If $a, b, c$ are the intercepts on coordinate axes respectively by a plane and its distance from the origin is $p$, then prove that $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{1}{p^2}$.

Question

Accepted Answer

>

> [!TLDR]
> Writing the plane in intercept form and applying the point-to-plane distance formula from the origin gives $\dfrac1{p^2}=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}$.
>
> **Concept.** A plane with intercepts $a,b,c$ on the axes is $\dfrac xa+\dfrac yb+\dfrac zc=1$. The distance of a point $(x_0,y_0,z_0)$ from $\ell x+m y+n z+d=0$ is $\dfrac{|\ell x_0+m y_0+n z_0+d|}{\sqrt{\ell^2+m^2+n^2}}$.
>
> Rewrite the plane as $\dfrac xa+\dfrac yb+\dfrac zc-1=0$. Distance from the origin $(0,0,0)$: …

Q.If $a, b, c$ are the intercepts on coordinate axes respectively by a plane and its distance from the origin is $p$ , then prove that $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{1}{p^2}$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.If a,b,ca, b, ca,b,c are the intercepts on coordinate axes respectively by a plane and its distance from the origin is ppp, then prove that 1a2+1b2+1c2=1p2\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{1}{p^2}a21​+b21​+c21​=p21​.

🎓Unlock everything free for 14 days

Q.If $a, b, c$ are the intercepts on coordinate axes respectively by a plane and its distance from the origin is $p$ , then prove that $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{1}{p^2}$ .

Unlock everything free for 14 days