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

If $\bar{\alpha}=3\bar{i}-\bar{j}\;and\;\bar{\beta}=2\bar{i}+\bar{j}-\bar{k}.$ Express $ \bar{\beta}$ as a sum of two vectors $ \bar{\beta_1}\;and\;\bar{\beta_2},\;where \; \bar{\beta_1}$ is parallel to $ \bar{\alpha}\;and\; \bar{\beta_2}$ is prependicular to $ \bar{\alpha}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If two vectors $\overrightarrow a\:and\:\overrightarrow b$ are parallel then $\overrightarrow a= \lambda\overrightarrow b.$ for some real constant $\lambda$.
  • If two vectors are $\perp$ then their dot product = 0.
Given $\large\overrightarrow \alpha=3\hat i-\hat j\:\:and\:\:\overrightarrow \beta=2\hat i+\hat j-\hat k$
We have to express $\large\overrightarrow \beta\:\:as\:\overrightarrow \beta_1+\overrightarrow \beta_2$
where $\large\overrightarrow \beta_1\:\:is\:\:parallel\:to\:\overrightarrow \alpha$ and $\large\overrightarrow \beta_2\:\:is\:\:\perp\:to\:\overrightarrow \alpha$.
Since $\large\overrightarrow \beta_1\:is\:||\:to\:\overrightarrow \alpha,\:\overrightarrow \beta_1=\lambda\overrightarrow \alpha$.
$\Rightarrow\:\large\overrightarrow \beta=\lambda(3\hat i-\hat j)+\overrightarrow \beta_2$
$\Rightarrow\:\large\:2\hat i+\hat j-\hat k=\lambda(3\hat i-\hat j)+\overrightarrow \beta_2$.
$\Rightarrow\:\large\overrightarrow \beta_2=(2\hat i+\hat j-\hat k)-\lambda(3\hat i-\hat j)$
$\Rightarrow\:\large\overrightarrow \beta_2=(2-3\lambda)\hat i+(1+\lambda)\hat j-\hat k$
Given that $\large\overrightarrow \beta_2\:is\:\perp\:\overrightarrow \alpha\:\Rightarrow\:\overrightarrow \beta_2.\overrightarrow \alpha=0$.
$\Rightarrow\:\large\big[(2-3\lambda)\hat i+(1+\lambda)\hat j-\hat k\big].\big[3\hat i-\hat j\big]=0$
$\Rightarrow\:\large(2-3\lambda)3-1(1+\lambda)+0=0$
$\Rightarrow\:\large6-9\lambda-1-\lambda=0$
$\Rightarrow\:\large\:5-10\lambda=0$
$\Rightarrow\:\large\lambda=\frac{1}{2}$
Substituting the value of $\large\lambda$ in $\large\overrightarrow \beta_1\:and\:\overrightarrow \beta_2$ we get
$\large\overrightarrow \beta_1=\frac{1}{2}(3\hat i-\hat j)$
$\large\overrightarrow \beta_2=(2-\frac{3}{2})\hat i+(1+\frac{1}{2})\hat j-\hat k$
$\large=\frac{1}{2}\hat i+\frac{3}{2}\hat j-\hat k$
answered Apr 17, 2013 by rvidyagovindarajan_1
 

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
...