Find the vector and Catesian equation of the plane that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $^i - 2^j +^k$

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Solution

The position vector of point $(1, 4, 6)$ is $\to a =^i + 4^j + 6^k$

The normal vector $\to n$ perpendicular to the plane is $\to n =^i - 2^j +^k$

The vector equation of the plane is given by $(\to r - \to a) . \to n = 0$

$\Rightarrow$ $[\to r - (^i + 4^j + 6^k)] . (^i - 2^j +^k) = 0....... (1)$

Where $\to r$ is the position vector of any point $P (x, y, z)$ in the plane.

And given by, $\to r = x^i + y^j + z^k$

Therefore, the equation(1) becomes

$\Rightarrow$ $[(x - 1)^i + (y - 4)^j + (z - 6)^k] . (^i - 2^j +^k) = 0$

$\Rightarrow$ $(x - 1) - 2 (y - 4) + (z - 6) = 0$

$\Rightarrow$ $x - 2 y + z + 1 = 0$

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