The vectors from origin to the points A and B | vedclass

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Question

(a) Given $\overrightarrow {OA} = \overrightarrow a = 3\hat i - 6\hat j + 2\hat k$ and $\overrightarrow {OB} = \overrightarrow b = 2\hat i + \hat j -

Vedclass · Accepted Answer

$\frac{5}{2}\sqrt {17} $ sq.unit

Vedclass · Answer

(a) Given $\overrightarrow {OA} = \overrightarrow a = 3\hat i - 6\hat j + 2\hat k$ and $\overrightarrow {OB} = \overrightarrow b = 2\hat i + \hat j - 2\hat k$

$	herefore \,\,\,(\overrightarrow a 	imes \overrightarrow b )\, $$= \left| {\begin{array}{*{20}{c}}{\hat i\,\,}&{\hat j\,\,}&{\hat k}\{\,3\,\,}&{ -6\,\,\,}&2\{\,\,\,2\,\,\,}&{1\,\,}&{ - 2\,\,\,}\end{array}}ight|\,$

$ = (12 - 2)\hat i + (4 + 6)\hat j + (3 + 12)\hat k$

$ = 10\hat i + 10\hat j + 15\hat k$$\Rightarrow \,\,|\overrightarrow a 	imes \overrightarrow b |\, = \,\sqrt {{{10}^2} + {{10}^2} + {{15}^2}} $

$ =\sqrt {425} $ $ = 5\sqrt {17} $

Area of $\Delta OAB = \frac{1}{2}|\overrightarrow a 	imes \overrightarrow b |\, =\frac{{5\sqrt {17} }}{2}\,$sq.unit.

The vectors from origin to the points $A$ and $B$ are $\overrightarrow A = 3\hat i - 6\hat j + 2\hat k$ and $\overrightarrow B = 2\hat i + \hat j - 2\hat k$ respectively. The area of the triangle $OAB$ be

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