Question

Refer the attached picture. Solve fast please.

3-D GEOMETRY. ​

Answer 1

Given:

• A variable plane remains at a constant distance 3p from the origin cut the coordinate axes at A, B and C.

To Prove:

• ${x}^{ - 2} + {y}^{ - 2} + {z}^{ - 2} = {p}^{ - 2}$

Proof:

Let,

the Equation of the variable plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Now,

This plane meets the X-axis, Y-axis and Z-axis at the points A(a,0,0) , B(0,b,0) and C(0,0,c) respectively.

Further,

Let,

(d,e,f) be the Coordinate of the centroid of ∆ABC.

Then,

We have,

$d = \frac{a + 0 + 0}{3} = \frac{a}{3} \\ \\ e = \frac{0 + b + 0}{3} = \frac{b}{3} \\ \\ e = \frac{0 + 0 + c}{3} = \frac{c}{3}$

Therefore,

We get,

$= > a = 3d \\ = > b = 3e \\ = > c =3f$

But,

We know that,

length of perpendicular from (0,0,0) to the given plane = 3p

Therefore,

We get,

$= > 3p = \frac{ | \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 | }{ \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} } } } \\ \\ = > 3p = \frac{1}{ \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} } } } \\ \\ = > \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} } } = \frac{1}{3p} \\ \\ = > \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} } = \frac{1}{9 {p}^{2} }$

Now,

Putting the respective values of a, b and c

We get,

$= > \frac{1}{ {(3d)}^{2} } + \frac{1}{ {(3e)}^{2} } + \frac{1}{ {(3f)}^{2} } = \frac{1}{9 {p}^{2} } \\ \\ = > \frac{1}{9 {d}^{2} } + \frac{1}{9 {e}^{2} } + \frac{1}{9 {f}^{2} } = \frac{1}{9 {p}^{2} } \\ \\ = > \frac{1}{ {d}^{2} } + \frac{1}{ {e}^{2} } + \frac{1}{ {f}^{2} } = \frac{1}{ {p}^{2} } \\ \\ = > {d}^{ - 2} + {e}^{ - 2} + {f}^{ - 2} = {p}^{ - 2}$

Hence,

the locus of the plane is,

• $\bold \red{{x}^{ - 2} + {y}^{ - 2} + {z}^{ - 2} = {p}^{ - 2} }$

Thus, Proved