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I'm trying to prove that the condition for distance of a point $(x_1,y_1,z_1)$ from a plane $ax+by+cz+d=0$ passing through a point $(\alpha,\beta,\gamma)$ to be maximum is $\dfrac{x_1-α}{a}=\dfrac{y_1-β}{b}=\dfrac{z_1-\gamma}{c}$ i.e. the line joining $(x_1,y_1,z_1)$ and $(\alpha,\beta,\gamma)$ must be normal to plane.

I know this can be done easily using geometric visualisation, and {hypotenuse> altitude} etc. which is very basic that it must be normal.

But i want to use cauchy schwarz inequality for proving this.

Tried to minimise $\dfrac{a}{(ax1+by1+cz1+d)^2}+ \dfrac{b}{(ax1+by1+cz1+d)^2}+\dfrac{c}{(ax1+by1+cz1+d)^2}$

By cauchy-schwarz inequality,

$$\scriptsize{( \frac{a}{(ax1+by1+cz1+d)^2} + \frac{b}{(ax1+by1+cz1+d)^2} + \frac{c}{(ax1+by1+cz1+d)^2})*{(\alpha^2+\beta^2+\gamma^2)}}$$

$$ \geq $$

$$( a \alpha /(ax1+by1+cz1+d)+ b\beta /(ax1+by1+cz1+d)+c\gamma /(ax1+by1+cz1+d))^2$$

Can't proceed further.

Anindya Prithvi
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Aman Kushwaha
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1 Answers1

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Using C-S inequality,

$$\ \frac { { (ax+by+cz+d) }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } =\frac { { (a(x-\alpha )+b(y-\beta )+c(z-\gamma )) }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \le \left( { \left( \frac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right) }^{ 2 }+{ \left( \frac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right) }^{ 2 }+{ \left( \frac { c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right) }^{ 2 } \right) \left( { \left( x-\alpha \right) }^{ 2 }+{ \left( y-\beta \right) }^{ 2 }+{ \left( z-\gamma \right) }^{ 2 } \right) ={ \left( x-\alpha \right) }^{ 2 }+{ \left( y-\beta \right) }^{ 2 }+{ \left( z-\gamma \right) }^{ 2 } $$ which is also the required answer. Pardon me for dropping the subscripts. Hope it helps.

Vishnu Kadiri
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