Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To prove that [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] are orthogonal given the condition [tex]\(|\vec{a} - 3 \vec{b}| = |\vec{a} + 3 \vec{b}|\)[/tex], we can follow these steps:
1. Square both sides of the equation:
Given:
[tex]\[ |\vec{a} - 3 \vec{b}| = |\vec{a} + 3 \vec{b}| \][/tex]
Squaring both sides to eliminate the absolute value:
[tex]\[ (\vec{a} - 3 \vec{b}) \cdot (\vec{a} - 3 \vec{b}) = (\vec{a} + 3 \vec{b}) \cdot (\vec{a} + 3 \vec{b}) \][/tex]
2. Expand both sides using the dot product property:
For the left side:
[tex]\[ (\vec{a} - 3 \vec{b}) \cdot (\vec{a} - 3 \vec{b}) = \vec{a} \cdot \vec{a} - 2 (\vec{a} \cdot 3 \vec{b}) + (3 \vec{b}) \cdot (3 \vec{b}) \][/tex]
Simplifying further:
[tex]\[ \vec{a} \cdot \vec{a} - 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
For the right side:
[tex]\[ (\vec{a} + 3 \vec{b}) \cdot (\vec{a} + 3 \vec{b}) = \vec{a} \cdot \vec{a} + 2 (\vec{a} \cdot 3 \vec{b}) + (3 \vec{b}) \cdot (3 \vec{b}) \][/tex]
Simplifying further:
[tex]\[ \vec{a} \cdot \vec{a} + 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
3. Set the expanded expressions equal and cancel common terms:
Equate the two expanded sides:
[tex]\[ \vec{a} \cdot \vec{a} - 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) = \vec{a} \cdot \vec{a} + 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
Simplifying the equation by cancelling out the common terms [tex]\(\vec{a} \cdot \vec{a}\)[/tex] and [tex]\(9 (\vec{b} \cdot \vec{b})\)[/tex] on both sides:
[tex]\[ -6 (\vec{a} \cdot \vec{b}) = 6 (\vec{a} \cdot \vec{b}) \][/tex]
4. Solve for the dot product [tex]\(\vec{a} \cdot \vec{b}\)[/tex]:
Add [tex]\(6 (\vec{a} \cdot \vec{b})\)[/tex] to both sides to isolate 0:
[tex]\[ -6 (\vec{a} \cdot \vec{b}) + 6 (\vec{a} \cdot \vec{b}) = 6 (\vec{a} \cdot \vec{b}) + 6 (\vec{a} \cdot \vec{b}) \][/tex]
Simplifies to:
[tex]\[ 0 = 12 (\vec{a} \cdot \vec{b}) \][/tex]
Thus:
[tex]\[ \vec{a} \cdot \vec{b} = 0 \][/tex]
5. Conclusion:
The dot product [tex]\(\vec{a} \cdot \vec{b} = 0\)[/tex] implies that [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] are orthogonal vectors.
Therefore, we have proven that [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] are orthogonal.
1. Square both sides of the equation:
Given:
[tex]\[ |\vec{a} - 3 \vec{b}| = |\vec{a} + 3 \vec{b}| \][/tex]
Squaring both sides to eliminate the absolute value:
[tex]\[ (\vec{a} - 3 \vec{b}) \cdot (\vec{a} - 3 \vec{b}) = (\vec{a} + 3 \vec{b}) \cdot (\vec{a} + 3 \vec{b}) \][/tex]
2. Expand both sides using the dot product property:
For the left side:
[tex]\[ (\vec{a} - 3 \vec{b}) \cdot (\vec{a} - 3 \vec{b}) = \vec{a} \cdot \vec{a} - 2 (\vec{a} \cdot 3 \vec{b}) + (3 \vec{b}) \cdot (3 \vec{b}) \][/tex]
Simplifying further:
[tex]\[ \vec{a} \cdot \vec{a} - 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
For the right side:
[tex]\[ (\vec{a} + 3 \vec{b}) \cdot (\vec{a} + 3 \vec{b}) = \vec{a} \cdot \vec{a} + 2 (\vec{a} \cdot 3 \vec{b}) + (3 \vec{b}) \cdot (3 \vec{b}) \][/tex]
Simplifying further:
[tex]\[ \vec{a} \cdot \vec{a} + 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
3. Set the expanded expressions equal and cancel common terms:
Equate the two expanded sides:
[tex]\[ \vec{a} \cdot \vec{a} - 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) = \vec{a} \cdot \vec{a} + 6 (\vec{a} \cdot \vec{b}) + 9 (\vec{b} \cdot \vec{b}) \][/tex]
Simplifying the equation by cancelling out the common terms [tex]\(\vec{a} \cdot \vec{a}\)[/tex] and [tex]\(9 (\vec{b} \cdot \vec{b})\)[/tex] on both sides:
[tex]\[ -6 (\vec{a} \cdot \vec{b}) = 6 (\vec{a} \cdot \vec{b}) \][/tex]
4. Solve for the dot product [tex]\(\vec{a} \cdot \vec{b}\)[/tex]:
Add [tex]\(6 (\vec{a} \cdot \vec{b})\)[/tex] to both sides to isolate 0:
[tex]\[ -6 (\vec{a} \cdot \vec{b}) + 6 (\vec{a} \cdot \vec{b}) = 6 (\vec{a} \cdot \vec{b}) + 6 (\vec{a} \cdot \vec{b}) \][/tex]
Simplifies to:
[tex]\[ 0 = 12 (\vec{a} \cdot \vec{b}) \][/tex]
Thus:
[tex]\[ \vec{a} \cdot \vec{b} = 0 \][/tex]
5. Conclusion:
The dot product [tex]\(\vec{a} \cdot \vec{b} = 0\)[/tex] implies that [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] are orthogonal vectors.
Therefore, we have proven that [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] are orthogonal.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
