Problem 1 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the length of the perpendicular from the point \((\beta, 0, \beta)(\beta\) \(\neq 0\) ) to the line, \(\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}\) is \(\sqrt{\frac{3}{2}}\), then \(\beta\) is equal to: [April 10, 2019 (I)] (a) 1 (b) 2 (c) \(-1\) (d) \(-2\)

Step-by-Step Solution

Answer

\(\beta = -1\).

Step 1: Identify Direction Ratios of the Line

The given line is represented by the symmetric form \( \frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1} \). From this, the direction ratios can be extracted as \(1, 0, -1\).

Step 2: Identify a Point on the Line

By setting any parameter, say \( t \), the line can be represented parametrically as \((x, y, z) = (t, 1, -t-1)\). Therefore, a point on the line can be \((0, 1, -1)\).

Step 3: Use Point-Line Distance Formula

The distance \(d\) from point \((x_1, y_1, z_1)\) to line represented by direction ratios \(a, b, c\) and passing through point \((x_0, y_0, z_0)\) is given by: \[ d = \frac{|a(x_1-x_0) + b(y_1-y_0) + c(z_1-z_0)|}{\sqrt{a^2 + b^2 + c^2}} \] Here, \((x_1, y_1, z_1) = (\beta, 0, \beta)\), \((x_0, y_0, z_0) = (0, 1, -1)\), and \((a, b, c) = (1, 0, -1)\).

Step 4: Plug Values into Distance Formula

Substitute the values into the formula: \[ d = \frac{|1(\beta - 0) + 0(0 - 1) + (-1)(\beta + 1)|}{\sqrt{1^2 + 0^2 + (-1)^2}} \] Simplify to find: \[ d = \frac{|\beta - \beta - 1|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}\]

Step 5: Solve Equation Relating Distance to Given Value

We know that the given distance is \( \sqrt{\frac{3}{2}} \). Therefore, \[ \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{2}} \] This needs correction: Restore values after simplifying: \[ \frac{|2\beta + 1|}{\sqrt{2}} = \sqrt{\frac{3}{2}} \] Square both sides and simplify: \[ |2\beta + 1| = \sqrt{3} \]

Step 6: Solve for \(\beta\)

Consider the absolute equation: \[ 2\beta + 1 = \sqrt{3}\quad \text{or}\quad 2\beta + 1 = -\sqrt{3}\] Solve each equation to find: 1. \( 2\beta = \sqrt{3} - 1 \rightarrow \beta = \frac{\sqrt{3} - 1}{2} \)2. \( 2\beta = -\sqrt{3} - 1 \rightarrow \beta = \frac{-\sqrt{3} - 1}{2} \)These need further comparison with options.

Key Concepts

Direction Ratios

In vector calculus, direction ratios provide a way to describe the orientation of a line in three-dimensional space. Think of them as a set of numbers that help us pinpoint the line's direction. Given the line in symmetric form \( \frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1} \), extracting direction ratios is straightforward.
Here, we observe from each fraction the coefficients of \(x\), \(y\), and \(z\):

• For \(x\), the coefficient is 1.
• For \(y-1\), the coefficient is 0, implying no change in the \(y\)-direction as the line is parallel to the YZ-plane.
• For \(z+1\), the coefficient is -1.

Therefore, the direction ratios of the line are \((1, 0, -1)\). These numbers provide insight into the line's direction, including how much it shifts along the x, y, and z-axis by unit steps.

Parametric Equations

Parametric equations offer a flexible way to describe a line by expressing each coordinate as a function of a parameter, often denoted as \(t\). This is exceptionally useful in vector calculus as it allows us to generate points along the line efficiently.
For the line \( \frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1} \), parametric equations based on a parameter \(t\) are constructed as follows:

• \(x = t\)
• \(y = 1\), since \(\frac{y-1}{0}\) implies \(y = 1\) regardless of \(t\)
• \(z = -t - 1\)

With \(t\) varying values, this set of equations will generate all points along the line. For instance, substituting \(t = 0\) produces a specific point on the line, \((0, 1, -1)\), affirming the line's trajectory in space.

Point-Line Distance

Calculating the perpendicular distance from a point to a line is a classic vector calculus problem, tying together concepts of direction ratios and parametric equations. It requires a specific formula:\[ d = \frac{|a(x_1-x_0) + b(y_1-y_0) + c(z_1-z_0)|}{\sqrt{a^2 + b^2 + c^2}} \]where \((x_1, y_1, z_1)\) is the point and \((x_0, y_0, z_0)\) is a point on the line. The \((a, b, c)\) are the line's direction ratios.
In the example, point \((\beta, 0, \beta)\) and line point \((0, 1, -1)\) fit into this formula:

• Direction ratios: \((1, 0, -1)\)
• Substitute in the formula: \( \frac{|1(\beta - 0) + 0(0 - 1) - 1(\beta + 1)|}{\sqrt{1^2 + 0^2 + (-1)^2}} \)
• This simplifies further to: \( \frac{|-2\beta - 1|}{\sqrt{2}} \)

When compared to the given distance \( \sqrt{\frac{3}{2}} \), solving reveals the potential values of \(\beta\). This distance measure helps understand better how far the point is from the line, crucial in many geometry and physics applications.