The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the poi

Q: Why is the vector $\vec{OM}$ the normal vector? Can't we just use the plane equation directly?

That is a fantastic conceptual question! By definition, the 'foot of the perpendicular' from a point to a plane is the point on the plane closest to the origin. The line segment connecting the origin to this point is, by construction, perpendicular to every line in the plane passing through that point. Therefore, the vector $\vec{OM}$ is the normal vector $\vec{n}$. Using this is the most direct path to defining the plane's orientation in 3D space.

Q: The volume equation looks like a nightmare to solve. Do I really need to expand $(20 + \alpha^2)^3$?

Absolutely not! In JEE Advanced, whenever you see a high-degree polynomial equation involving a variable defined as a natural number ($\alpha \in \mathbb{N}$), it is a massive signal to stop and test values. Expanding that cubic would lead to a complex polynomial that is unnecessary to solve. Always look for the constraint—here, $\alpha$ being a natural number is your greatest weapon.

Q: Why did we divide the final equation by 2? Is it necessary?

It is not strictly necessary for the math to work, but it is a mark of a seasoned problem solver. Simplifying the equation $2x + 2y + 4z = 24$ to $x + y + 2z = 12$ reduces the risk of arithmetic errors when you are checking the options. Always simplify your final equations; it makes the verification step much cleaner and faster.

Question

The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the points A, B, C is $(2, \alpha, 4), \alpha \in \mathbb{N}$. If the volume of the tetrahedron OABC is 144 unit$^3$, then which of the following points is NOT on P?

Options:

(A) $(2, 2, 4)$

(B) $(0, 4, 4)$

(C) $(3, 0, 4)$

(D) $(0, 6, 3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In 3D geometry, the most powerful tool you possess is the normal vector. It defines the orientation of the plane. Because the point $M(2, \alpha, 4)$ is the foot of the perpendicular from the origin $O(0, 0, 0)$, the vector $\vec{OM}$ is, by definition, perpendicular to the plane.

This is our "Aha!" moment. The components of this vector, which are simply the coordinates of $M$, become the coefficients of our plane equation. Thus, our normal vector is $\vec{n} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$.

With a point on the plane and the normal vector, we invoke the standard equation of a plane: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting our values, we get:

$$2(x - 2) + \alpha(y - \alpha) + 4(z - 4) = 0$$

Expanding this, we arrive at the skeleton of our plane:

$$2x + \alpha y + 4z = 20 + \alpha^2$$

This equation is the DNA of our plane; everything else follows from here.

### The Volume Challenge

The plane cuts the axes at points $A, B,$ and $C$. These are the intercepts. To find them, we set two variables to zero at a time.

For the $x$-intercept $A$, we set $y=0$ and $z=0$, yielding $x = \frac{20 + \alpha^2}{2}$. Similarly, for $B$, we find $y = \frac{20 + \alpha^2}{\alpha}$, and for $C$, we find $z = \frac{20 + \alpha^2}{4}$.

Now, we face the volume of the tetrahedron $OABC$. The formula is elegant and simple: $V = \frac{1}{6} |x_A \cdot y_B \cdot z_C|$. Substituting our intercepts, we get:

$$V = \frac{1}{6} \left( \frac{20 + \alpha^2}{2} ight) \left( \frac{20 + \alpha^2}{\alpha} ight) \left( \frac{20 + \alpha^2}{4} ight)$$

This simplifies to the following expression for the volume:

$$V = \frac{(20 + \alpha^2)^3}{48\alpha}$$

### The Intuitive Leap

We are given that $V = 144$. Setting our expression equal to this, we get:

$$\frac{(20 + \alpha^2)^3}{48\alpha} = 144 \implies (20 + \alpha^2)^3 = 6912\alpha$$

Many students would panic here, trying to expand the cubic term. But remember, we are in a JEE Advanced setting. We are told $\alpha \in \mathbb{N}$. This is a massive hint!

We don't need to solve the cubic algebraically; we need to test values. Let's test $\alpha = 2$. The left side becomes $(20 + 2^2)^3 = 24^3 = 13824$. The right side becomes $6912 	imes 2 = 13824$.

It matches perfectly! The beauty of mathematics often lies in these moments of symmetry where the complexity collapses into a simple integer.

### Final Verification

With $\alpha = 2$, our plane equation becomes $2x + 2y + 4z = 20 + 2^2$, which simplifies to $2x + 2y + 4z = 24$. Dividing by 2, we get the clean, final equation:

$$x + y + 2z = 12$$

Now, we simply test the options. For the point $(3, 0, 4)$, we calculate $3 + 0 + 2(4) = 11$. Since $11 
eq 12$, this point does not lie on the plane.

We have successfully navigated the geometry, conquered the algebra, and verified our result. Keep this confidence, and remember: every complex problem is just a series of simple, logical steps waiting to be connected.

Select Answer:

Visualized Solution

Visualizing the Setup

Finding the Normal Vector

Equation of Plane $P$

Simplifying the Equation

Finding the Intercepts

Volume of Tetrahedron

Simplifying Volume Expression

Equating to Given Volume

Solving for $α$

Final Plane Equation

Checking the Options

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Volume Challenge

The Intuitive Leap

Final Verification

Similar Questions

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + 4 \hat{k}) = 16$ and $r \cdot (- \hat{i} + \hat{j} + \hat{k}) = 6$ . Then which of the following points does NOT lie on $P$ ?

Let the lines $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on $P$ ?

A plane $P$ contains the line $x + 2 y + 3 z + 1 = 0 = x - y - z - 6$ , and is perpendicular to the plane $- 2 x + y + z + 8 = 0$ . Then which of the following points lies on $P$ ?

The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ is :

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Let $Q$ be the foot of perpendicular drawn from the point $P (1, 2, 3)$ to the plane $x + 2 y + z = 14$ . If $R$ is a point on the plane such that $∠ P R Q = 6 0^{\circ}$ , then the area of $△ P QR$ is equal to:

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals