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 $A,B,C,D$ are the points with position vectors $\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k},2\overrightarrow{i}+\overrightarrow{j}+3\overrightarrow{k},2\overrightarrow{i}-3\overrightarrow{k},3\overrightarrow{i}+2\overrightarrow{i}+\overrightarrow{k},$ respectively ,find the projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}.$

$\begin{array}{1 1} 3 \\ \sqrt 3 \\ 9 \\ 27 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Projection of $ \overrightarrow a \: along \: \overrightarrow b\: is \: \large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|} $
Given $\overrightarrow{OA}=\hat i+\hat j+\hat k$, $\overrightarrow{OB}=2\hat i+\hat j+3\hat k$, $\overrightarrow{OC}=2\hat i-3\hat k$ and $\overrightarrow{OD}=3\hat i+2\hat j+\hat k$
Let us determine $ \overrightarrow{AB}$ and $ \overrightarrow{CD}$
$ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad=(2\hat i+\hat j+3\hat k)-(\hat i+\hat j+\hat k)$
$\qquad=\hat i+2\hat k$
$ \overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$
$\qquad=(3\hat i+2\hat j+\hat k)-(2\hat i-3\hat k)$
$\qquad=\hat i+2\hat j+ 4 \hat k$
Projection of $ \overrightarrow {AB} $ along $ \overrightarrow {CD}$ is
$\large\frac{\overrightarrow {AB}.\overrightarrow {CD}}{|\overrightarrow {CD}|} =\large\frac{(\hat i+2 \hat k).(\hat i+2 \hat j +4 \hat k)}{\sqrt {1^2+2^2+(-2)^2}}$
$=\large\frac{1+8}{\sqrt {1+4+4}}=\frac{9}{3}$
$=3$
Hence the projection of $ \overrightarrow{AB} \: along \: \overrightarrow{CD}$ is 3
answered May 28, 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
...