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3-1.Vectors
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$ \hat i + 2\hat j + 3\hat k $ અને $ 3\hat i - 2\hat j + \hat k $ થી બનતા સમાંતરબાજુ ચતુષ્કોણનું ક્ષેત્રફળ કેટલું થશે?
A
$8$
B
$ 8\sqrt 3 $
C
$ 3\sqrt 8 $
D
$192$
Solution
(b)Area of parallelogram $ = \overrightarrow A \times \overrightarrow B $
$ = (\hat i + 2\hat j + 3\hat k) \times 3\hat i – 2\hat j + \hat k)$
$ = \left|{\,\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\1&2&3\\3&{ – 2}&1\end{array}\,} \right|$
$ = (8)\hat i + (8)\hat j – (8)\hat k$
Magnitude $ = \sqrt {64 + 64 + 64} $=$8\sqrt 3 $
Standard 11
Physics
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