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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.
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