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

Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Two or more vectors are said to be collinear if their magnitudes are equal or proportional.
Step 1:
Let the position vectors of points $A,B$ and $C$ be
$\overrightarrow{OA}=\hat i+2\hat j+7\hat k$
$\overrightarrow{OB}=2\hat i+6\hat j+3\hat k$
$\overrightarrow{OC}=3\hat i+10\hat j-\hat k$
Step 2:
Now $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad\quad\;\;=(2\hat i+6\hat j+3\hat k)-(\hat i+2\hat j+7\hat k)$
$\qquad\quad\;\;=\hat i+4\hat j-4\hat k$
Step 3:
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(2\hat i+6\hat j+3\hat k)$
$\quad\;\;\;=\hat i+4\hat j-4\hat k$
Step 4:
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(\hat i+2\hat j+7\hat k)$
$\quad\;\;\;=2\hat i+8\hat j-8\hat k$
$\quad\;\;\;=2(\hat i+4\hat j-4\hat k)$
Step 5:
$|\overrightarrow{AB}|=\sqrt{1^2+4^2+(-4)^2}$
$\qquad=\sqrt{1+16+16}$
$\qquad=\sqrt{33}$
Step 6:
$|\overrightarrow{BC}|=\sqrt{1^2+4^2+(-4)^2}$
$\qquad=\sqrt{1+16+16}$
$\qquad=\sqrt{33}$
Step 7:
$|\overrightarrow{AC}|=\sqrt{2^2+8^2+(-8)^2}$
$\qquad=\sqrt{4+64+64}$
$\qquad=\sqrt{132}$
$\qquad=2\sqrt{33}$
Step 8:
$AB=\sqrt{33},BC=\sqrt{33},AC=2\sqrt{33}$
Thus $AB+BC=AC$
Hence A,B,C are collinear.
answered May 21, 2013 by sreemathi.v
 

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
...