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A vector $\overrightarrow{r}$ is inclined at equal angles to the three axes.If the magnitude of $\overrightarrow{r}$ is $2\sqrt 3$ units,find $\overrightarrow{r}.$

$\begin{array}{1 1} (A)\;2\hat i-2\hat j+2\hat k \\ (B)\;2(\pm \hat i\pm \hat j\pm \hat k) \\(C)\;4(\pm \hat i\pm \hat j\pm \hat k) \\ (D)\;2(\pm \hat i+ \hat j+ \hat k)\end{array} $

1 Answer

  • $\overrightarrow{r}=|\overrightarrow{r}|(l \hat i+m \hat j+ n \hat k)$
  • Sum of the square of the direction cosines is one
Given $ |\overrightarrow r|=2 \sqrt 3 units $
Also $ |\overrightarrow r| $ is equally inclined with $OX,Oy$ and $OZ$
Hence this direction ratios are equal
Let the direction ratios be l,m and n
Since the direction ration are equal
$ l=m=n$
=>$ \cos \alpha=\cos \beta=\cos \gamma$
We know tha sum of squares of the dirction cosines is one
Therefore $l^2+m^2+n^2=1$
$=>3l^2=1$ therefore $ l=\pm \large\frac{1}{\sqrt 3}$
Hence the direction cosines are $\pm \large\frac{1}{\sqrt 3},$$\pm \large\frac{1}{\sqrt 3},$$\pm \large\frac{1}{\sqrt 3}$
It is given $ |\overrightarrow r|=2 \sqrt 3 units $
Therefore $ \overrightarrow r=|\overrightarrow r| (l \hat i+m \hat j+n \hat k)$
$=2 \sqrt {3}\bigg(\pm \large\frac{1}{\sqrt 3} \hat i$$\pm \large\frac{1}{\sqrt 3} \hat j$$\pm \large\frac{1}{\sqrt 3}\hat k\bigg)$
$=(\pm 2 \hat i\pm2 \hat j\pm 2 \hat k)$
$=2(\pm \hat i\pm \hat j\pm \hat k)$
answered May 27, 2013 by meena.p

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