First slide

Planes in 3D

Question

Distance of point $P (\vec{p})$ from the plane $\vec{r} . \vec{n}$ is

Moderate

A

$| \vec{p} \cdot \vec{n} |$
B

$\frac{| \vec{p} \times \vec{n} |}{| \vec{n} |}$
C

$\frac{| \vec{∣ p} \cdot \vec{n} |}{| \vec{n} |}$
D

none of these

Solution

Let $Q (\vec{q})$ . the foot of altitude drawn from P to the plane $\vec{r} . \vec{n} = 0$ Then
$\begin{array}{l} \vec{q} - \vec{p} = λ \vec{n} or \vec{q} = \vec{p} + λ \vec{n} \\ also \vec{q} \cdot \vec{n} = 0 \Rightarrow (\vec{p} + λ \vec{n}) \cdot \vec{n} = 0 \\ or λ = - \frac{\vec{p} \cdot \vec{n}}{| \vec{n} |^{2}} or \vec{q} - \vec{p} = - \frac{(\vec{p} \cdot \vec{n})}{| \vec{n} |^{2}} \vec{n} \end{array}$
Thus, required distance
$= | \vec{q} - \vec{p} | = \frac{| \vec{p} \cdot \vec{n} |}{| \vec{n} |} = | \vec{p} \cdot \hat{n} |$

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