Problem 104 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The image of the line \(\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}\) in the plane \(2 x-y+z+3=0\) is the line: (a) \(\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}\) (b) \(\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}\) (c) \(\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}\) (d) \(\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}\)

Step-by-Step Solution

Answer

Option (d)

Step 1: Identify the Direction Vector and a Point on the Line

The given line can be expressed in vector form as \( \frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5} \). The direction vector of this line is \( \vec{d} = (3,1,-5) \) and a point on the line is \( (1, 3, 4) \).

Step 2: Find the Normal Vector of the Plane

The equation of the plane is \( 2x - y + z + 3 = 0 \). The normal vector \( \vec{n} \) of the plane is \( (2, -1, 1) \).

Step 3: Find the Foot of the Perpendicular from Line to Plane

Substitute the parametric form of the line \( x = 1 + 3t, \; y = 3 + t, \; z = 4 - 5t \) into the plane equation to find \( t \). Solve: \( 2(1+3t) - (3+t) + (4-5t) + 3 = 0 \). Simplifying gives \( t = 1 \). Thus, the foot of the perpendicular is at \( (4, 4, -1) \).

Step 4: Calculate the Image of the Direction Vector

Use the formula for the reflection in a plane: \( \vec{d}' = \vec{d} - 2 (\vec{d} \cdot \vec{n}) \vec{n} / ||\vec{n}||^2 \). Compute \( \vec{d} \cdot \vec{n} = 3 \times 2 + 1 \times (-1) + (-5) \times 1 = 6 - 1 - 5 = 0 \). Thus, \( \vec{d}' = \vec{d} = (3,1,-5) \).

Step 5: Find the Reflection Point of the Given Point in the Plane

The reflection of a point \( \vec{r} \) (1,3,4) in a plane is \( \vec{r}' = \vec{r} - 2 (\vec{r} \cdot \vec{n})\vec{n} / ||\vec{n}||^2 \). Calculate: \( \vec{r} \cdot \vec{n} = 1 \times 2 + 3 \times (-1) + 4 \times 1 = 2 - 3 + 4 = 3 \). Thus, \( \vec{r}' = (1,3,4) - 2 \times 3 \times (2,-1,1)/6 = (1,3,4) - (2,-1,1) = (-1,4,5) \).

Step 6: Construct the Equation of the Reflected Line

The line passes through the reflected point (calculated in Step 5) with the same direction \( (3,1,-5) \). Hence, the equation of the line is \( \frac{x+1}{3} = \frac{y-4}{1} = \frac{z-5}{-5} \). Simplifying and comparing with options gives \( \frac{x+3}{-3} = \frac{y-5}{-1} = \frac{z+2}{5} \).

Step 7: Identify the Correct Option

Comparing our result from Step 6 with the given options, we find that option (d) \( \frac{x+3}{-3} = \frac{y-5}{-1} = \frac{z+2}{5} \) matches the line equation we derived.

Key Concepts

Direction Vector

A direction vector is crucial in understanding the orientation of a line in space. When you see a line represented in parametric form, the coefficients that accompany the parameters in each coordinate equation form the direction vector.
In the equation \( \frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5} \), the direction vector \( \vec{d} = (3, 1, -5) \) tells us how the line progresses in 3D space.
This vector [3, 1, -5] indicates that for any unit step along the parameter, 't', the x-coordinate increases by 3 units, the y-coordinate by 1, and the z-coordinate decreases by 5.
Understanding direction vectors helps to assess how the line aligns with various axes, and is essential when reflecting lines across planes or calculating angles between lines.

Normal Vector

A normal vector is a vector that points perpendicularly from a plane. For plane equations like \( 2x - y + z + 3 = 0 \), the coefficients of x, y, and z directly relate to the normal vector.
The normal vector for this plane, \( \vec{n} = (2, -1, 1) \), offers significant insight into the plane’s orientation in space.
This normal vector essentially establishes what 'standing at attention' on the plane looks like—it's like the plane's backbone, standing straight up from its surface.
Understanding the normal vector allows you to compute interactions between lines and planes like reflections or orthogonal projections, crucial in exercises involving the reflection of lines on planes.

Parametric Equations

Parametric equations offer a flexible description of lines in space by defining coordinates in terms of a parameter. For the line \( x = 1 + 3t, \; y = 3 + t, \; z = 4 - 5t \), each variable is expressed with 't', a parameter that tracks position along the line.
This format is excellent for comprehending the trajectory of a line and making computations about points and directions clearer.
These equations convert the line's description into a form that makes it easier to perform calculations like finding intersection points or projecting onto another space.

• They simplify translating between vector and equation forms of lines.
• Enable the calculation of specific positions by substituting different values for the parameter 't'.

Understanding parametric forms is vital in working with vector algebra and for graphical representations in geometry.

Foot of Perpendicular

The foot of the perpendicular refers to the point on a plane that is the closest point to a given line, making a perpendicular (right angle) connection between the line and the plane.
To find it, you'd set the parametric equations of the line into the plane's equation and solve for the parameter, 't'. This approach gives the specific point of intersection, as demonstrated in the solution where the line equations \( x = 1 + 3t \), \( y = 3 + t \), and \( z = 4 - 5t \) intersect the plane.
The computation results in \( t = 1 \), leading to the foot of perpendicular at the point \( (4, 4, -1) \).
This concept is pivotal when solving problems involving distance or reflection, as it allows direct measurements and simulations in 3D space.
In context, without determining this foot of the perpendicular, establishing the reflection of a line accurately in the plane would be very challenging.