Problem 21
Question
Use the dot product to compute the length of \([1,2,3,4]\) '.
Step-by-Step Solution
Verified Answer
The length of the vector is \( \sqrt{30} \).
1Step 1: Understand the Vector Length Formula
The length (or magnitude) of a vector \( \mathbf{v} = [v_1, v_2, v_3, v_4] \) is calculated using the formula: \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2} \). Importantly, this formula can also be derived using the dot product, where the dot product of \( \mathbf{v} \) with itself is equal to the squared length of the vector.
2Step 2: Compute the Dot Product
Compute the dot product of \( \mathbf{v} \) with itself: \( \mathbf{v} \cdot \mathbf{v} = 1^2 + 2^2 + 3^2 + 4^2 \). Compute each term: \( 1^2 = 1 \), \( 2^2 = 4 \), \( 3^2 = 9 \), \( 4^2 = 16 \). Thus, \( \mathbf{v} \cdot \mathbf{v} = 1 + 4 + 9 + 16 = 30 \).
3Step 3: Calculate the Length of the Vector
The length of the vector \( \mathbf{v} \) is the square root of the dot product we calculated. Therefore: \( \|\mathbf{v}\| = \sqrt{30} \).
4Step 4: Simplification
\( \sqrt{30} \) is the most simplified form as \( 30 \) is not a perfect square. This is the final length of the vector.
Key Concepts
Vector LengthMagnitudeVector Algebra
Vector Length
When we talk about the length of a vector, we're essentially referring to how long the vector is when represented in space. This concept is mathematically described as the "magnitude" of the vector. For a vector \( \mathbf{v} = [v_1, v_2, v_3, v_4] \), the vector length is determined using the formula:
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2} \]
This formula arises from the Pythagorean theorem extended to multiple dimensions. It gives a single positive number that represents the vector’s size, irrespective of its direction.
To compute this using the dot product, you take the dot product of the vector with itself:
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2} \]
This formula arises from the Pythagorean theorem extended to multiple dimensions. It gives a single positive number that represents the vector’s size, irrespective of its direction.
To compute this using the dot product, you take the dot product of the vector with itself:
- The dot product \( \mathbf{v} \cdot \mathbf{v} \) equals \( v_1^2 + v_2^2 + v_3^2 + v_4^2 \).
- Then, the vector length is the square root of this sum.
Magnitude
Magnitude in the context of vectors is another term for a vector's length. It provides insight into the "size" or "extent" of the vector without considering its direction. Calculating the magnitude of a vector involves using the same formula as for vector length:
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2} \]
This value is always non-negative, as it is the result of a square root applied to a sum of squared terms, which are themselves always non-negative.
The magnitude is extremely useful because it allows us to
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2} \]
This value is always non-negative, as it is the result of a square root applied to a sum of squared terms, which are themselves always non-negative.
The magnitude is extremely useful because it allows us to
- Assess the vector's contribution in various applications, from physics to computer graphics.
- Compare vectors to determine which is larger or more significant in a given context.
Vector Algebra
Vector algebra refers to the operations that can be performed on vectors, making it a fundamental part of vector mathematics and physics. Important aspects of vector algebra include dot products and the computation of vector lengths.
The dot product, an essential operation in vector algebra, involves multiplying corresponding components of two vectors and summing up these products:
The dot product, an essential operation in vector algebra, involves multiplying corresponding components of two vectors and summing up these products:
- For vectors \( \mathbf{a} = [a_1, a_2, a_3, a_4] \) and \( \mathbf{b} = [b_1, b_2, b_3, b_4] \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is \( a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 \).
- It measures the cosine of the angle between the two vectors when both are normalized.
Other exercises in this chapter
Problem 20
Suppose that \(A\) is an \(m \times k\) matrix and \(B\) is a \(k \times n\) matrix. Show that $$ (A B)^{\prime}=B^{\prime} A^{\prime} $$
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