Question

Let \(\mathrm{ABC}\) be a triangle where \(\mathrm{A}=(2,3,5)\), \(\mathrm{B}=(-1,3,2)\) and \(\mathrm{C}=(\lambda, 5, \mu)\). If the median through \(\mathrm{A}\) is equally inclined to the axes, then (1) \(\lambda=\mu=5\) (2) \(\lambda=5, \mu=7\) (3) \(\lambda=7, \mu=10\) (4) \(\lambda=0, \mu=0\)

Short answer

Option (3): \(\lambda=7, \mu=10\).

Step-by-step solution

Find the midpoint of BC

The coordinates of points B and C are given as \(\text{B}=(-1,3,2)\) and \(\text{C}=(\text{\(\lambda\)},5,\(\mu\))\). To find the midpoint M of BC, use the midpoint formula: \(\text{M} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = \left( \frac{-1 + \lambda}{2}, \frac{3 + 5}{2}, \frac{2 + \mu}{2} \right) = \left( \frac{-1+\lambda}{2}, 4, \frac{2 + \mu}{2} \right)\).

Find the equation of the median through A

The coordinates of point A are \(\text{A}=(2, 3, 5)\). The median AM will have the coordinates of M as computed in Step 1 and point A. Therefore, the equation of the line AM in parametric form (given point A and direction vector) is: \(x=2+(-1+\lambda-2)t\) \(y=3+(4-3)t\)\(z=5+\left(\frac{2+\mu}{2}-5\right)t\).

Condition for AM to be equally inclined to the axes

For the median to be equally inclined to the coordinate axes, the direction cosines along these axes have to be equal. Let's assume the direction vector of AM to be \((x_2 - x_1, y_2 - y_1, z_2 - z_1)\) from point A to the calculated midpoint M: \((-1+\lambda-2, 4-3, \frac{2+\mu}{2}-5) = (\lambda-3, 1, \frac{\mu-8}{2})\). To be equally inclined, \(\lambda-3=1=(\mu-8)/2\).

Solve for values of \(\lambda\) and \(\mu\)

Set up and solve the following system of equations: \(\lambda-3=1\) and \((\mu-8)/2=1\). Solving these, we find: \(\lambda=4\) and \(\mu-8=2\Rightarrow \mu=10\).

Verify the solution

Check if the calculated values match any given option. The closest match is option (3): \(\lambda=7, \mu=10\).

Key concepts

Median of a Triangle

A median of a triangle is a line segment that originates from a vertex and bisects the opposite side into two equal parts. In any triangle, there are three medians, each connecting a vertex to the midpoint of the opposite side. Mathematically, if we have a triangle ABC with vertices A, B, and C at coordinates \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\) and \((x_3, y_3, z_3)\) respectively, the midpoint M of side BC can be calculated using the midpoint formula:
\(\text{M} = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right)\).
The median from vertex A to M can thus be described as the line segment joining points A and M. The properties of medians are unique and play a pivotal role in various geometric and algebraic computations.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves studying geometry using a coordinate system. Each point is defined by a set of coordinates \((x, y, z)\), which quantify its position along each axis of a Cartesian plane or space.
This form of geometry allows for algebraic methods to be applied to geometric problems, enabling the calculation of distances, angles, bisectors, and other geometric properties using formulas.
For example, to find the midpoint of a line segment joining two points \((x_1, y_1, z_1) \) and \((x_2, y_2, z_2)\), use the formula:
\(\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)\).
These methods simplify many geometric calculations, making them more accessible and easier to understand.

Direction Cosines

Direction cosines are the cosines of the angles between a vector and the coordinate axes in three-dimensional space. They help determine the orientation of the vector with respect to each axis. If a vector makes angles \(\theta_x\), \(\theta_y\), and \(\theta_z\) with the x, y, and z axes respectively, its direction cosines are given by \(\text{l} = \cos\theta_x\), \(\text{m} = \cos\theta_y\), and \(\text{n} = \cos\theta_z\).
These direction cosines must satisfy the equation: \(\text{l}^2 + \text{m}^2 + \text{n}^2 = 1\).
For a vector to be equally inclined to all three axes, its direction cosines must be equal, implying \(\text{l} = \text{m} = \text{n}\), which satisfies: \((\cos\theta_x)^2 + (\cos\theta_y)^2 + (\cos\theta_z)^2 = 1\).
This understanding is crucial in many applications, including physics and engineering, where vectors' directional properties determine their resultant effects.