Question

If P=2i+3j-4k and Q=5i+2j+4k find the angle between the two vectors ​

Answer 1

GIVEN :–

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$\: \: \: \bf \to \overrightarrow{P}=2 \hat{i}+3 \hat{j}-4 \hat{k}$

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$\: \: \: \bf \to \overrightarrow{Q}=5 \hat{i}+2 \hat{j} + 4 \hat{k}$

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TO FIND :–

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• Angle between vectors = ?

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SOLUTION :–

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• We know that –

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$\implies \large{ \boxed{ \bf \cos( \theta) = \dfrac{\overrightarrow{P}. \overrightarrow{Q}}{ | \overrightarrow{P}| | \overrightarrow{Q}| }}}$

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$\bf \implies \cos( \theta) = \dfrac{(2 \hat{i}+3 \hat{j}-4 \hat{k} ). (5 \hat{i}+2 \hat{j} + 4 \hat{k})}{ |2 \hat{i}+3 \hat{j}-4 \hat{k} | |5 \hat{i}+2 \hat{j} + 4 \hat{k}|}$

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$\bf \implies \cos( \theta) = \dfrac{(2 \times 5) + (3 \times 2) - (4 \times 4)}{ \sqrt{{(2)}^{2} + {(3)}^{2} + {( - 4)}^{2} } \sqrt{ {(5)}^{2} + {(2)}^{2} + {(4)}^{2}}}$

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$\bf \implies \cos( \theta) = \dfrac{10+6-16}{ \sqrt{4 +9+16} \sqrt{25+4+4}}$

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$\bf \implies \cos( \theta) = \dfrac{16-16}{ \sqrt{29} \sqrt{33}}$

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$\bf \implies \cos( \theta) = \dfrac{0}{ \sqrt{29} \sqrt{33}}$

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$\bf \implies \cos( \theta) = 0$

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$\bf \implies \cos( \theta) = \cos \left( \dfrac{\pi}{2} \right )$

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$\bf \large \implies{ \boxed{ \bf \theta =\dfrac{\pi}{2} }}$

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• Hence , Angle between vectors is 90°.