Problem 48
Question
Find each quantity if \(\mathbf{v}=3 \mathbf{i}-5 \mathbf{j}\) and \(\mathbf{w}=-2 \mathbf{i}+3 \mathbf{j}\) \(\|\mathbf{v}\|+\|\mathbf{w}\|\)
Step-by-Step Solution
Verified Answer
\( \sqrt{34} + \sqrt{13} \)
1Step 1 - 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 \( \| \mathbf{v} \| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \).
2Step 2 - 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 \( \| \mathbf{w} \| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \).
3Step 3 - Add the magnitudes
Finally, add the magnitudes of the two vectors: \( \| \mathbf{v} \| + \| \mathbf{w} \| = \sqrt{34} + \sqrt{13} \).
Key Concepts
Vector OperationsMagnitude of VectorsPythagorean Theorem
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.
Mastering these operations makes solving complex problems more manageable.
- 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:
\[ \textbar \textbf{v} \textbar = \sqrt{v_1^2 + v_2^2} \]\
Calculating the magnitude helps us understand the length of the vector. It's a fundamental step in many vector-related problems.
\[ \textbar \textbf{v} \textbar = \sqrt{v_1^2 + v_2^2} \]\
- 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:
\[ \textbar \textbf{v} \textbar = \sqrt{a^2 + b^2} \]
The theorem isn't just for triangles; it's a critical tool for vector analysis.
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.
Other exercises in this chapter
Problem 47
Identify and graph each polar equation. $$ r=1+2 \sin \theta $$
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Polar coordinates of a point are given. Find the rectangular coordinates of each point. $$ \left(5, \frac{5 \pi}{3}\right) $$
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Write each expression in rectangular form \(x+\) yi and in exponential form \(r e^{i \theta} .\) $$ \left[\sqrt{2}\left(\cos \frac{5 \pi}{16}+i \sin \frac{5 \pi
View solution Problem 48
Identify and graph each polar equation. $$ r=1-2 \sin \theta $$
View solution