find \(|\mathbf{u}|\)

The magnitude of vector \(\mathbf{u} = -2\mathbf{i} + 3\mathbf{j}\) is given by the formula \(|\mathbf{u}| = \sqrt{(-2)^2 + 3^2}\). Evaluate this expression: \(|\mathbf{u}| = \sqrt{4 + 9} = \sqrt{13}\).

Step 2: Find \(|\mathbf{v}|\)

The magnitude of vector \(\mathbf{v} = \mathbf{i} - 2\mathbf{j}\) is given by the formula \(|\mathbf{v}| = \sqrt{1^2 + (-2)^2}\). Evaluate this expression: \(|\mathbf{v}| = \sqrt{1 + 4} = \sqrt{5}\).

Step 3: Find \(|2 \mathbf{u}|\)

To find the magnitude of \(2\mathbf{u}\), use the property \(|c\mathbf{u}| = |c| \cdot |\mathbf{u}|\). So, \(|2 \mathbf{u}| = 2 \cdot |\mathbf{u}| = 2\sqrt{13}\).

Step 4: Find \(\left|\frac{1}{2} \mathbf{v}\right|\)

To find the magnitude of \(\frac{1}{2}\mathbf{v}\), use the property \(|c\mathbf{v}| = |c| \cdot |\mathbf{v}|\). So, \[\left|\frac{1}{2} \mathbf{v}\right| = \frac{1}{2} \cdot |\mathbf{v}| = \frac{1}{2}\sqrt{5}\].

Step 5: Find \(|\mathbf{u} + \mathbf{v}|\)

First, calculate \(\mathbf{u} + \mathbf{v} = (-2\mathbf{i} + 3\mathbf{j}) + (\mathbf{i} - 2\mathbf{j}) = (-2+1)\mathbf{i} + (3-2)\mathbf{j} = -\mathbf{i} + \mathbf{j}\). The magnitude \(|\mathbf{u} + \mathbf{v}| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}\).

Step 6: Find \(|\mathbf{u} - \mathbf{v}|\)

First, calculate \(\mathbf{u} - \mathbf{v} = (-2\mathbf{i} + 3\mathbf{j}) - (\mathbf{i} - 2\mathbf{j}) = (-2-1)\mathbf{i} + (3+2)\mathbf{j} = -3\mathbf{i} + 5\mathbf{j}\). The magnitude \(|\mathbf{u} - \mathbf{v}| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}\).

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

Using the magnitudes found earlier, \(|\mathbf{u}| - |\mathbf{v}| = \sqrt{13} - \sqrt{5}\).

Key Concepts

Vectors

Vectors are powerful mathematical objects used to represent both a quantity and its direction. They are often depicted as arrows starting from one point to another and can describe things like velocity, force, and displacement. Each vector is expressed in terms of its components, which are values along the x-axis (horizontal) and y-axis (vertical).

For example, if a vector is \((-2 \mathbf{i} + 3 \mathbf{j})\), \(-2\) indicates the movement in the horizontal direction, while \(3\) is the movement in the vertical direction. Vectors can be added together or scaled using scalar multiplication, which alters the vector's magnitude or direction.

Magnitude Formula

The magnitude of a vector illustrates its size or length but not its direction. To find this, we use the Pythagorean Theorem, which applies because vectors create right triangles with the coordinate axes.

To calculate the magnitude of a vector \(\mathbf{u} = a\mathbf{i} + b\mathbf{j}\), where \(a\) and \(b\) are real numbers, use the formula:

• \(|\mathbf{u}| = \sqrt{a^2 + b^2}\).

This formula stems from the understanding that each vector component represents a side of a right triangle, and thus the magnitude is equivalent to the hypotenuse in coordinate space.

For instance, with a vector \((-2\mathbf{i} + 3\mathbf{j})\), the magnitude \(|\mathbf{u}|\) becomes \(|\mathbf{u}| = \sqrt{(-2)^2 + 3^2} = \sqrt{13}\). This value highlights how long the vector is regardless of its direction.

Vector Addition

Vector addition involves combining two vectors to form a new resultant vector. This is done by adding their corresponding components, much like simple arithmetic for numbers.

When adding vectors \(\mathbf{u}\) and \(\mathbf{v}\), such as \(-2 \mathbf{i} + 3 \mathbf{j}\) and \(\mathbf{i} - 2 \mathbf{j}\), the operation is performed as follows:

• Add the components of \(-2\mathbf{i}\) and \(+\mathbf{i}\) to get \(-\mathbf{i}\).
• Similarly, add \(3\mathbf{j}\) and \(-2\mathbf{j}\) to get \[\mathbf{j}\].

This results in a new vector \(-\mathbf{i} + \mathbf{j}\).

After forming the resultant vector, you can calculate its magnitude to understand the actual effect of adding vectors in terms of direction and strength. The process is handy for various applications in physics, such as finding resultant forces.

Scalar Multiplication

Scalar multiplication in vectors involves changing the vector's magnitude while maintaining its direction. This is achieved by multiplying each of the vector's components by a scalar (a real number).

Suppose you have a vector \(\mathbf{u} = -2\mathbf{i} + 3\mathbf{j}\) and you want to find the magnitude of \(2\mathbf{u}\). To perform scalar multiplication, multiply both components by \(2\), yielding \(-4\mathbf{i} + 6\mathbf{j}\). The resulting vector has a magnitude of \(|2\mathbf{u}| = 2\sqrt{13}\).

Similarly, scaling a vector like \(\mathbf{v}\) with a fraction, such as \(\frac{1}{2}\), results in smaller component values. If \(\mathbf{v} = \mathbf{i} - 2\mathbf{j}\), the scalar \(|\frac{1}{2}\mathbf{v}|\) becomes \[\frac{1}{2}\sqrt{5}\], effectively halving its length without altering its direction. Scalar multiplication is essential in vector calculations, enabling control over vector length for balancing and scaling in fields like engineering and graphics.