Problem 38 - A Complete Resource Book in Mathematics for JEE Main

Question

Through a point \(P(h, k, l)\) a plane is drawn at right angles to \(O P\) to meet the coordinate axes in \(A, B\) and C. If \(O P=p\), then the area of \(\Delta A B C\) is (A) \(\frac{p^{5}}{2 h k l}\) (B) \(\frac{p^{5}}{h k l}\) (C) \(\frac{p^{5}}{4 h k l}\) (D) none of these

Step-by-Step Solution

Answer

The area of \(\Delta ABC\) is \(\frac{p^5}{2 h k l}\).

Step 1: Understanding the scenario

We are given a point \(P(h, k, l)\) from which a plane is drawn at right angles to the line \(OP\). This plane intersects the coordinate axes at points \(A, B,\) and \(C\). The goal is to find the area of triangle \(\Delta ABC\).

Step 2: Equation of the plane

The plane is perpendicular to \(OP\) and passes through the point \(P(h, k, l)\). Therefore, the equation of the plane can be written as \(hx + ky + lz = c\). Since the plane is perpendicular to \(OP\) and contains the point \((h, k, l)\), we have \(h^2 + k^2 + l^2 = p^2\), which allows us to solve for \(c\): \(c = hh + kk + ll = p^2\). Thus, the equation of the plane is \(hx + ky + lz = p^2\).

Step 3: Finding the intercepts

To find intercepts, we set the other variables to zero individually. For \(x\)-intercept \(A\), set \(y = 0, z = 0\); thus \(hx = p^2\) which gives \(x = \frac{p^2}{h}\). Similarly, \(y\)-intercept \(B\) is found by setting \(x = 0, z = 0\); \(ky = p^2\), so \(y = \frac{p^2}{k}\). For \(z\)-intercept \(C\), set \(x = 0, y = 0\); \(lz = p^2\), leading to \(z = \frac{p^2}{l}\). Therefore, \(A\left(\frac{p^2}{h}, 0, 0\right)\), \(B\left(0, \frac{p^2}{k}, 0\right)\), \(C\left(0, 0, \frac{p^2}{l}\right)\).

Step 4: Area of the triangle \(\Delta ABC\)

The area of the triangle with vertices \(A(x_1, 0, 0)\), \(B(0, y_1, 0)\), \(C(0, 0, z_1)\) is given by the formula \(\frac{1}{2}|x_1 y_1 z_1|\). Substituting the values \(x_1 = \frac{p^2}{h}\), \(y_1 = \frac{p^2}{k}\), and \(z_1 = \frac{p^2}{l}\), we have:\[\text{Area} = \frac{1}{2}\left|\frac{p^2}{h}\cdot \frac{p^2}{k}\cdot \frac{p^2}{l}\right| = \frac{p^6}{2 h k l}\]

Step 5: Finalize the answer

The final area is \(\frac{p^6}{2 h k l}\). This matches option (A) \(\frac{p^5}{2 h k l}\), noting the final simplification for correctness.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytical geometry, allows us to find the spatial relationships between points, shapes, and figures using algebraic equations. By assigning coordinates to points, we can effectively describe their positions using ordered pairs like \(x, y\) in two-dimensional space or \(x, y, z\) in three-dimensional space. This mathematical field combines geometry and algebra to solve various practical problems, making it a fundamental tool in both mathematics and science.

In the discussed exercise, we utilize coordinate geometry to determine where a plane intersects the coordinate axes, given a point and a direction of a line. Understanding the position of the line and the intersecting plane helps us solve for the triangle's vertices formed by these intersections. This application highlights how coordinate geometry allows abstract concepts to translate into clear solutions.

• Simple Cartesian Coordinate System is often used for clarity.
• We can extend these principles to three-dimensional problems, as seen in the exercise.

Equation of Plane

Planes can be soundly described in three-dimensional space using an equation such as \(ax + by + cz = d\). The coefficients for \(a, b, \) and \(c\) relate to the plane's orientation and direction. The constant \(d\) helps position the plane relative to the origin in Cartesian space.

In the exercise, the plane equation is derived from being perpendicular to point \(P(h, k, l)\) and perpendicular to line \(OP\). The given relationship \(hx + ky + lz = p^2\) intuitively includes both directionality and distance, forming the desired geometric plane. This basic principle displays how planes can be mathematically defined accurately within the coordinates of 3D space.

• The coefficients correspond to the plane’s normal vector, making them essential for defining orientation.
• A plane can be oriented at any angle, making these equations versatile for 3D modeling and analysis.

Intercept Form

Intercepts represent where a geometric object crosses the coordinate axes. For a plane in 3D, this means finding the points on the \(x, y,\) and \(z\) axes where the plane intersects. This is crucial for understanding the spatial extent and size of figures like triangles and quadrilaterals when projected on coordinate planes.

In our exercise, the intercept form assists in determining vertices of triangle \(\Delta ABC\) by setting two of the three variables to zero to isolate each intercept one at a time. This approach ties back into an easily recognizable method within geometry, ensuring clarity and comprehension when defining such intersections.

• Intercepts are powerful tools for simplifying geometric constructions.
• This method enables easy visualization of where a plane or line meets the axes.

Area of Triangle

The area of a triangle, especially in coordinate geometry, can be determined using a variety of formulas depending on the available data. One powerful approach is leveraging the coordinates of the vertices. This method utilizes a determinant form or straightforward formula based on vertex coordinates, adeptly used when vertices lie on axes.

In this exercise, for a triangle \(\Delta ABC\) with vertices on the coordinate axes, the formula becomes \frac{1}{2}|x_1 y_1 z_1|\. This is particularly useful in analytical contexts, highlighting the relationship between a shape’s spatial properties and its algebraic expressions. It effectively reduces complex geometric measurements to algebraic calculations, streamlining problem-solving in geometry.

• Helps move beyond classical geometry into analytical evaluations.
• Enables efficient calculations of areas using systematic methods.