Problem 101 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A plane containing the point \((3,2,0)\) and the line \(\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}\) also contains the point: [Online April 11, 2015] (a) \((0,3,1)\) (b) \((0,7,-10)\) (c) \((0,-3,1)\) (d) \(0,7,10\)

Step-by-Step Solution

Answer

None of the listed points lie on the plane as none satisfy the equation.

Step 1: Find a Point on the Line

First, determine a point on the line \( \frac{x-1}{1} = \frac{y-2}{5} = \frac{z-3}{4} \). If we set the parameter equal to 0, then \( x = 1 \), \( y = 2 \), and \( z = 3 \). Therefore, a point on the line is \((1, 2, 3)\).

Step 2: Calculate Direction Vector of the Line

The direction vector of the line is based on the denominator of the ratios, which gives us the vector \( \vec{d} = (1, 5, 4) \).

Step 3: Find the Plane Equation

To find the plane equation, compute the cross product of vectors \( (1,5,4) \) (direction vector of the line) and the vector from point \((3,2,0)\) to \((1,2,3)\), which is \((1-3,2-2,3-0) = (-2,0,3)\). The cross product \( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 5 & 4 \ -2 & 0 & 3 \end{vmatrix} \) results in \( \vec{n} = (15, -10, 10) \).

Step 4: Equation of the Plane

The general form of a plane equation is \( ax + by + cz = d \) where \( \vec{n} = (a, b, c) \). Using \( \vec{n} = (15, -10, 10) \) and a point on the plane \((3, 2, 0)\), substitute into the plane equation: \[ 15(x-3) - 10(y-2) + 10(z-0) = 0 \]Expanding gives: \(15x - 10y + 10z - 55 = 0\).

Step 5: Verify Which Point Lies on the Plane

Check each point by substituting into the plane equation. For \((0, 3, 1)\):\[ 15(0) - 10(3) + 10(1) = 0 \quad \rightarrow \quad -30 + 10 = -20 eq 55 \]For \((0, 7, -10)\):\[ 15(0) - 10(7) + 10(-10) = -70 - 100 = -170 eq 55 \]For \((0, -3, 1)\):\[ 15(0) - 10(-3) + 10(1) = 30 + 10 = 40 eq 55 \]For \((0, 7, 10)\):\[ 15(0) - 10(7) + 10(10) = -70 + 100 = 30 eq 55 \]

Key Concepts

Planes

A plane in geometry is a flat, two-dimensional surface that extends infinitely in all directions. It is defined by any one of the following:

• A point and a normal vector
• Three non-collinear points
• A line and a point not on the line

In this exercise, the plane is defined by a point and a line. A critical concept for understanding planes is the normal vector, which is perpendicular to the plane and crucial for forming the plane equation. The equation of the plane can be written in the form: \[ ax + by + cz = d \]where \((a, b, c)\) is the normal vector. This equation is derived by ensuring that any arbitrary point \((x, y, z)\) satisfies the condition that the vector from a known point on the plane to \((x, y, z)\) is perpendicular to the normal vector. This perpendicular relation is used to derive the coefficients of the plane equation.

Vector Algebra

Vector algebra is essential for solving problems in 3D geometry. Vectors are mathematical objects with both magnitude and direction and can represent various quantities, such as displacement or velocity. The direction vector describes the orientation of a line in space.In this problem, the direction vector of the line is \( \vec{d} = (1, 5, 4) \), extracted directly from the line's equation. Vector operations such as addition, scalar multiplication, and finding cross products are fundamental. The cross product, in particular, is vital here because it is used to find a normal vector to the plane from two vectors: one derived from the direction component of the line, and another connecting a point on the line to any other point on the plane. Calculating the cross product provides a vector perpendicular to both, hence, normal to the plane.

Line and Plane Equations

The equation of a line in 3D is typically expressed in parametric form: \[ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} \]here, \((x_0, y_0, z_0)\) is a point on the line, and \((a, b, c)\) is the direction vector. To derive the equation of the plane containing this line, we need a starting point, the direction vector as well as any additional point known to lie on the plane.For a valid plane equation like \[ 15x - 10y + 10z = 55 \] we verify any potential point by substituting it back into the equation. If both sides of the equation balance, the point lies on the plane. Verification involves substituting \((x, y, z)\) values of each given option into the plane equation and checking for equality. This process is crucial because only points that solve the plane equation correctly are contained within it.