Problem 124 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The line passing through the points \((5,1, a)\) and \((3, b, 1)\) crosses the yz-plane at the point \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\). Then \([2008]\) (a) \(a=2, b=8\) (b) \(a=4, b=6\) (c) \(a=6, b=4\) (d) \(a=8, b=2\)

Step-by-Step Solution

Answer

The correct values are \(a=6\) and \(b=4\), option (c).

Step 1: Identify the Direction Vector

To find the equation of a line through two points, we first need to determine the direction vector. The direction vector \( \vec{d} \) from point \((5,1,a)\) to \((3,b,1)\) is calculated as follows: \[ \vec{d} = (3-5, b-1, 1-a) = (-2, b-1, 1-a) \].

Step 2: Equation of the Line in Parametric Form

The parametric equations for the line can be expressed as: \[ x = 5 - 2t, \] \[ y = 1 + (b-1)t, \] \[ z = a + (1-a)t \]. This uses the point \((5,1,a)\) and the direction vector \(\vec{d}=(-2,b-1,1-a)\).

Step 3: Find the Point where the Line Crosses the yz-plane

The line crosses the yz-plane when \(x = 0\). Setting \(x = 5 - 2t = 0\), we solve for \(t\):\[ t = \frac{5}{2} \].

Step 4: Substitute \(t\) to Find \(y\) and \(z\) Coordinates

Using \(t = \frac{5}{2}\), substitute into the parametric equations for \(y\) and \(z\):\[ y = 1 + (b-1)\left(\frac{5}{2}\right) \] \[ z = a + (1-a)\left(\frac{5}{2}\right) \].

Step 5: Equate \(y\) and \(z\) to Given Coordinates

Equating to \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\), we set:\[ 1 + (b-1)\left(\frac{5}{2}\right) = \frac{17}{2} \] \[ a + (1-a)\left(\frac{5}{2}\right) = \frac{-13}{2} \].

Step 6: Solve for \(b\)

Solving \[ 1 + \frac{5}{2}(b-1) = \frac{17}{2} \]:1. Express the left side: \[ 1 + \frac{5}{2}b - \frac{5}{2} = \frac{-3}{2} + \frac{5}{2}b \].2. Equate to \(\frac{17}{2}\):\[ \frac{-3}{2} + \frac{5}{2}b = \frac{17}{2} \].3. Solve for \(b\):\[ \frac{5}{2}b = \frac{20}{2} \Rightarrow 5b = 20 \Rightarrow b = 4 \].

Step 7: Solve for \(a\)

Solving \[ a + (1-a)\left(\frac{5}{2}\right) = \frac{-13}{2} \]:1. Express the left side: \[ a + \frac{5}{2} - \frac{5}{2}a \].2. Simplify and equate: \[ \frac{2a - 5a + 5}{2} = \frac{-13}{2} \Rightarrow -\frac{3a}{2} + \frac{5}{2} = \left(\frac{-13}{2}\right) \].3. Solve for \(a\): \[-3a + 5 = -13 \Rightarrow -3a = -13 - 5 \Rightarrow -3a = -18 \Rightarrow a = 6 \].

Step 8: Verify the Solution

The solutions \(a = 6\) and \(b = 4\) match option (c). Substitute back into the line equation to ensure correctness, confirming this choice with the given yz-plane crossing point.

Key Concepts

Direction Vector

When finding the equation of a line that passes through two given points, an essential component is determining the direction vector. This vector effectively "points the way" from one point to the other. In our example, the direction vector \( \vec{d} \) is calculated using two points: \((5,1,a)\) and \((3,b,1)\).
The formula to find the direction vector is:

• Subtract the x-coordinates: \(3 - 5\) which gives \(-2\)
• Subtract the y-coordinates: \(b - 1\)
• Subtract the z-coordinates: \(1 - a\)

Putting these together, we arrive at the direction vector, \( \vec{d} = (-2, b-1, 1-a) \). This vector helps form the basis for expressing the line's equation in a linear form, showing how the line extends through the 3D space from those two points.

Parametric Form

To describe a line in three-dimensional space, parametric equations come quite handy. These equations express the coordinates \(x\), \(y\), and \(z\) as functions of a parameter \(t\), which ideally traverses the entire line. Given our direction vector \( \vec{d} = (-2, b-1, 1-a) \) and a point the line passes through \((5,1,a)\), we set up the following parametric equations:

• \( x = 5 - 2t \)
• \( y = 1 + (b-1)t \)
• \( z = a + (1-a)t \)

These equations are significant as they translate the geometric representation of a line into algebraic expressions, allowing us to calculate specific coordinates along the line by varying \(t\). This form is incredibly flexible and useful in distinguishing various points the line reaches, especially when solving for specific conditions like crossing the yz-plane.

yz-plane Intersection

The yz-plane is a specific plane in the coordinate system where the x-coordinate is always zero. Therefore, a line crosses the yz-plane when \(x = 0\). To find this special point of intersection for our line, we use the parametric equation for \(x\) which is \( x = 5 - 2t \).
Setting this equation to zero, \(5 - 2t = 0\), we can solve for the parameter \(t\):
\[ t = \frac{5}{2} \]
With this value of \(t\), we substitute it back into the parametric equations for \(y\) and \(z\):

• \( y = 1 + (b-1)\left(\frac{5}{2}\right) \)
• \( z = a + (1-a)\left(\frac{5}{2}\right) \)

These calculations will yield the \(y\) and \(z\) coordinates of the yz-plane intersection point, allowing us to determine \(a\) and \(b\) through given constraints, ensuring that our line meets precisely at the provided point on this plane, namely \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\).