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If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of ABC is :

Option 1)
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1$

Option 2)
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=3$

Option 3)
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{1}{9}$

Option 4)
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9$

Answers (1)

best_answer

As we learnt

Intercept form -

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

- wherein

Let the plane cuts the coordinate axis at $A\left ( a,0,0 \right ),B\left ( 0,b,0 \right )\, and \: C\left ( 0,0,c \right )$

Centroid of triangle -

$\left ( \frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3} \right )$

- wherein

Distance of a point from plane (Cartesian form) -

The length of perpendicular from $P(x_{1},y_{1},z_{1})$ to the plane

$ax+by+cz+d= 0$ is given by $\frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}$

-

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

where, $\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}= 3$

$\Rightarrow {\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}= \frac{1}{9}$

$G=\left (h_3k,l \right )= \left ( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right )$

So, $\frac{1}{9h^2}+\frac{1}{9k^2}+\frac{1}{9l^2}=\frac{1}{9}$

$\Rightarrow \frac{1}{h^2}+\frac{1}{k^2}+\frac{1}{l^2}=1$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1$

Option 1)

$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1$

Option 2)

$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=3$

Option 3)

$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{1}{9}$

Option 4)

$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9$

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Aadil

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