SATHEE: 3D Geometry 3 Question 48

3D Geometry 3 Question 48

####48. A variable plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ at a unit distance from origin cuts the coordinate axes at $A, B$ and $C$ . Centroid $(x, y, z)$ satisfies the equation $\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = K$ . The value of $K$ is

(2005, 2M)

(a) 9

(b) 3

(d) $1 / 3$

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Answer:

Correct Answer: 48. (a)

Solution:

Since, $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ cuts the coordinate axes at $A (a, 0, 0), B (0, b, 0), C (0, 0, c)$ .

And its distance from origin $= 1$

where, $P$ is centroid of triangle.

$\begin{aligned} ∴ & P (x, y, z) & = \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \\ x & = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3} \end{aligned}$

From Eqs. (i) and (ii),

$\begin{aligned} \frac{1}{9 x^{2}} + \frac{1}{9 y^{2}} + \frac{1}{9 z^{2}} & = 1 \\ or & \frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = 9 & = K \\ ∴ K & = 9 \end{aligned}$

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3D Geometry 3 Question 48

Answer:

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