# Find the vector and cartesian equations of the planes $(b) \quad \text{that passes through the point} \; (1, 4, 6) \; \text{and the normal to the plane is } \hat i-2\hat j + \hat k$

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

Toolbox:
• Vector equation of the plane passing through a point and normal to the plane is given by $(\overrightarrow r-\overrightarrow a).\overrightarrow N=0$
• The Cartesian equation is given by $a(x-x_1)+b(y-y_1)+c(z-z_1)$=0
Step 1:
Let the position vector of the point $(1,4,1)$ be $\overrightarrow a=\hat i+4\hat j+6\hat k$
The normal vector $\overrightarrow N$ perpendicular to the plane is $\overrightarrow N=\hat i-2\hat j+\hat k$
The vector equation of the plane is given by $(\overrightarrow r-\overrightarrow a).\overrightarrow N=0$
Now substituting for $\overrightarrow a$ and $\overrightarrow N$ we get
$[\overrightarrow r-(\hat i+4\hat j+6\hat k)].(\hat i-2\hat j+\hat k)=0$-----(1)
Step 2:
But we know $\overrightarrow r=x\hat i+y\hat j+z\hat k$
Substituting this in equ(1) we get,
$[\overrightarrow (x\hat i+y\hat j+z\hat k)-(\hat i+4\hat j+6\hat k)].(\hat i-2\hat j+\hat k)=0$
$[(x-1)\hat i+(y-4)\hat j+(z-6)\hat k].(\hat i-2\hat j+\hat k)=0$
But we know $\hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1$
Step 3:
On simplifying we get
$\Rightarrow (x-1)+(y-4)(-2)+(z-6).1=0$
$x-1-2y+8+z-6=0$
$x-2y+z+1=0$
This is the required Cartesian equation of the required plane.