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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra

If ${\mid\overrightarrow{a}\times\overrightarrow{b}\mid}^2$=${\mid\overrightarrow{a}.\overrightarrow{b}\mid}^2=144$ and$\mid\overrightarrow{a}\mid=4,$ then $\mid\overrightarrow{b}\mid$ is equal to ________.

$\begin{array}{1 1} 3 \\ 3 \sqrt 2 \\ 16 \\ 9 \end{array} $

1 Answer

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  • $(\overrightarrow a \times \overrightarrow b)^2+(\overrightarrow a . \overrightarrow b)^2=|\overrightarrow a|^2 |\overrightarrow b|^2$
$(\overrightarrow a \times \overrightarrow b)^2+(\overrightarrow a . \overrightarrow b)^2=144$
$|a|=4$
We know $(\overrightarrow a \times \overrightarrow b)^2+(\overrightarrow a . \overrightarrow b)^2=|\overrightarrow a|^2 |\overrightarrow b|^2$
Now substituting the values we get,
$144=16 \times |\overrightarrow b|^2$
Therefore $|\overrightarrow b|^2=\large\frac{144}{16}$$=9$
Therefore $|\overrightarrow b|=3$
answered May 30, 2013 by meena.p
 

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