Question

Find each quantity if \(\mathbf{v}=3 \mathbf{i}-5 \mathbf{j}\) and \(\mathbf{w}=-2 \mathbf{i}+3 \mathbf{j}\) \(\|\mathbf{v}\|+\|\mathbf{w}\|\)

Short answer

\( \sqrt{34} + \sqrt{13} \)

Step-by-step solution

- Find the magnitude of vector \mathbf{v}

First, calculate the magnitude of the vector \( \mathbf{v} = 3 \mathbf{i} - 5 \mathbf{j} \). The magnitude is given by \( \| \mathbf{v} \| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \).

- Find the magnitude of vector \mathbf{w}

Next, calculate the magnitude of the vector \( \mathbf{w} = -2 \mathbf{i} + 3 \mathbf{j} \). The magnitude is given by \( \| \mathbf{w} \| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \).

- Add the magnitudes

Finally, add the magnitudes of the two vectors: \( \| \mathbf{v} \| + \| \mathbf{w} \| = \sqrt{34} + \sqrt{13} \).

Key concepts

Vector Operations

Vectors are essential tools in physics and engineering. They help us represent physical quantities having both magnitude and direction. Operations on vectors include addition, subtraction, and multiplication. These operations are easy to perform, making vectors a versatile mathematical tool.

• Addition: Simply add the corresponding components of the vectors. For example, given two vectors \(\textbf{a} = a_1 \textbf{i} + a_2 \textbf{j}\) and \( \textbf{b} = b_1 \textbf{i} + b_2 \textbf{j} \), the resultant vector is \( \textbf{r} = (a_1 + b_1) \textbf{i} + (a_2 + b_2) \textbf{j} \).
• Subtraction: Similar to addition, but you subtract corresponding components. For vectors \( \textbf{a} \textbf{b}\), the difference is \( \textbf{r} = (a_1 - b_1)\textbf{i} + (a_2 - b_2)\textbf{j} \).

Mastering these operations makes solving complex problems more manageable.

Magnitude of Vectors

Understanding the magnitude of a vector is crucial, as it indicates the vector's length. For a 2D vector \( \textbf{v} = v_1 \textbf{i} + v_2 \textbf{j} \), the magnitude is calculated using the Pythagorean theorem. The formula to find the magnitude \( \textbar \textbf{v} \textbar \) is:
\[ \textbar \textbf{v} \textbar = \sqrt{v_1^2 + v_2^2} \]\

• Example: For the vector given in the exercise \( \textbf{v}=3 \textbf{i}-5 \textbf{j} \), the magnitude is \[ \textbar \textbf{v} \textbar = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \].

Calculating the magnitude helps us understand the length of the vector. It's a fundamental step in many vector-related problems.

Pythagorean Theorem

The Pythagorean theorem is a cornerstone of geometry, establishing a relationship between the sides of a right triangle. It states that in a right triangle, the square of the hypotenuse length (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Mathematically, it is expressed as:

\[ c^2 = a^2 + b^2 \]

For vectors, the theorem helps find the magnitude. When dealing with a vector \( \textbf{v} = a \textbf{i} + b \textbf{j} \), we form a right triangle with components \( a \) and \( b \). Applying the Pythagorean theorem, the magnitude (hypotenuse) \( \textbar \textbf{v} \textbar \) is:

\[ \textbar \textbf{v} \textbar = \sqrt{a^2 + b^2} \]

• In practice, this enables us to calculate the length of any vector once its components are known. The exercise does this for both \( \textbf{v} \) and \( \textbf{w} \), giving us substantial information about their sizes using simple algebraic steps.

The theorem isn't just for triangles; it's a critical tool for vector analysis.