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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find \(| a × b |\), if \(a = \hat i - 7\hat j + 7\hat k\) and \(b = 3\hat i - 2\hat j + 2\hat k\)

$\begin{array}{1 1} \sqrt {405} \\ 38 \\ 2 \sqrt{19} \\ 19 \sqrt 2 \end{array} $

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1 Answer

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Toolbox:
  • $ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $
  • Where $a=a_1\hat i+a_2\hat j+a_3\hat k$ and $b=b_1\hat i+b_2\hat j+b_3\hat k$
  • (i.e)$\overrightarrow a \times\overrightarrow b=\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1)$
Step 1:
Let $\overrightarrow a=\hat i-7\hat j+7\hat k$ and $\overrightarrow b=3\hat i-2\hat j+2\hat k$
We know $\overrightarrow a\times \overrightarrow b=\begin{vmatrix} \hat i & \hat j & \hat k\\1 & -7 & 7\\3 &-2 &2\end{vmatrix}$
Step 2:
On expanding along $R_1$ we get
$\qquad\qquad\qquad=\hat i(-7\times 2-7\times -2)-\hat j(1\times 2-7\times 3)+\hat k(1\times -2-3\times-7)$
$\qquad\qquad\qquad=\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)$
$\qquad\qquad\qquad=\hat i(0)-\hat j(-19)+\hat k(19)$
$\qquad\qquad\qquad=19\hat j+19\hat k$
Step 3:
$|\overrightarrow a\times\overrightarrow b|=\sqrt{19^2+19^2}$
$\qquad\;\;\;\;\;=19\sqrt 2 \approx 26.87$
answered May 22, 2013 by sreemathi.v
edited Feb 20, 2014 by balaji.thirumalai
 

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