Problem 32 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The projection of the line segment joining the points \((1,-1,3)\) and \((2,-4,11)\) on the line joining the points \((-1,2,3)\) and \((3,-2,10)\) is \(\quad\). [NA Jan. 9, \(\mathbf{2 0 2 0}\) (I)]

Step-by-Step Solution

Answer

The projection length is 8.

Step 1: Determine the Vector of the Line Segment

First, we find the vector representing the line segment joining the points \((1, -1, 3)\) and \((2, -4, 11)\). The vector \(\mathbf{A}\) from point \((1, -1, 3)\) to \((2, -4, 11)\) is calculated as follows:\[\mathbf{A} = (2 - 1, -4 + 1, 11 - 3) = (1, -3, 8)\]

Step 2: Determine the Vector of the Line

Next, we find the vector representing the line joining the points \((-1, 2, 3)\) and \((3, -2, 10)\). The vector \(\mathbf{B}\) from point \((-1, 2, 3)\) to \((3, -2, 10)\) is calculated as follows:\[\mathbf{B} = (3 + 1, -2 - 2, 10 - 3) = (4, -4, 7)\]

Step 3: Calculate the Dot Product

The projection of vector \(\mathbf{A}\) onto vector \(\mathbf{B}\) is given by the dot product of \(\mathbf{A}\) and \(\mathbf{B}\) divided by the magnitude of \(\mathbf{B}\). Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\):\[\mathbf{A} \cdot \mathbf{B} = (1 \cdot 4) + ((-3) \cdot (-4)) + (8 \cdot 7) = 4 + 12 + 56 = 72\]

Step 4: Find the Magnitude of the Line Vector

Next, calculate the magnitude of vector \(\mathbf{B}\):\[\|\mathbf{B}\| = \sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9\]

Step 5: Calculate the Projection Length

The length of the projection of vector \(\mathbf{A}\) onto \(\mathbf{B}\) is given by:\[\text{Projection length} = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{B}\|} = \frac{72}{9} = 8\]

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra, especially useful when working with vectors in space. The dot product, denoted as \(\mathbf{A} \cdot \mathbf{B}\), is calculated as the sum of the products of the corresponding components of two vectors. It is a scalar quantity, which means it has only magnitude and no direction. Given two vectors \(\mathbf{A} = (a_1, a_2, a_3)\) and \(\mathbf{B} = (b_1, b_2, b_3)\), the dot product is calculated as:

• \(\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3\)

The dot product can tell you if two vectors are perpendicular: if their dot product is zero, they are orthogonal. In the given problem, the dot product \(\mathbf{A} \cdot \mathbf{B} = 72\) is not zero, indicating that vectors \(\mathbf{A}\) and \(\mathbf{B}\) are not perpendicular. Utilizing the dot product is essential when calculating the vector projection, as it plays a key role in projecting one vector onto another.

Magnitude of a Vector

The magnitude of a vector, also known as the vector's length or norm, is a measure of how long the vector is in a geometrical sense. In a 3D coordinate system, the magnitude is determined using the Pythagorean theorem generalized to three dimensions. For a vector \(\mathbf{B} = (b_1, b_2, b_3)\), its magnitude \(\|\mathbf{B}\|\) is calculated as follows:

• \(\|\mathbf{B}\| = \sqrt{b_1^2 + b_2^2 + b_3^2}\)

In our example with the vector \(\mathbf{B} = (4, -4, 7)\), the magnitude is calculated as \(\|\mathbf{B}\| = \sqrt{16 + 16 + 49} = 9\). Knowing the magnitude of a vector is crucial when dealing with vector projections, as it allows us to normalize vectors or determine their relative sizes. When one vector projects onto another, the magnitude of the vector in the denominator acts as a normalizing factor in the projection formula.

3D Coordinate System

A 3D coordinate system allows us to represent vectors that have three dimensions, corresponding to the three spatial axes: x, y, and z. Here, each point in space is identified by three coordinates \((x, y, z)\). This system is particularly useful for solving problems in physics, engineering, and other fields that involve three-dimensional space.

• Each axis is perpendicular to the others.
• The position of a point or a vector is determined by its distance along each of these axes.

In a 3D coordinate system, vectors can be visualized as arrows pointing from the origin or between any two points. For instance, given points \((1, -1, 3)\) and \((2, -4, 11)\), a vector can be formed by calculating the difference along each axis: \(\mathbf{A} = (2-1, -4+1, 11-3) = (1, -3, 8)\). The 3D coordinate system simplifies the understanding of spatial relationships, vector operations, and projections, as seen in this exercise where vectors and points are integral to the solution.