Let a = -27 +53 + 4k. Find a.

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Vector Problem Example and Solution**

**Problem Statement:**

Let \(\vec{a} = -2\vec{i} + 5\vec{j} + 4\vec{k}\). Find \(|\vec{a}|\).

**Solution:**

To find the magnitude of the vector \(\vec{a}\), we use the formula for the magnitude of a vector in three-dimensional space:

\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]

Here, \(a_x\), \(a_y\), and \(a_z\) are the components of the vector \(\vec{a}\).

Given \( \vec{a} = -2\vec{i} + 5\vec{j} + 4\vec{k} \), we have:
- \(a_x = -2\)
- \(a_y = 5\)
- \(a_z = 4\)

Now, we can calculate the magnitude:

\[ |\vec{a}| = \sqrt{(-2)^2 + 5^2 + 4^2} \]
\[ |\vec{a}| = \sqrt{4 + 25 + 16} \]
\[ |\vec{a}| = \sqrt{45} \]
\[ |\vec{a}| = 3\sqrt{5} \]

Therefore, the magnitude of \(\vec{a}\) is \( 3\sqrt{5} \).
Transcribed Image Text:**Vector Problem Example and Solution** **Problem Statement:** Let \(\vec{a} = -2\vec{i} + 5\vec{j} + 4\vec{k}\). Find \(|\vec{a}|\). **Solution:** To find the magnitude of the vector \(\vec{a}\), we use the formula for the magnitude of a vector in three-dimensional space: \[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \] Here, \(a_x\), \(a_y\), and \(a_z\) are the components of the vector \(\vec{a}\). Given \( \vec{a} = -2\vec{i} + 5\vec{j} + 4\vec{k} \), we have: - \(a_x = -2\) - \(a_y = 5\) - \(a_z = 4\) Now, we can calculate the magnitude: \[ |\vec{a}| = \sqrt{(-2)^2 + 5^2 + 4^2} \] \[ |\vec{a}| = \sqrt{4 + 25 + 16} \] \[ |\vec{a}| = \sqrt{45} \] \[ |\vec{a}| = 3\sqrt{5} \] Therefore, the magnitude of \(\vec{a}\) is \( 3\sqrt{5} \).
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