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Find direction Vector a that has the same as (-6,5,67 but has length 4

Find direction Vector a that has the same as (-6,5,67 but has length 4

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
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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Can anyone please help me to solve this problem? I am stuck please help me!
The image contains handwritten notes on finding a vector. Here is the transcription: --- **Objective:** Find a vector **a** that has the same direction as \(\langle -6, 6, 6 \rangle\) but has length 4. --- **Solution Steps:** 1. **Find the Unit Vector:** Calculate the magnitude of the vector \(\langle -6, 6, 6 \rangle\). Magnitude \( = \sqrt{(-6)^2 + 6^2 + 6^2}\) Solve to find: Magnitude \( = \sqrt{36 + 36 + 36}\) Magnitude \( = \sqrt{108}\) Magnitude \( = 6\sqrt{3}\) 2. **Normalize the Vector:** Divide each component of the vector by its magnitude to find the unit vector in the same direction. Unit Vector \( = \frac{1}{6\sqrt{3}}\langle -6, 6, 6 \rangle\) Simplify to: Unit Vector \( = \langle -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle\) 3. **Scale the Unit Vector to Length 4:** Multiply the unit vector by 4 to obtain a vector of the desired length. Vector **a** \( = 4 \times \langle -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle\) Resulting Vector **a** \( = \langle -\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \rangle\) 4. **Final Answer:** The vector **a** with length 4 in the same direction as \(\langle -6, 6, 6 \rangle\) is: Vector **a** \( = \langle -\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \rangle\) ---
Transcribed Image Text:The image contains handwritten notes on finding a vector. Here is the transcription: --- **Objective:** Find a vector **a** that has the same direction as \(\langle -6, 6, 6 \rangle\) but has length 4. --- **Solution Steps:** 1. **Find the Unit Vector:** Calculate the magnitude of the vector \(\langle -6, 6, 6 \rangle\). Magnitude \( = \sqrt{(-6)^2 + 6^2 + 6^2}\) Solve to find: Magnitude \( = \sqrt{36 + 36 + 36}\) Magnitude \( = \sqrt{108}\) Magnitude \( = 6\sqrt{3}\) 2. **Normalize the Vector:** Divide each component of the vector by its magnitude to find the unit vector in the same direction. Unit Vector \( = \frac{1}{6\sqrt{3}}\langle -6, 6, 6 \rangle\) Simplify to: Unit Vector \( = \langle -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle\) 3. **Scale the Unit Vector to Length 4:** Multiply the unit vector by 4 to obtain a vector of the desired length. Vector **a** \( = 4 \times \langle -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle\) Resulting Vector **a** \( = \langle -\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \rangle\) 4. **Final Answer:** The vector **a** with length 4 in the same direction as \(\langle -6, 6, 6 \rangle\) is: Vector **a** \( = \langle -\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \rangle\) ---
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