logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

A vector $\overrightarrow{r}$ has magnitude 14 and direction ratios 2,3,-6.Find the direction cosines and components of $\overrightarrow{r}$,given that $\overrightarrow{r}$ makes an acute angle with x-axis.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\overrightarrow{r}=|\overrightarrow{r} \;|(l \hat i+m \hat j+ n \hat k)$
  • The direction cosines of a vector are proportional to a,b,c then its direction cosines are $\large\pm\frac{a}{\sqrt{a^2+b^2+c^2}},\pm\frac{b}{\sqrt{a^2+b^2+c^2}},\pm\frac{c}{\sqrt{a^2+b^2+c^2}}$
The direction cosines of a vector are proportional to a,b,c then its direction cosines are $\large\pm\frac{a}{\sqrt{a^2+b^2+c^2}},\pm\frac{b}{\sqrt{a^2+b^2+c^2}},\pm\frac{c}{\sqrt{a^2+b^2+c^2}}$
But $a=2,b=3$ and $c=-6$
Dirction cosines of $\overrightarrow {r}$ are $\large \pm\frac{2}{\sqrt{2^2+3^2+(-6)^2}}\pm\frac{3}{\sqrt{2^2+3^2+(-6)^2}}\pm\frac{-6}{\sqrt{2^2+3^2+(-6)^2}}\bigg)$
$=\large\pm\frac{2}{7},\pm\frac{3}{7},\pm\frac{-6}{7}$
But $\overrightarrow (r)$ makes an actue angle with x-axis.
Hence $ \cos \alpha >0\;(ie)\;l>0$
Hence the direction cosines are $\bigg(\large\frac{2}{7},\frac{3}{7},\frac{-6}{7}\bigg)$
Therefore $\overrightarrow r=14 \bigg(\large\frac{2}{7} $$\hat i+\large\frac{3}{7}$$ \hat j-\large\frac{6}{7}$$ \hat k\bigg)$
$\overrightarrow{r}=|\overrightarrow{r}| \;(l \hat i+m \hat j+ n \hat k)$
Hence the components of $\overrightarrow r$ along $OX,OY,and\; OZ$
$4 \hat i+6 \hat j-12 \hat k$
answered May 27, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...