A plane at a unit distance from the origin in | vedclass

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(A) Let the equation of the plane be $\frac{X}{a} + \frac{Y}{b} + \frac{Z}{c} = 1$. Since the plane is at a unit distance from the origin $(0, 0, 0)$,

Vedclass · Accepted Answer

$9$

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(A) Let the equation of the plane be $\frac{X}{a} + \frac{Y}{b} + \frac{Z}{c} = 1$. Since the plane is at a unit distance from the origin $(0, 0, 0)$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$. The coordinates of the points are $P(a, 0, 0)$,$Q(0, b, 0)$,and $R(0, 0, c)$. The centroid $(x, y, z)$ of $\Delta PQR$ is given by $x = \frac{a+0+0}{3} = \frac{a}{3}$,$y = \frac{0+b+0}{3} = \frac{b}{3}$,and $z = \frac{0+0+c}{3} = \frac{c}{3}$. Thus,$a = 3x$,$b = 3y$,and $c = 3z$. Substituting these into the plane equation condition: $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = 1$. $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = 1$. Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$. Therefore,$k = 9$.

$A$ plane at a unit distance from the origin intersects the three axes at points $P, Q,$ and $R$. If the centroid of $\Delta PQR$ is $(x, y, z)$,and it satisfies the equation $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,then find the value of $k$.

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