If two vectors and having equal magnitude are inclined at an angle , then

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Q.
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $θ$ , then [2024]

1 $| \vec{A} - \vec{B} |$ $= \sqrt{2} R \sin (\frac{θ}{2})$

2 $| \vec{A} + \vec{B} |$ $= 2 R \sin (\frac{θ}{2})$

3 $| \vec{A} + \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

4 $| \vec{A} - \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

Ans.
(3)

The magnitude of resultant vector

$R^{'} = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$

Here $a = b = R$

Then $R^{'} = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos θ}$

$= R \sqrt{2} \sqrt{1 + \cos θ} = \sqrt{2} R \sqrt{2 \cos^{2} \frac{θ}{2}} = 2 R \cos \frac{θ}{2}$

Solution

Q.
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $θ$ , then [2024]

1 $| \vec{A} - \vec{B} |$ $= \sqrt{2} R \sin (\frac{θ}{2})$

2 $| \vec{A} + \vec{B} |$ $= 2 R \sin (\frac{θ}{2})$

3 $| \vec{A} + \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

4 $| \vec{A} - \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

Ans.
(3)

The magnitude of resultant vector

$R^{'} = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$

Here $a = b = R$

Then $R^{'} = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos θ}$

$= R \sqrt{2} \sqrt{1 + \cos θ} = \sqrt{2} R \sqrt{2 \cos^{2} \frac{θ}{2}} = 2 R \cos \frac{θ}{2}$

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Solution

Q. If two vectors A→ and B→ having equal magnitude R are inclined at an angle θ, then [2024]

1 |A→-B→|=2Rsin(θ2)

2 |A→+B→|=2Rsin(θ2)

3 |A→+B→|=2Rcos(θ2)

4 |A→-B→|=2Rcos(θ2)

Ans. (3) The magnitude of resultant vector R'=a2+b2+2abcosθ Here a=b=R Then R'=R2+R2+2R2 cosθ =R21+cosθ=2R2cos2θ2=2Rcosθ2

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Q.
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $θ$ , then [2024]

1 $| \vec{A} - \vec{B} |$ $= \sqrt{2} R \sin (\frac{θ}{2})$

2 $| \vec{A} + \vec{B} |$ $= 2 R \sin (\frac{θ}{2})$

3 $| \vec{A} + \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

4 $| \vec{A} - \vec{B} |$ $= 2 R \cos (\frac{θ}{2})$

Ans.
(3)

The magnitude of resultant vector

$R^{'} = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$

Here $a = b = R$

Then $R^{'} = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos θ}$

$= R \sqrt{2} \sqrt{1 + \cos θ} = \sqrt{2} R \sqrt{2 \cos^{2} \frac{θ}{2}} = 2 R \cos \frac{θ}{2}$