Problem 72 - A Complete Resource Book in Mathematics for JEE Main

Question

The distance of the point \((1,0,2)\) from the point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane \(x-y+z=16\), is: (A) 8 (B) \(3 \sqrt{21}\) (C) 13 (D) \(2 \sqrt{14}\)

Step-by-Step Solution

Answer

The distance is 13, corresponding to option (C).

Step 1: Parametrize the Line

The given line equation is \( \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t \). Parametrize this line so that it passes through the point (2, -1, 2) and has direction ratios (3, 4, 12). Thus, \( x=3t+2 \), \( y=4t-1 \), \( z=12t+2 \).

Step 2: Substitute into Plane Equation

The plane equation is \( x-y+z=16 \). Substitute the parametric equations into the plane equation: \( (3t+2) - (4t-1) + (12t+2) = 16 \).

Step 3: Simplify and Solve for t

Simplify the expression: \( 3t + 2 - 4t + 1 + 12t + 2 = 16 \), which reduces to \( 11t + 5 = 16 \). Solve for \( t \) to get \( 11t = 11 \), hence \( t = 1 \).

Step 4: Find Intersection Point

Substitute \( t = 1 \) back into the parametric equations: \( x = 3(1) + 2 = 5 \), \( y = 4(1) - 1 = 3 \), \( z = 12(1) + 2 = 14 \). So the point of intersection is (5, 3, 14).

Step 5: Calculate Distance

Find the distance from the point (1, 0, 2) to the intersection point (5, 3, 14) using the distance formula: \( \sqrt{(5-1)^2 + (3-0)^2 + (14-2)^2} = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13. \)

Step 6: Identify the Correct Answer

The calculated distance of 13 corresponds to option (C). Hence, the correct answer is 13.

Key Concepts

Distance Formula

The distance formula is a crucial concept in coordinate geometry. It's used to determine the distance between two points in a plane or 3-dimensional space. The formula helps translate the Pythagorean theorem into a usable form. It is expressed as:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]Here,

• \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the coordinates of the two points we're measuring.
• \(D\) is the distance between those points.

This versatile formula is your go-to tool for measuring straight-line distances, no matter the dimensions. It's a straightforward computation, yet it underpins many geometric and physical applications, offering a bridge between abstract math and practical problems.

Parametric Equations

Understanding parametric equations will expand your grasp of geometry. These equations represent a line or curve through a set of equations, each defining one of the coordinates. In the context of a line, you might encounter equations such as:- \(x = at + b\)- \(y = ct + d\)- \(z = et + f\)Here,

• \(t\) is the parameter, which can be any real number.
• The coefficients \(a, c, e\) define the direction ratios of the line.
• The constants \(b, d, f\) specify a point that the line passes through.

These equations are powerful as they allow you to manipulate lines and curves in 3-D space by varying the parameter \(t\). This approach simplifies finding intersections and tangents, and it models motion smoothly and naturally.

Plane Equation

Plane equations represent a flat surface extending infinitely in three-dimensional space. They are crucial in solving geometric problems involving lines and points in space. The general form is:\[ax + by + cz = d\]Where

• \(a, b,\) and \(c\) are constants that define the plane's orientation (often derived from the normal vector).
• \(x, y,\) and \(z\) represent any point located on the plane.
• \(d\) is a constant that positions the plane in space.

Planes are useful in dividing space or finding intersections with lines. For example, substituting parametric line equations into the plane equation helps identify where a line crosses the plane. This method plays a significant role in computer graphics and calculus, sending a gentle nod to the wonderful interactions in the geometry world.

Line Intersection

In geometry, finding where a line intersects a plane is a process that brings together line and plane equations. This intersection can provide specific points that solve complex spatial problems. To find an intersection:1. Start with both the line's parametric equations and the plane's equation.2. Substitute the line's equations into the plane's, then solve for the parameter \(t\).3. Use the value of \(t\) to find the coordinates of the intersection point by plugging it back into the line's equations.An intersection point confirms where a line meets the plane, revealing the particular coordinate satisfying both geometrical entities. This is essential in fields such as engineering, computer animation, and architectural design, providing the grounded reality that many abstract and practical models rely upon.