Problem 24 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The length of the perpendicular from the point \((2,-1,4)\) on the straight line, \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}\) is [April \(\mathbf{0 8}, \mathbf{2 0 1 9}\) (I)] (a) greater than 3 but less than 4 (b) less than 2 (c) greater than 2 but less than 3 (d) greater than 4

Step-by-Step Solution

Answer

The distance is greater than 2 but less than 3 (option (c)).

Step 1: Find the Direction Ratios of the Line

The given line equation is \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=k\). The line can be described parametrically as: \(x = 10k - 3, y = -7k + 2, z = k\). The direction ratios of the line are \(10, -7, 1\).

Step 2: Find a Point on the Line

Choose a value of \(k\) to find a point on the line. For simplicity, take \(k = 0\). Substituting into the parametric equations gives the point on the line as \((-3, 2, 0)\).

Step 3: Write the Vector from the Point to the Line's Point

Let point \( A(2, -1, 4) \) be the point from which we need to find the perpendicular distance to the line. The vector from point \(A\) to point \(B(-3, 2, 0)\) is \((2 - (-3), -1 - 2, 4 - 0) = (5, -3, 4)\).

Step 4: Calculate the Cross Product for Perpendicular Distance

To find the perpendicular distance, calculate the cross product between the direction ratio vector of the line \(\langle 10, -7, 1 \rangle\) and the vector from \(A\) to \(B\): \[\vec{AB} \times \text{{direction vector}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -3 & 4 \ 10 & -7 & 1 \end{vmatrix}\]This results in \( (-27, 30, 5) \).

Step 5: Magnitude of the Cross Product

Calculate the magnitude of the cross product: \[\|\vec{AB} \times \text{{direction vector}}\| = \sqrt{(-27)^2 + 30^2 + 5^2}\]This results in \(\sqrt{944}\).

Step 6: Magnitude of the Direction Ratios

Calculate the magnitude of the direction ratios: \[\|\langle 10, -7, 1 \rangle\| = \sqrt{10^2 + (-7)^2 + 1^2} = \sqrt{150}\]The simplified result is \(5\sqrt{6}\).

Step 7: Calculate the Perpendicular Distance

Use the formula for the perpendicular distance of a point from a line:\[\text{Distance} = \frac{\|\vec{AB} \times \text{direction vector}\|}{\|\text{direction vector}\|} = \frac{\sqrt{944}}{5\sqrt{6}} \]Simplifying, we get the distance as approximately 2.90.

Key Concepts

3D coordinate geometry

In 3D coordinate geometry, each point is represented by three coordinates: \(x, y, z\). These coordinates describe the point's position in a three-dimensional space. For example, the point \(2, -1, 4\) is located 2 units along the x-axis, -1 units along the y-axis, and 4 units along the z-axis.
3D geometry encompasses various objects such as lines, planes, and surfaces. It also involves calculating distances and angles between these elements. In the given exercise, we determine the perpendicular distance from a point to a line, which is a common geometric problem in 3D space.
The relationships between points and lines in 3D geometry can often be expressed using vectors and mathematical equations. Understanding these relationships is crucial for solving problems involving distances and intersections.

Direction ratios

Direction ratios are a set of numbers that are proportional to the direction cosines of a line. They represent the line's orientation in space. For the line given in the equation \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}\), the direction ratios are \(10, -7, 1\).
These values show the direction of the line in relation to each axis, providing essential information on how the line is situated in 3D space. Knowing the direction ratios helps in finding the direction vector of the line. The direction vector is crucial when calculating the perpendicular or shortest distance from a point to a line.
Understanding direction ratios is essential for problems involving vector algebra and geometry, as it allows one to describe lines not only in terms of points but also orientation, which plays a vital role in several geometric calculations.

Parametric equations

Parametric equations are a way of expressing the coordinates of the points that make up a geometric object using a parameter. In the line \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=k\), the equations \(x = 10k - 3, y = -7k + 2, z = k\) are parametric forms of the line equation.
This means each coordinate is expressed in terms of \(k\), a parameter that can take various real numbers, generating different points on the line. By choosing specific values for \(k\), one can find specific points on the line.
Parametric equations are powerful tools in both 2D and 3D geometry because they allow for easy manipulation and calculation of points along a geometric entity. They are particularly useful in computer graphics, physics simulations, and any context requiring a dynamic representation of geometric objects.

Cross product

The cross product is a vector operation used to find a vector perpendicular to two given vectors. It is only defined in three-dimensional space. In the problem, we find the cross product of the direction vector of the line \(\langle 10, -7, 1 \rangle\) and the vector from point \(A(2, -1, 4)\) to point \(B(-3, 2, 0)\), resulting in \((-27, 30, 5)\).
To compute the cross product, one can use the determinant of a 3x3 matrix, as shown in the original solution. The result gives us a vector that is perpendicular to both initial vectors, crucially helping in calculating the perpendicular distance from the point to the line.
The magnitude of this cross product vector, combined with the understanding of the direction ratios of the line, is used to determine the exact perpendicular distance. Cross products are extensively used in physics, engineering, and computer graphics, whenever an understanding of rotational force or perpendicular vectors is necessary.