Get Answers to all your Questions

header-bg qa

Find the vector equation of the line passing through (1,2,3) and parallel to each of the
planes \vec{r}\cdot \left ( \hat{i}-\hat{j}+\hat{k} \right )= 5 and \vec{r}\cdot \left ( 3\hat{i}+\hat{j}+\hat{k} \right )= 6. Also find the point of intersection of the line thus obtained with the plane \vec{r}\cdot \left ( 2\hat{i}+\hat{j}+\hat{k} \right )= 4.

 

 

 

 
 
 
 
 

Answers (1)

Let the equation of the line passing through (1,2,3)
be  \frac{x-1}{a}= \frac{y-2}{b}= \frac{z-3}{c}---(i) where a,b,c are d.r's of the required line
This line is parallel to the planes \vec{r}\cdot \left ( \hat{i}-\hat{j}+2\hat{k} \right )= 5 and \vec{r}\cdot \left ( 3\hat{i}+\hat{j}+\hat{k} \right )= 6\cdot
so clearly the normals of there two plane will be perpandicular to the line (i)
\therefore a-b+2c= 0\; \S \; 3a+b+c= 0
On adding there two , we get 4a+3c= 0
   \Rightarrow a= \frac{-3c}{4}\; \; \therefore b= \frac{5c}{4}
so the d.r's of line (i) are = \frac{-3c}{4},\frac{5c}{4}\, ie \; -3,5,4
By (i) the required line is : \frac{x-1}{-3}= \frac{y-2}{5}= \frac{z-3}{4}= \lambda \; \left ( say \right )
Now consider the random pair on line
(i) as : p\left ( -3\lambda +1,5\lambda +2,4\lambda +3 \right ) for the intersection of line and the plane .
\vec{r}\cdot \left ( 2\hat{i}+\hat{j}+\hat{k} \right )= 4\; ie\; 2x+y+z= 4 point p must satisfy the eq.of plane
so 2\left ( -3\lambda +1 \right )+\left ( 5\lambda +2 \right )+\left ( 4\lambda +3 \right )= 4\Rightarrow \lambda = -1
Hence the point of intersection is  p(4,-3,-1).

Posted by

Ravindra Pindel

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads