Problem 37 - 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 \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = \sqrt{5}, |\mathbf{v}| = \sqrt{13}, |2\mathbf{u}| = 2\sqrt{5}, \left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2}\sqrt{13}, |\mathbf{u}+\mathbf{v}| = \sqrt{26}, |\mathbf{u}-\mathbf{v}| = \sqrt{10}, |\mathbf{u}| - |\mathbf{v}| = \sqrt{5} - \sqrt{13}\).

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

The magnitude of a vector \(\mathbf{u} = a\mathbf{i} + b\mathbf{j}\) is given by \(|\mathbf{u}| = \sqrt{a^2 + b^2}\). Here, \(\mathbf{u} = 2\mathbf{i} + \mathbf{j}\), so \(|\mathbf{u}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\).

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

Similarly, for \(\mathbf{v} = 3\mathbf{i} - 2\mathbf{j}\), its magnitude is \(|\mathbf{v}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\).

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

Scaling a vector by a constant \(k\) scales its magnitude by \(|k|\). Thus, \(|2\mathbf{u}| = 2|\mathbf{u}| = 2\sqrt{5}\).

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

Similarly, scaling by \(\frac{1}{2}\) gives \(\left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2}|\mathbf{v}| = \frac{1}{2}\sqrt{13}\).

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

The sum of vectors \(\mathbf{u} + \mathbf{v} = (2 + 3)\mathbf{i} + (1 - 2)\mathbf{j} = 5\mathbf{i} - \mathbf{j}\). The magnitude is \(|\mathbf{u} + \mathbf{v}| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}\).

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

The difference of vectors \(\mathbf{u} - \mathbf{v} = (2 - 3)\mathbf{i} + (1 + 2)\mathbf{j} = -\mathbf{i} + 3\mathbf{j}\). The magnitude is \(|\mathbf{u} - \mathbf{v}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}\).

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

The difference in magnitudes is \(\sqrt{5} - \sqrt{13}\). This cannot be simplified further.

Key Concepts

Addition of Vectors

Vectors are like direction-aware numbers. They can point in any direction and have a specific length. Adding vectors combines both their directions and magnitudes into a new vector.

Imagine you have two vectors, \( \mathbf{u} \) and \( \mathbf{v} \). To add them, simply add their corresponding components. For example:

• \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \)
• \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \)

When adding these vectors together, we add the \( \mathbf{i} \) components and the \( \mathbf{j} \) components separately:

• \( \mathbf{u} + \mathbf{v} = (2 + 3)\mathbf{i} + (1 - 2)\mathbf{j} = 5\mathbf{i} - \mathbf{j} \)

This new vector \( 5\mathbf{i} - \mathbf{j} \) is the result of the addition, pointing in a direction that combines the influence of both \( \mathbf{u} \) and \( \mathbf{v} \). When dealing with vector addition, always handle each component separately.

Scaling Vectors

Scaling vectors is all about changing their size. Imagine a vector as an arrow; scaling stretches or shrinks this arrow but does not change its direction unless the scaling factor is negative.

To scale a vector, multiply each of its components by the scaling factor. For instance, if you have vector \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \) and want to double its size:

• \( 2\mathbf{u} = 2(2\mathbf{i} + \mathbf{j}) = 4\mathbf{i} + 2\mathbf{j} \)

Similarly, to reduce the size of vector \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \) by half, use:

• \( \frac{1}{2}\mathbf{v} = \frac{1}{2}(3\mathbf{i} - 2\mathbf{j}) = \frac{3}{2}\mathbf{i} - \mathbf{j} \)

Scaling affects the magnitude of the vector linearly. If you double the vector, its magnitude becomes twice as large. Conversely, halving the vector reduces the magnitude by half. This concept allows for flexible adjustments to the vector's reach while keeping its direction consistent.

Magnitude of a Vector

The magnitude of a vector is similar to measuring its length. It tells you how long the vector is, from start to finish, without considering direction.

To find the magnitude of a vector like \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \), use the formula:

• \( |\mathbf{u}| = \sqrt{2^2 + 1^2} = \sqrt{5} \)

This process involves squaring each component, adding them together, and finally taking the square root of the sum. The result, \( \sqrt{5} \), represents the length of vector \( \mathbf{u} \).

The concept also applies to differences and sums of vectors. For \( \mathbf{u} + \mathbf{v} = 5\mathbf{i} - \mathbf{j} \), the magnitude is:

• \( |\mathbf{u} + \mathbf{v}| = \sqrt{5^2 + (-1)^2} = \sqrt{26} \)

The magnitude of a vector is always a non-negative number, reflecting the distance from the origin to the point represented by the vector in a coordinate system.