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Question

If \(p\) is the perpendicular distance from origin to the plane passing through \((1, 0, 0)\), \((0, 1, 0)\) and \((0, 0, 1)\), then what is \(3p^2\) equal to ?

This question was previously asked in
NDA 2 2024 GAT Question Paper (01-Sep-2024)
The correct answer is
1

Plane Equation Derivation

The problem asks us to find the value of \(3p^2\), where \(p\) is the perpendicular distance from the origin \((0, 0, 0)\) to a plane passing through three specific points: \(A(1, 0, 0)\), \(B(0, 1, 0)\), and \(C(0, 0, 1)\).

These points are easily recognizable as the points where the plane intersects the coordinate axes. They represent the x-intercept, y-intercept, and z-intercept, respectively.

We can use the intercept form of the equation of a plane, which is given by:

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

Here, \(a\), \(b\), and \(c\) are the intercepts on the x, y, and z axes. From the given points, we have \(a = 1\), \(b = 1\), and \(c = 1\).

Substituting these values into the intercept form, we get the equation of the plane:

\(\frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1\)

This simplifies to:

\(x + y + z = 1\)

To use the distance formula, we rewrite the plane equation in the general form \(Ax + By + Cz + D = 0\):

\(x + y + z - 1 = 0\)

From this form, we identify the coefficients: \(A = 1\), \(B = 1\), \(C = 1\), and \(D = -1\).

Distance Calculation

The perpendicular distance (\(p\)) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is calculated using the formula:

\(p = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\)

In this problem, the point is the origin, so \((x_0, y_0, z_0) = (0, 0, 0)\). The plane is \(x + y + z - 1 = 0\), with \(A = 1\), \(B = 1\), \(C = 1\), and \(D = -1\).

Plugging these values into the distance formula:

\(p = \frac{|1(0) + 1(0) + 1(0) + (-1)|}{\sqrt{1^2 + 1^2 + 1^2}}\) \(p = \frac{|0 + 0 + 0 - 1|}{\sqrt{1 + 1 + 1}}\) \(p = \frac{|-1|}{\sqrt{3}}\) \(p = \frac{1}{\sqrt{3}}\)

So, the perpendicular distance from the origin to the plane is \(\frac{1}{\sqrt{3}}\).

Final Calculation of \(3p^2\)

The question asks for the value of \(3p^2\). We have calculated \(p = \frac{1}{\sqrt{3}}\).

First, let's find \(p^2\):

\(p^2 = \left( \frac{1}{\sqrt{3}} \right)^2\) \(p^2 = \frac{1^2}{(\sqrt{3})^2}\) \(p^2 = \frac{1}{3}\)

Now, we multiply this result by 3:

\(3p^2 = 3 \times p^2\) \(3p^2 = 3 \times \frac{1}{3}\) \(3p^2 = 1\)

Therefore, the value of \(3p^2\) is 1.

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  1. Which of the following are the direction ratios of S?
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