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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