Problem 14 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the vectors \(\overline{\mathrm{AB}}=3 \hat{i}+4 \hat{k}\) and \(\overline{\mathrm{AC}}=5 \hat{i}-2 \hat{j}+4 \hat{k}\) are the sides of a triangle \(\mathrm{ABC}\), then the length of the median through \(\mathrm{A}\) is (a) \(\sqrt{18}\) (b) \(\sqrt{72}\) (c) \(\sqrt{33}\) (d) \(\sqrt{45}\)

Step-by-Step Solution

Answer

The length of the median through A is \( \sqrt{33} \) (option c).

Step 1: Calculate vector BC

To find the vector \( \overline{\mathrm{BC}} \), we use the equation \( \overline{\mathrm{BC}} = \overline{\mathrm{AC}} - \overline{\mathrm{AB}} \). Given \( \overline{\mathrm{AB}} = 3 \hat{i} + 4 \hat{k} \) and \( \overline{\mathrm{AC}} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k} \), we subtract the components: \( \overline{\mathrm{BC}} = (5 - 3) \hat{i} + (-2 - 0) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i} - 2 \hat{j} \).

Step 2: Find midpoint of BC

To find the midpoint \( \mathrm{M} \) of \( \overline{\mathrm{BC}} \), we calculate \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \). Using \( \overline{\mathrm{B}} = (3, 0, 4) \) and \( \overline{\mathrm{C}} = (5, -2, 4) \), the midpoint is \( \mathrm{M} = \left( \frac{3+5}{2}, \frac{0-2}{2}, \frac{4+4}{2} \right) = (4, -1, 4) \).

Step 3: Determine vector AM

Vector \( \overline{\mathrm{AM}} \) is the median through \( \mathrm{A} \), calculated as \( \overline{\mathrm{M}} - \overline{\mathrm{A}} \). Given \( \overline{\mathrm{A}} = (0, 0, 0) \) and \( \mathrm{M} = (4, -1, 4) \), we have \( \overline{\mathrm{AM}} = 4 \hat{i} - 1 \hat{j} + 4 \hat{k} \).

Step 4: Calculate length of AM

The length of the vector \( \overline{\mathrm{AM}} = 4 \hat{i} - 1 \hat{j} + 4 \hat{k} \) is computed as \( \sqrt{(4)^2 + (-1)^2 + (4)^2} = \sqrt{16 + 1 + 16} = \sqrt{33} \).

Key Concepts

Median of a Triangle

In geometry, the concept of a median is fundamental when dealing with triangles. The median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side. Medians are key in understanding the balance and symmetry within a triangle.
For triangle ABC, the median from vertex A, labelled as median AM, connects A to the midpoint of side BC. This means median AM effectively divides the triangle into two smaller triangles of equal area. In a 3D space, the calculation of the median involves understanding vector points that include spatial coordinates defined by components like \( \hat{i}, \hat{j}, \text{and} \hat{k} \). In the example problem, we determined the midpoint of BC by averaging the components of B and C coordinates. This is vital in calculating the median's vector, which is a crucial step in deriving its length.
Understanding medians helps one appreciate both the conceptual structure of triangles and their practical mathematical applications.

Vector Subtraction

Vector subtraction is a fundamental operation in vector algebra used to find a resultant vector. When subtracting vectors, you are essentially finding the vector that connects the tip of one vector to the tip of another.
In the exercise, vector subtraction is used to find vector \(\overline{BC}\) by subtracting \(\overline{AB}\) from \(\overline{AC}\). The equation \(\overline{BC} = \overline{AC} - \overline{AB}\) highlights this concept perfectly. For each component of the vectors, we perform subtraction separately:

• For the \( \hat{i} \) components, it's \(5 - 3\).
• For the \( \hat{j} \) components, it's \(-2 - 0\).
• For the \( \hat{k} \) components, it's \(4 - 4\).

Ultimately, vector subtraction allows students to move from abstract concepts of directional quantities to precise numerical values, crucial for solving many geometric problems.

Distance Formula in 3D Geometry

The distance formula in 3D geometry is an extension of the 2D distance formula, formulated to incorporate a third dimension. This is pivotal in computing the magnitude or length of a vector in 3D space.
Imagine a vector defined by coordinates \((x, y, z)\), the length or magnitude of this vector is given by the formula: \[ \sqrt{x^2 + y^2 + z^2}\]This gives the direct line or "straight-line" distance from the origin to the vector’s endpoint.
In the context of the problem, this principle is applied to compute the length of vector \( \overline{AM} \), where \( \overline{AM} = 4 \hat{i} - 1 \hat{j} + 4 \hat{k} \). Plugging these into the formula: \( \sqrt{4^2 + (-1)^2 + 4^2} = \sqrt{33}\)
Understanding and applying the distance formula is crucial for students as it ties together algebraic expressions with geometrical meanings, providing a clear understanding of distances in three dimensions.