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Hindi
3-1.Vectors
medium

બે સદિશો $\overrightarrow A = 2\hat i + 4\hat j + 4\hat k$ અને $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ વચ્ચેનો ખૂણો ....... $^o$ મેળવો.

A

$0$

B

$45$

C

$60$

D

$90$

Solution

$\cos \theta = \frac{{\overrightarrow {\rm A} \,.\,\overrightarrow B }}{{|\overrightarrow {\rm A} \,|\,.\,|\,\overrightarrow B |}} = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{|\overrightarrow {\rm A} \,|\,.\,|\overrightarrow B |\,}}$$ = \frac{{2 \times 4 + 4 \times 2 – 4 \times 4}}{{|\overrightarrow A |\,.\,|\overrightarrow B |}} = 0$

$\theta = {\cos ^{ – 1}}(0^\circ )$ $ \Rightarrow \,\,\,\theta = 90^\circ $

Standard 11
Physics

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