Problem 54 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let a plane \(P\) contain two lines \(\vec{r}=\hat{i}+\lambda(\hat{i}+\hat{j}), \lambda \in \mathbf{R}\) and \(\vec{r}=-\hat{j}+\mu(\hat{j}-\hat{k}), \mu \in \mathbf{R}\) If \(Q(\alpha, \beta, \gamma)\) is the foot of the perpendicular drawn from the point \(M(1,0,1)\) to \(P\), then \(3(\alpha+\beta+\gamma)\) equals

Step-by-Step Solution

Answer

5

Step 1: Identify Plane's Normal Vector

The plane contains the given lines \( \vec{r}=\hat{i}+\lambda(\hat{i}+\hat{j}) \) and \( \vec{r}=-\hat{j}+\mu(\hat{j}-\hat{k}) \). To find a normal vector to the plane, we find the direction vectors of these lines: \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = \hat{j} - \hat{k} \). The cross product \( \vec{a} \times \vec{b} \) gives us the normal vector \( \vec{n} \) for the plane.

Step 2: Calculate Cross Product

Calculate \( \vec{a} \times \vec{b} \) where \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = \hat{j} - \hat{k} \). Using the determinant method: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 0 & 1 & -1 \end{vmatrix} = \hat{i} (1) - \hat{j} (1) + \hat{k} (1) = \hat{i} - \hat{j} + \hat{k}. \] So, the normal vector \( \vec{n} = \hat{i} - \hat{j} + \hat{k}. \)

Step 3: Formulate Plane Equation

Using the normal vector \( \hat{i} - \hat{j} + \hat{k} \) and a point on one of the lines, say \( \vec{r}_0 = \hat{i} \), the equation is: \[ (\hat{i} - \hat{j} + \hat{k}) \cdot ((x, y, z) - (1, 0, 0)) = 0. \] Simplifying, we get: \[ (x - 1) - y + z = 0 \] which simplifies further to \( x - y + z = 1. \)

Step 4: Find Perpendicular Foot Point

The foot of the perpendicular from point \( M(1,0,1) \) onto the plane \( x-y+z=1 \) is found using the formula for projection of a point onto a plane: \[ d = \frac{(1-1)-(0)+1}{1^2 + (-1)^2 + 1^2} = \frac{1}{3}. \] The coordinates of the foot point \( Q(\alpha, \beta, \gamma) \) are obtained by shifting \( M \) along the normal vector viz.: \( Q = \left(1-\frac{1}{3}, 0+\frac{1}{3}, 1-\frac{1}{3}\right) = \left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right). \)

Step 5: Calculate Expression

Finally, calculate the required expression: \[ 3(\alpha + \beta + \gamma) = 3\left(\frac{2}{3} + \frac{1}{3} + \frac{2}{3}\right) = 3 \times \frac{5}{3} = 5. \]

Key Concepts

Vector Algebra

Vector Algebra is a branch of mathematics that deals with quantities having both magnitude and direction. It is useful for describing objects in physics and engineering, such as forces and velocities. In 3D geometry, vector algebra helps to simplify the representation and manipulation of geometrical shapes.

Vectors are often represented with components such as \( \hat{i}, \hat{j}, \hat{k} \), which correspond to the x, y, and z-axis respectively. These components indicate the vector's direction in a 3-dimensional space. For example, if we have a vector \( \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} \), then 'a', 'b', and 'c' are the scalar components of the vector in each of the three dimensions.

In the context of the given exercise, vector algebra is used to represent the direction of lines within a plane. By expressing these lines in terms of vectors, we can find relationships between them, such as whether they are parallel or intersecting.

Cross Product

The Cross Product is an operation on two vectors in three-dimensional space, producing another vector that is perpendicular to the plane containing the original vectors. In vector algebra, it's denoted as \( \vec{a} \times \vec{b} \), where \( \vec{a} \) and \( \vec{b} \) are vectors.

The result of the cross product points in a direction given by the right-hand rule. It is often used to find a vector normal (perpendicular) to a surface. The magnitude of the cross product vector equals the area of the parallelogram that the original vectors span. This makes it particularly useful for calculating surface normals in physics and computer graphics.

In the exercise, the cross product of \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = \hat{j} - \hat{k} \) is computed to ascertain a normal vector to the plane. The resulting vector \( \hat{i} - \hat{j} + \hat{k} \) is orthogonal to both \( \vec{a} \) and \( \vec{b} \), which assists in forming the equation of the plane.

Plane Equation

A Plane Equation represents a flat, two-dimensional surface extending infinitely in three-dimensional space. It can be derived if we know a vector normal to the plane and a point on the plane. The equation takes the form \( nx + my + oz = d \), where \( n \), \( m \), and \( o \) are the components of the normal vector.

In our exercise, we use the normal vector \( \vec{n} = \hat{i} - \hat{j} + \hat{k} \), found through the cross product, and a point on one of the given lines, \( \vec{r}_0 = \hat{i} \). This leads to the equation \( (x-1) - y + z = 0 \), which can be rearranged to \( x - y + z = 1 \).

This equation describes every point \( (x, y, z) \) that lies on the plane, and is fundamental in geometry for understanding the spatial layout of planes relative to other objects.

Perpendicular Projection

Perpendicular Projection is used to find the closest point on a plane or line to another point in space by drawing a line perpendicular to the surface. This point is known as the 'foot of the perpendicular'.

The formula for finding the perpendicular foot from a point \( M(x_1, y_1, z_1) \) onto a plane with the equation \( ax + by + cz = d \) involves calculating the distance from the point to the plane and shifting \( M \) by this distance along the normal vector. This formula can be expressed as:

• Compute the distance \( d \) using \( d = \frac{ax_1 + by_1 + cz_1 - d}{a^2 + b^2 + c^2} \).
• Use \( d \) to adjust each coordinate of \( M \) accordingly.

In our case, point \( M(1, 0, 1) \) is shifted along the plane's normal vector to find point \( Q(\alpha, \beta, \gamma) \). Thus, \( Q \) has coordinates \( \left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right) \), successfully giving us the foot of the perpendicular.