Problem 52 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The foot of the perpendicular drawn from the point \((4,2,3)\) to the line joining the points \((1,-2,3)\) and \((1,1,0)\) lies on the plane: [Sep. 03, 2020 (I)] (a) \(2 x+y-z=1\) (b) \(x-y-2 z=1\) (c) \(x-2 y+z=1\) (d) \(x+2 y-z=1\)

Step-by-Step Solution

Answer

The foot of the perpendicular lies on the plane (c) \( x-2y+z=1 \).

Step 1: Identify the direction vector of the line

The line is determined by the points \( (1, -2, 3) \) and \( (1, 1, 0) \). Calculate the direction vector by subtracting the coordinates of the first point from the second: \( (1-1, 1-(-2), 0-3) = (0, 3, -3) \).

Step 2: Find the vector from a point on the line to the given point

Use the point \( (1, -2, 3) \) on the line. Calculate the vector to the point \( (4, 2, 3) \): \( (4-1, 2-(-2), 3-3) = (3, 4, 0) \).

Step 3: Find the perpendicular vector

The foot of the perpendicular is found where the projection of the vector \( (3, 4, 0) \) onto the direction vector \( (0, 3, -3) \) is zero. Calculate the cross product to ensure orthogonality.

Step 4: Determine if the foot of the perpendicular lies on given planes

Check each plane by using potential coordinates that satisfy the equations in step 3. Plug them into the equations: \(2x+y-z=1\), \(x-y-2z=1\), \(x-2y+z=1\), \(x+2y-z=1\).

Step 5: Validate which plane contains the foot of the perpendicular

Substitute the perpendicular foot coordinates obtained into the plane equations. The correct option will satisfy the equation exactly.

Key Concepts

Direction Vector

In coordinate geometry, understanding the direction vector is crucial for analyzing lines in space. A direction vector indicates the orientation of a line, based on the difference between two points on it. For a line passing through points \( (1, -2, 3) \) and \( (1, 1, 0) \), we calculate the direction vector by subtracting coordinates:

• \( x \)-component: \((1 - 1) = 0\)
• \( y \)-component: \((1 - (-2)) = 3\)
• \( z \)-component: \((0 - 3) = -3\)

This gives us the direction vector \( (0, 3, -3) \). It tells us that the line strides parallel to the y-axis and has a negative trend along the z-axis. By understanding this vector, we can determine how this line behaves in a three-dimensional space.

Perpendicular Foot

The perpendicular foot from a point to a line or plane is the closest point on that line or plane. Imagine dropping a perpendicular from the point to land exactly on the line or plane.To find this point when given a point like \( (4, 2, 3) \) and a line with a direction vector of \( (0, 3, -3) \), we calculate the projection of the vector connecting a given point to a line onto the direction vector.

• Start with the vector from the line point \( (1, -2, 3) \) to the given point: \( (3, 4, 0) \).
• The perpendicular foot exists where the projection of this vector is zero.

This ensures that the line from the point to the foot is perfectly perpendicular to the original line.

Projection

Projection plays a key role in finding perpendicularly closest points to lines and planes. In vector terms, projecting one vector onto another reduces finding a piece of the first that is parallel to the second. For our case, we project the vector \( (3, 4, 0) \) onto the direction vector \( (0, 3, -3) \):

• Perform a dot product of both vectors.
• Divide by the magnitude squared of the direction vector.
• Multiply the result by the direction vector components.

Through this projection, we assure that our perpendicular vector is orthogonal, meaning it forms a right angle with the direction vector at the perpendicular foot. This step is crucial for ensuring correct geometric relations in space.

Plane Equation

Plane equations allow us to represent flat surfaces in 3D space mathematically. These equations typically take the form \( ax + by + cz = d \), where \(a\), \(b\), and \(c\) are coefficients that describe the plane's orientation.In our scenario, the foot of the perpendicular, the closest point on the line to a point, needs to be checked against several plane equations to identify which plane it lies upon. By substituting the coordinates of the perpendicular foot into these equations:

• \(2x + y - z = 1\)
• \(x - y - 2z = 1\)
• \(x - 2y + z = 1\)
• \(x + 2y - z = 1\)

The correct plane equation is the one where this substitution holds true. This validation process is essential for problems involving projections onto planes, as it confirms the precise location of the perpendicular foot.