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If |vec(a)+vec(b)|ge|vec(a)-vec(b)| then...

If `|vec(a)+vec(b)|ge|vec(a)-vec(b)|` then angle between `vec(a)` and `vec(b)` may be

A

`50^(@)`

B

`90^(@)`

C

`60^(@)`

D

`120^(@)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the inequality given: \[ |\vec{a} + \vec{b}| \geq |\vec{a} - \vec{b}| \] ### Step 1: Square both sides of the inequality Since both sides are non-negative, we can square them without changing the inequality: \[ |\vec{a} + \vec{b}|^2 \geq |\vec{a} - \vec{b}|^2 \] ### Step 2: Expand both sides using the formula for the square of a vector's magnitude Using the property that \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \), we expand both sides: \[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \geq (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \] This gives us: \[ |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 \geq |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 \] ### Step 3: Simplify the inequality Now, we can simplify the inequality by canceling out the common terms: \[ |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 - |\vec{a}|^2 - |\vec{b}|^2 \geq -2\vec{a} \cdot \vec{b} \] This simplifies to: \[ 2\vec{a} \cdot \vec{b} \geq -2\vec{a} \cdot \vec{b} \] ### Step 4: Combine like terms Combining the terms gives us: \[ 4\vec{a} \cdot \vec{b} \geq 0 \] ### Step 5: Divide by 4 Dividing both sides by 4 (which is positive) does not change the inequality: \[ \vec{a} \cdot \vec{b} \geq 0 \] ### Step 6: Relate the dot product to the angle The dot product can be expressed in terms of the angle \( \theta \) between the vectors: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \] Thus, we have: \[ |\vec{a}| |\vec{b}| \cos \theta \geq 0 \] Since \( |\vec{a}| \) and \( |\vec{b}| \) are both positive (assuming non-zero vectors), we can conclude that: \[ \cos \theta \geq 0 \] ### Step 7: Determine the angle range The condition \( \cos \theta \geq 0 \) implies that the angle \( \theta \) between the vectors \( \vec{a} \) and \( \vec{b} \) can be: \[ 0^\circ \leq \theta \leq 90^\circ \] ### Final Conclusion The angle between \( \vec{a} \) and \( \vec{b} \) may be between \( 0^\circ \) and \( 90^\circ \) inclusive. ---

To solve the problem, we need to analyze the inequality given: \[ |\vec{a} + \vec{b}| \geq |\vec{a} - \vec{b}| \] ### Step 1: Square both sides of the inequality ...
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