logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
0 votes

If $P$ is the point of intersection of the line joining the points $Q(2,3,5)\:\:and\:\:R(1,-1,4)$ with the plane $5x-4y-z-1=0$. If $S$ is foot of $\perp$ drawn from the point $T(2,1,4)$ to the line $QR$, then the length of segment $\overrightarrow {PS}$ = ?

$\begin{array}{1 1} (a)\:\large\frac{1}{\sqrt 2}\:\:\:\qquad\:(b)\:\sqrt 2\:\:\:\qquad\:(c)\:2\:\:\:\qquad\:(d)\:2\sqrt 2. \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Eqn. of the line through the points $(x_1,y_1,z_1)\:and\:(x_2,y_2,z_2)$ is $\large\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}=\frac{z-z_1}{z_1-z_2}=\lambda$
  • If two lines are $\perp$,  then the dot product of their  direction ratios  $(d.r.) =0$
Given: $Q(2,3,5)\:\:and\:\:R(1,-1,4)$.
Eqn. $\overline {QR}$ is $\large\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$......(i)
Any point $(x,y,z)$ on the line $QR$ can be expressed in terms of $\lambda$,
$\therefore$ Let foot of $\perp$ from $T(2,1,4)$ on the line $\overline {QR}\:\:\:\: be\:\:\:S (\lambda+2,4\lambda+3,\lambda+5)$ (from (i))
$\Rightarrow\:\:\:\overline {TS}$ is $\perp$ to the line $\overline { QR}$
$\therefore\:$ $(d.r.\:\: of\:\:\:\overline {QR}).(d.r.\:\:of \:\:\overline {TS})=0$
$\Rightarrow\:(1,4,1).(\lambda,4\lambda+2,\lambda+1)=0$
$\Rightarrow\:\lambda+16\lambda+8+\lambda+1=0$ or $ \lambda=-\large\frac{1}{2}$
$\therefore\:S(\large\frac{3}{2},$$1,\large\frac{9}{2})$
Given the line $\overline{QR}$ cuts the plane $5x-4y-z-1=0$ at the point $ P$
Let $P$ be assumed in terms of $\lambda$ from (i) as $P(\lambda+2,4\lambda+3,\lambda+5)$
Since $P$ is the intersecting point of the line (i) and the plane, $p$ satisfies the eqn. of the plane.
$\therefore \:5(\lambda+2)-4(4\lambda+3)-(\lambda+5)-1=0$
$\Rightarrow\:5\lambda-16\lambda-\lambda+10-12-5-1=0$
$\Rightarrow\:\lambda=-\large\frac{2}{3}$
$\therefore P(\large\frac{4}{3},\frac{1}{3},\frac{13}{3})$
$\therefore$ distance $PS$ = $\sqrt {\frac{1}{36}+\frac{16}{36}+\frac{1}{36}}$
$=\frac{1}{\sqrt 2}$

 

answered Dec 25, 2013 by rvidyagovindarajan_1
edited Dec 25, 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
...