This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

$\begin{array}{1 1} x+y-z=2 \\ x+y+z=-2 \\ x+y+z=2 \\ x+y-z=-2 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- For any arbitrary point $P(x,y,z)$ on the plane,the position vector $\overrightarrow r$ is given by $\overrightarrow r=x\hat i+y\hat j+z\hat k$
- The cartesian equation of the plane is of the form $lx+my+nz=d$

Step 1:

The given equation of the plane is $\overrightarrow r.(\hat i+\hat j-\hat k)=2$-----(1)

But we know $\overrightarrow r=x\hat i+y\hat j+z\hat k$

Substituting for $\overrightarrow r$ in equation (1) we get

$(x\hat i+y\hat j+z\hat k)(\hat i+\hat j-\hat k)=2$

Step 2:

Apply the dot product rule we get

$x+y-z=2$

This equation is the required Cartesian equation of the plane.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...