If the distance between the plane Ax – 2y + z = d and the plane containing the lines and is , then | d

Engineering

Mathematics

Distance from a Plane

Question

If the distance between the plane Ax – 2y + z = d and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $\sqrt{6}$ , then | d | is

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$\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} | = \hat{i} (- 1) - \hat{j} (- 2) + \hat{k} (- 1) = \hat{i} - 2 \hat{j} + \hat{k} \Rightarrow A = 1$

Plane: (x – 1) – 2(y – 2) + (z – 3) = 0

x – 2y + 2 – 1 + 4 – 3 = 0

x – 2y + z = 0

$\frac{| d - 0 |}{\sqrt{6}} = \sqrt{6} \Rightarrow | d | = 6$