Tag: maths

Equation of plane passes through $(2,2,1)$ is given by,
$a(x-2)+b(y-2)+c(z-1) = 0.......(A)$
Given it also passes through $(9,3,6)$
$\Rightarrow 7a+b+5c=0 ......(1)$
and this plane is perpendicular to plane $2x+6y+6z-1=0$
$\Rightarrow 2a+6a+6c=0 .....(2)$
Solving (1) and (2), we get $ a= \dfrac{-3a}{5}, b = \dfrac{-4a}{5}$
Putting these values in (A) our required plane is,
$3x+4y-5z=9$

We have $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$

Let $ax+by+cz=1$ be the desired plane.
Since, it is perpendicular to $x+2y+3z=7$ & $2x-3y+4z=0$
Therefore, $a+2b+3c=0$        .... (1)
and $2a-3b+4c=0$       ...(2)
and it passes through $(1,1,1)$
Therefore, $a+b+c=1$      ...(3)
Solving $(1),(2)$ and $(3)$ simultaneously, we get

Since $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is the proof any vector $(x,y,z)$on the plane. The given equation can be written as.

$x\hat{i}+y\hat{j}++z\hat{k}=(s—2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\\x=(s-2t)\quad y=(3-t)\quad z=2s+t$

Similarly We get $x-2y=s-6$ and $y+z=3+2s$

Now eliminating $s$ we get $2(x-2y)-(y+z)=-15$

$2x-5y-z+15=0$ is the required form of the equation.

Since, the direction ratios of the normal are $ \alpha , \beta , \gamma $
The equation of the plane will be of the form, $ \alpha x + \beta y + \gamma z  = d$
And the plane passes through the point $(2,1,4)$.
Hence, equation of plane is $ \alpha x + \beta y + \gamma z $ = $ 2 \alpha + \beta  + 4 \gamma  $.