logo

Ask Questions, Get Answers

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

Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from the point (2,-1,5) to the line \( \large\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11} \).

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
Let L be the foot of the perpendicular drawn from the point $P(2,-1,5)$ to the given line.
The coordinates of a general point on $\large\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$ are given by
$\large\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$$=k$
$x=10k+11,y=-4k-2,z=-11k-8$
Let the coordinates of L be $(10k+11,-4k-2,-11k-8)$
$\therefore$ Direction ratios of PL are proportional to $10k+11-2,-4k-2+1,-11k-8-5$
(ie)$10k+9,-4k-1,-11k-13$
Direction ratios of the given line are proportional to $(10,-4,-11)$
Step 2:
Since $PL$ is $\perp$ to the given line
$\therefore 10(10k+9)-4(-4k-1)-11(-11k-13)=0$
On simplifying we get,
$90+100k+16k+4+121k+143=0$
$237k=-237$
$\Rightarrow k=-1$
Hence the coordinates of L are (1,2,3)
$PL=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$\;\;\;\;\;=\sqrt{(2-1)^2+(-1-2)^2+(5-3)^2}$
$\;\;\;\;\;=\sqrt{1^2+(-3)^2+(2)^2}$
$\;\;\;\;\;=\sqrt{14}$units.
answered Sep 25, 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
...