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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Compute the magnitude of the following vector $\large\frac{1}{\sqrt 3}\hat i + \frac{1}{\sqrt 3}\hat j - \frac{1}{\sqrt 3}\hat k. $

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Toolbox:
  • The distance between the initial point and the terminal point of a vector is the magnitude (or length) of the vector $\overrightarrow{AB}$.It is denoted by $\mid\overrightarrow{AB}\mid$ .
  • $\mid\overrightarrow{AB}\mid=\sqrt{a_1^2+a_2^2+a_3^2}$
  • Where $\overrightarrow{AB}=a_1\hat i+a_2\hat j+a_3\hat k.$
Step 1:
$\overrightarrow a = \large\frac{1}{\sqrt 3}$$\hat i +\large \frac{1}{\sqrt 3}$$\hat j - \large\frac{1}{\sqrt 3}$$\hat k$
Here $a_1=\large\frac{1}{\sqrt{3}}$$,a_2=\large\frac{1}{\sqrt{3}}$ and $a_3=-\large\frac{1}{\sqrt{3}}$
$\mid \overrightarrow{a}\mid=\sqrt{a_1^2+a_2^2+a_3^2}$
Step 2:
Hence $\mid\overrightarrow{a}\mid=\sqrt{\big(\large\frac{1}{\sqrt{3}}\big)^2+\big(\large\frac{1}{\sqrt{3}}\big)^2+\big(-\large\frac{1}{\sqrt{3}}\big)^2}$
$\qquad\qquad\;\;=\sqrt{\large\frac{1}{3}+\large\frac{1}{3}+\large\frac{1}{3}}$
$\qquad\qquad\quad=\large\sqrt{\frac{3}{3}}$$=1$
$\mid\overrightarrow{a}\mid=1$
answered Oct 3, 2013 by sreemathi.v
 

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