Find the distance of the point A(1,2,-1) from the plane $¯ r (¯ i - 2^j + 4^k) = 10$

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Solution

$\begin{matrix} x - 2 y + 4 z = 10, A (1, 2. - 1) x - 2 y + 4 z = 10 d i s tan c e o f t h e p o i n t = \frac{| l x_{1} + m y_{1} + n z_{1} + d |}{\sqrt{l^{2} + m^{2} + n^{2}}} = \frac{| 17 |}{\sqrt{21}} = 3.7 H e n c e, t h e d i s tan c e o f t h e p o i n t f r o m t h e p l a n e i s 3.7 \end{matrix}$

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