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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the unit vector in the direction of the vector \( \overrightarrow a = \hat i + \hat j + 2\hat k\) .

$\begin{array}{1 1}(A) \large\frac{1}{6}\hat i+\large\frac{1}{6}\hat j+\large\frac{1}{6}\hat k \\(B) 6\hat i+6\hat j+12 \hat k \\(C) \large\frac{1}{\sqrt 6}\hat i+\large\frac{1}{\sqrt 6}\hat j+\large\frac{2}{\sqrt 6}\hat k \\ (D) \sqrt 6\hat i+\sqrt 6\hat j+\sqrt 6 \hat k \end{array} $

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1 Answer

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Toolbox:
  • Unit vector $\hat a$ : Let $\lambda=\large\frac{1}{|\overrightarrow{a}|}$$,|\lambda\overrightarrow a|=|\lambda||\overrightarrow{a}|$
  • $|\lambda\overrightarrow{a}|=\large\frac{1}{|\overrightarrow{a}|}$$\times |\overrightarrow{a}|=1$
  • Therefore $\large\frac{1}{|\overrightarrow{a}|}$$\times |\overrightarrow{a}|=\hat a$
  • Where $|\overrightarrow a|\neq 0$ and $\hat a$ is the unit vector.
Step 1:
Let $\overrightarrow a=\hat i+\hat j+2\hat k$
$|\overrightarrow{a}|=\sqrt{1^2+1^2+2^2}$=$\sqrt 6$
Hence the magnitude of $\overrightarrow a=\sqrt 6$
Step 2:
Now the unit vector in the direction of $\overrightarrow{a}=\large\frac{1}{|\overrightarrow a|}$$\overrightarrow a$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\large\frac{1}{\sqrt 6}$$(\hat i+\hat j+2\hat k)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\large\frac{1}{\sqrt 6}$$\hat i+\large\frac{1}{\sqrt 6}$$\hat j+\large\frac{2}{\sqrt 6}$$\hat k$
Step 3:
Hence the unit vector in the direction of $\overrightarrow{a}$ is $\large\frac{1}{\sqrt 6}$$\hat i+\large\frac{1}{\sqrt 6}$$\hat j+\large\frac{2}{\sqrt 6}$$\hat k$
answered May 17, 2013 by sreemathi.v
 

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