Question

Two vectors A and B have magnitudes 2 units and 4 units respectively. Find A. B is angle between these two vectors is (a) 0^(@) (b) 60^(@) (c) 90^(@) (d) 120^(@) .

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

As we know that :

$\large{\boxed{\boxed{\sf{\vec{A} . \vec{B} \: = \: |\vec{A}| |\vec{B}| \cos \theta}}}}$

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➠ Put θ = 0°

$\implies {\sf{\vec{A}. \vec{B} \: = \: |2| \: \times \: |4| \: \times \: \cos 0^{\circ}}} \\ \\ \implies {\sf{\vec{A} . \vec{B} \: = \: 8 \: \times \: \cos 0^{\circ}}} \\ \\ \implies {\sf{\vec{A} \vec B \: = \: 8 \: \times \: 1}} \\ \\ \implies {\sf{\vec A . \vec B \: = \: 8 \: units}}$

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➠ Put θ = 60°

$\implies {\sf{\vec A . \vec B \: = \: |2| \: |4| \: \times \: \cos 60^{\circ}}} \\ \\ \implies {\sf{\vec A. \vec B \: = \: 8 \: \times \: \dfrac{1}{2}}} \\ \\ \implies {\sf{\vec A . \vec B \: = \: 4 \: units}}$

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➠ Put θ = 90°

$\implies {\sf{\vec A \vec B \: = \: |2| \: |4| \: \times \: \cos 90^{\circ}}} \\ \\ \implies {\sf{\vec A. \vec B \: = \: 8 \: \times \: 0}} \\ \\ \implies {\sf{\vec A. \vec B \: = \: 0 \: units}}$

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➠ Put θ = 120°

$\implies {\sf{\vec A. \vec B \: = \: |2| \: |4| \: \times \: \cos 120^{\circ}}} \\ \\ \implies {\sf{\vec A. \vec B \: = \: 8 \: \times \: \dfrac{-1}{2}}} \\ \\ \implies {\sf{\vec A . \vec B \: = \: -4 \: units}}$

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

We know that,

$\Large{\star{\boxed{\rm{\overrightarrow{A}. \overrightarrow{B} = |\overrightarrow{A}| \: |\overrightarrow{B}| \times Cos\theta}}}}$

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★ Puuting $\bold{\theta}$ as 0°.

$\tt{→\overrightarrow{A}. \overrightarrow{B} = |2| \times |4| \times cos\: 0^{\circ}} \\ \\ \sf{→\overrightarrow{A}. \overrightarrow{B} = 8 \times 1} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = 8 \: units}$

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★ Putting $\bold{\theta}$ as 60°

$\tt{→\overrightarrow{A}. \overrightarrow{B} = |2| \times |4| \times Cos \: 60^{\circ}} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = \cancel{2} \times 4 \times \frac{1}{\cancel{2}}} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = 4 \: units}$

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★ Putting $\theta$ as 90°

$\tt{→|2| \times |4| \times Cos \: 90^{\circ}} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = 2 \times 4 \times 0} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = 0 \: units}$

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★ Putting $\theta$ as 120°.

$\tt{→\overrightarrow{A}. \overrightarrow{B} = |2| \times |4| \times \times Cos \: 120^{\circ}} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = \cancel{2} \times 4 \times \frac{-1}{\cancel{2}}} \\ \\ \tt{→\overrightarrow{A}. \overrightarrow{B} = -4 \: units}$