Ask Questions, Get Answers

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

Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Can you answer this question?

1 Answer

0 votes
  • The position vector of a point which divides the line in the ratio $m:n$ internally is given by $\overrightarrow r=\large\frac{m\overrightarrow a+n\overrightarrow b}{m+n}$
Step 1:
Let the position vectors $\overrightarrow{OA}=\hat i-2\hat j-8\hat k$ and $\overrightarrow {OC}=11\hat i+3\hat j+7\hat k$ respectively.
Now let the point B divides the line AC in the ratio $\lambda:1$
Hence the position vector of $B$ is $\overrightarrow {OB}=\large\frac{\lambda(11\hat i+3\hat j+7\hat k)+1(\hat i-2\hat j-8\hat k)}{\lambda+1}$
Step 2:
But we know $\overrightarrow{OB}=5\hat i-2\hat k$
Therefore $5\hat i-2\hat k=\large\frac{\lambda(11\hat i+3\hat j+7\hat k)+1(\hat i-2\hat j-8\hat k)}{\lambda+1}$
$\Rightarrow 5\hat i-2\hat k=\hat i\large\frac{(11\lambda+1)}{\lambda+1}$$+\hat j\large\frac{(3\lambda-2)}{\lambda+1}$$+\large\frac{(7\lambda-8)}{\lambda+1}$
Step 3:
Now equating the coefficients of $\hat j$ on both sides we get
$ 3\lambda=2$
Therefore $\lambda=\large\frac{2}{3}$
Step 4:
Now let us check if $A,B$ and $C $ are collinear.
Similarly $\large\frac{7\lambda-8}{\lambda+1}=\frac{7\times\Large\frac{2}{3}-8}{\Large\frac{2}{3}+1}$
Step 5:
This proves $A,B$ and $C$ are collinear and B divides it in the ratio $2:3$
answered May 24, 2013 by sreemathi.v

Related questions

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