Question:
If \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \) and \( |\vec{a}| = 10 |\vec{b}| \), then :
If \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \) and \( |\vec{a}| = 10 |\vec{b}| \), then :
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The vector identity \( (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 \) works identically to the algebraic difference of squares formula \( (x+y)(x-y) = x^2 - y^2 \).
Updated On: Jul 4, 2026
- \( |\vec{a}| = \sqrt{2} \)
- \( |\vec{b}| = \sqrt{2} \)
- \( |\vec{b}| = 10\sqrt{2} \)
- \( |\vec{a}| = \frac{10}{\sqrt{2}} \)
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The Correct Option is B
Solution and Explanation
Concept:
This problem uses the distributive property of the dot product over vector addition and subtraction, along with the magnitude relationship \( \vec{x} \cdot \vec{x} = |\vec{x}|^2 \).
Step 1: Expand the dot product equation.
We are given: \[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \] Expanding using the distributive law: \[ \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 198 \] Since the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)), the middle terms cancel out: \[ |\vec{a}|^2 - |\vec{b}|^2 = 198 \]
Step 2: Substitute the given magnitude relationship.
We are given that \( |\vec{a}| = 10 |\vec{b}| \). Substituting this into our simplified equation: \[ (10 |\vec{b}|)^2 - |\vec{b}|^2 = 198 \] \[ 100 |\vec{b}|^2 - |\vec{b}|^2 = 198 \] \[ 99 |\vec{b}|^2 = 198 \]
Step 3: Solve for \( |\vec{b}| \) and \( |\vec{a}| \).
Divide both sides by 99: \[ |\vec{b}|^2 = \frac{198}{99} = 2 \] Taking the positive square root (since magnitude is always non-negative): \[ |\vec{b}| = \sqrt{2} \] This matches option (B).
Step 1: Expand the dot product equation.
We are given: \[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \] Expanding using the distributive law: \[ \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 198 \] Since the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)), the middle terms cancel out: \[ |\vec{a}|^2 - |\vec{b}|^2 = 198 \]
Step 2: Substitute the given magnitude relationship.
We are given that \( |\vec{a}| = 10 |\vec{b}| \). Substituting this into our simplified equation: \[ (10 |\vec{b}|)^2 - |\vec{b}|^2 = 198 \] \[ 100 |\vec{b}|^2 - |\vec{b}|^2 = 198 \] \[ 99 |\vec{b}|^2 = 198 \]
Step 3: Solve for \( |\vec{b}| \) and \( |\vec{a}| \).
Divide both sides by 99: \[ |\vec{b}|^2 = \frac{198}{99} = 2 \] Taking the positive square root (since magnitude is always non-negative): \[ |\vec{b}| = \sqrt{2} \] This matches option (B).
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