Find the equation of the plane passing through the point $a = 2 i + 3 j - k$ and perpendicular to the vector $3 i - 2 j - 2 k$ and the distance of the plane from the origin?

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Solution

Equation of plane is given by
$\to r . (2 ˆ i + 3 ˆ j - ˆ k) = 2$
Distance of plane from origin $= 2$ units.
Now, to find the equation of the plane passing through the point $a = 2 ˆ i + 3 ˆ j - ˆ k$ and perpendicular to a vectors $3 ˆ i - 2 ˆ j - 2 ˆ k$ and the distance of this plane from the origin.
We know that equation of a plane in vector form is given by $\to r . \to n = d$
here plane is passing through
$\begin{matrix} a = 2 ˆ i + 3 ˆ j - ˆ k \to n = 3 ˆ i - 2 ˆ j - 2 ˆ k \end{matrix}$
So, the plane passes through vector $a$ and perpendicular to vector $n$ is given by equation.
$\begin{matrix} (\to r - \to a) . \to n = 0 \Rightarrow \to r \to n = \to a \to n \Rightarrow \to r . (2 ˆ i + 3 ˆ j - ˆ k) = (2 ˆ i + 3 ˆ j - ˆ k) . (3 ˆ i - 2 ˆ j - 2 ˆ k) \Rightarrow \to r . (2 ˆ i + 3 ˆ j - ˆ k) = 6 - 6 + 2 \Rightarrow \to r . (2 ˆ i + 3 ˆ j - ˆ k) = 2 \end{matrix}$
is the required equation of plane.
Distance of plane from origin $= 2$ units.

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