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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Compute the magnitude of the following vectors: $(iii)\;\overrightarrow c = \large\frac{1}{\sqrt 3}$$\hat i +\large \frac{1}{\sqrt 3}$$\hat j - \large\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 $\mid\overrightarrow{AB}\mid$ or simply $AB$.
  • $\mid\overrightarrow{AB}\mid=\sqrt{a_1^2+a_2^2+a_3^2}$
  • Where $\overrightarrow{AB}=a_1\hat i+a_2\hat j+a_3\hat k.$
Step 1:
$\overrightarrow c = \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 $a_3=-\large\frac{1}{\sqrt{3}}$
$\mid \overrightarrow{c}\mid=\sqrt{a_1^2+a_2^2+a_3^2}$
Step 2:
Hence $\mid\overrightarrow{c}\mid=\sqrt{\big(\large\frac{1}{\sqrt{3}}\big)^2+\big(\large\frac{1}{\sqrt{3}}\big)^2+\big(-\large\frac{1}{\sqrt{3}}\big)^2}$
$\qquad\qquad=\sqrt{\large\frac{1}{3}+\large\frac{1}{3}+\large\frac{1}{3}}=\large\sqrt{\frac{3}{3}}$$=1$
$\mid\overrightarrow{c}\mid=1$
answered May 16, 2013 by sreemathi.v
 

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