Math Problem Statement

If \(\vec{v} = (1, -2, 2, 4)\), find \(|\vec{v}|\).

Solution

To find v|\vec{v}| for the vector v=(1,2,2,4)\vec{v} = (1, -2, 2, 4), we calculate the magnitude of v\vec{v} using the formula:

v=x12+x22+x32+x42|\vec{v}| = \sqrt{x_1^2 + x_2^2 + x_3^2 + x_4^2}

where x1=1x_1 = 1, x2=2x_2 = -2, x3=2x_3 = 2, and x4=4x_4 = 4.

Calculating step-by-step:

  1. x12=12=1x_1^2 = 1^2 = 1
  2. x22=(2)2=4x_2^2 = (-2)^2 = 4
  3. x32=22=4x_3^2 = 2^2 = 4
  4. x42=42=16x_4^2 = 4^2 = 16

Summing these values:

v=1+4+4+16=25=5|\vec{v}| = \sqrt{1 + 4 + 4 + 16} = \sqrt{25} = 5

Thus, v=5|\vec{v}| = 5.

Would you like further details on the process or have any questions?

Here are some related questions that could expand your understanding:

  1. How do you find the magnitude of a vector in three-dimensional space?
  2. What are some real-world applications of finding the magnitude of a vector?
  3. How does the concept of vector magnitude apply in physics?
  4. Can vectors have negative magnitudes, and why or why not?
  5. How would the magnitude change if one of the vector components was zero?

Tip: Remember that the magnitude of a vector is always a non-negative value.

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Math Problem Analysis

Mathematical Concepts

Vector Magnitude
Euclidean Distance
Linear Algebra

Formulas

Magnitude of a vector formula: |\vec{v}| = \sqrt{x_1^2 + x_2^2 + x_3^2 + x_4^2}

Theorems

Pythagorean Theorem (for finding distances in multidimensional space)

Suitable Grade Level

Grade 10-12