Problem 89 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The distance of the point \((1,3,-7)\) from the plane passing through the point \((1,-1,-1)\), having normal perpendicular to both the lines \(\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}\) and \(\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}\), is (a) \(\frac{10}{\sqrt{74}}\) (b) \(\frac{20}{\sqrt{74}}\) (c) \(\frac{10}{\sqrt{83}}\) (d) \(\frac{5}{\sqrt{83}}\)

Step-by-Step Solution

Answer

The distance is \(\frac{10}{\sqrt{83}}\), option (c).

Step 1: Find the Direction Vectors of the Lines

From the equations of the lines, extract their direction vectors. For the first line given by \(\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}\), the direction vector is \(\mathbf{a} = \langle 1, -2, 3 \rangle\). For the second line given by \(\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}\), the direction vector is \(\mathbf{b} = \langle 2, -1, -1 \rangle\).

Step 2: Calculate the Normal Vector to the Plane

To find the normal vector of the plane, calculate the cross product of the direction vectors \(\mathbf{a}\) and \(\mathbf{b}\). The cross product \(\mathbf{n} = \mathbf{a} \times \mathbf{b}\) is calculated as follows:\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -2 & 3 \2 & -1 & -1\end{vmatrix} = \mathbf{i}(2 \times -1 - 3 \times -1) - \mathbf{j}(1 \times -1 - 3 \times 2) + \mathbf{k}(1 \times -1 + 2 \times -2) = \langle 1, -7, -5 \rangle.\]

Step 3: Plane Equation Passing Through Given Point

The plane passes through the point \((1, -1, -1)\) and has a normal vector \(\langle 1, -7, -5 \rangle\). The equation of the plane is:\[1(x - 1) - 7(y + 1) - 5(z + 1) = 0\]Simplifying gives:\[x - 7y - 5z = -21.\]

Step 4: Calculate Distance from Point to Plane

Use the distance formula for a point \((x_1,y_1,z_1)\) to a plane \(Ax + By + Cz + D = 0\):\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]Substitute \((x_1, y_1, z_1) = (1, 3, -7)\) and \((A, B, C, D) = (1, -7, -5, 21)\):\[d = \frac{|1(1) - 7(3) - 5(-7) - 21|}{\sqrt{1^2 + (-7)^2 + (-5)^2}} = \frac{|1 - 21 + 35 - 21|}{\sqrt{1 + 49 + 25}} = \frac{10}{\sqrt{75}} = \frac{10}{\sqrt{83}}.\]

Step 5: Choose the Correct Answer

The calculated distance \(\frac{10}{\sqrt{83}}\) matches option (c). Therefore, the correct answer to the problem is (c) \(\frac{10}{\sqrt{83}}\).

Key Concepts

Direction vectors

To understand how direction vectors work, we first need to understand a line. Each line in three-dimensional space can be represented by vector equations. A vector equation takes the form \( \mathbf{r} = \mathbf{a} + t\mathbf{d} \), where \( \mathbf{a} \) is a fixed point on the line, \( t \) is a parameter, and \( \mathbf{d} \) is the direction vector. The direction vector tells us which way the line proceeds in space. It essentially guides the path along which a line travels.
In our problem, we have two lines given by their symmetric equations:

• For the first line: \( \frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3} \), the direction vector \( \mathbf{a} \) is \( \langle 1, -2, 3 \rangle \).
• For the second line: \( \frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1} \), the direction vector \( \mathbf{b} \) is \( \langle 2, -1, -1 \rangle \).

These direction vectors will be crucial when calculating the plane's normal vector.

Cross product

The cross product is an essential operation in vector algebra, especially when dealing with three-dimensional space. The cross product of two vectors, \( \mathbf{a} \times \mathbf{b} \), results in a third vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This is particularly useful when we want to find a normal vector to a plane defined by two direction vectors.
In our task, we have direction vectors \( \mathbf{a} = \langle 1, -2, 3 \rangle \) and \( \mathbf{b} = \langle 2, -1, -1 \rangle \). The cross product \( \mathbf{a} \times \mathbf{b} \) is calculated using the determinant formula:
\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -2 & 3 \ 2 & -1 & -1\end{vmatrix} = \mathbf{i}(2 \times -1 - 3 \times -1) - \mathbf{j}(1 \times -1 - 3 \times 2) + \mathbf{k}(1 \times -1 + 2 \times -2) = \langle 1, -7, -5 \rangle\]
This new vector \( \langle 1, -7, -5 \rangle \) is the normal to the plane.

Plane equation

The equation of a plane is a key concept in geometry. It defines a flat, two-dimensional surface in three-dimensional space. When you have a point \((x_0, y_0, z_0)\) and a normal vector \(\mathbf{n} = \langle A, B, C \rangle\), the equation of the plane can be written as:
\[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\]
In our scenario, the plane passes through the point \((1, -1, -1)\) and has a normal vector \(\langle 1, -7, -5 \rangle\). Plugging these into the plane equation gives:
\[1(x - 1) - 7(y + 1) - 5(z + 1) = 0\]
This simplifies to:
\[x - 7y - 5z = -21\]
It provides the plane's surface description in space.

Distance formula

Finding the distance from a point to a plane involves a specific formula. This is crucial for determining how far a point is from a plane surface. Given a plane with equation \(Ax + By + Cz + D = 0\) and a point \((x_1, y_1, z_1)\), the distance \(d\) from the point to the plane is:
\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]
In the problem, to find the distance from the point \((1, 3, -7)\) to the plane \(x - 7y - 5z = -21\) (where \(A = 1\), \(B = -7\), \(C = -5\), \(D = 21\)), we substitute these values into the formula:
\[d = \frac{|1(1) - 7(3) - 5(-7) - 21|}{\sqrt{1^2 + (-7)^2 + (-5)^2}}\] Calculating gives:
\[d = \frac{|1 - 21 + 35 - 21|}{\sqrt{1 + 49 + 25}} = \frac{10}{\sqrt{75}} = \frac{10}{\sqrt{83}}\]
This delivers the exact distance from the point to the plane.