Question

The orthocentre of triangle formed by points: \( (2,1,5) \), \( (3,2,3) \) and \( (4,0,4) \) is:

• A. \( (3,1,2) \)
• B. \( (3,2,3) \)
• C. \( (3,1,4) \)
• D. \( (1,4,0) \)

Hint:

The orthocentre is found by solving the system of equations formed by the altitudes of the triangle.

Answer 1

The Correct Option is C

We are given the points: \[ A = (2, 1, 5), \quad B = (3, 2, 3), \quad C = (4, 0, 4) \] We need to find the orthocentre of the triangle formed by these points. Step 1: Finding the Equation of the Plane Containing \( A, B, C \) To find the plane containing these points, we need the normal vector to the plane. \[ \text{Normal Vector} = \mathbf{AB} \times \mathbf{AC} \] \[ \mathbf{AB} = (3 - 2, 2 - 1, 3 - 5) = (1, 1, -2) \] \[ \mathbf{AC} = (4 - 2, 0 - 1, 4 - 5) = (2, -1, -1) \] Now compute the cross product: \[ \mathbf{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & -2
2 & -1 & -1 \end{vmatrix} \] \[ \mathbf{N} = \hat{i}[(1)(-1) - (-2)(-1)] - \hat{j}[(1)(-1) - (1)(2)] + \hat{k}[(1)(-1) - (1)(2)] \] \[ \mathbf{N} = \hat{i}(-1 - 2) - \hat{j}(-1 - 2) + \hat{k}(-1 - 2) \] \[ \mathbf{N} = -3\hat{i} + 3\hat{j} - 3\hat{k} \] Step 2: Equation of the Plane Using the normal vector \( \mathbf{N} = (-3, 3, -3) \) and point \( A = (2, 1, 5) \), the plane equation is: \[ -3(x - 2) + 3(y - 1) - 3(z - 5) = 0 \] Expanding, \[ -3x + 6 + 3y - 3 - 3z + 15 = 0 \] \[ -3x + 3y - 3z + 18 = 0 \] Dividing the entire equation by -3: \[ x - y + z = 6 \] Step 3: Finding Altitudes and Their Intersection An altitude from vertex \( A \) is perpendicular to the line \( BC \). The direction vector of \( BC \) is: \[ \mathbf{BC} = (4 - 3, 0 - 2, 4 - 3) = (1, -2, 1) \] The parametric equation of line \( BC \) is: \[ (x, y, z) = (3, 2, 3) + t(1, -2, 1) \] Step 4: Equation of the Altitude from \( A \) to \( BC \) The altitude from \( A \) must satisfy: \[ (x - 2, y - 1, z - 5) \cdot (1, -2, 1) = 0 \] \[ (x - 2) - 2(y - 1) + (z - 5) = 0 \] Expanding, \[ x - 2 - 2y + 2 + z - 5 = 0 \] \[ x - 2y + z - 5 = 0 \] Step 5: Finding the Intersection (Orthocentre) By solving the system of equations: \[ x - y + z = 6 \] \[ x - 2y + z = 5 \] Subtract the second equation from the first: \[ (x - y + z) - (x - 2y + z) = 6 - 5 \] \[ y = 1 \] Now substitute \( y = 1 \) back into one of the equations: \[ x - 1 + z = 6 \implies x + z = 7 \] From the second equation: \[ x - 2(1) + z = 5 \implies x + z = 7 \] Choosing \( x = 3 \), \( z = 4 \). Thus, the orthocentre is: \[ (3, 1, 4) \] Step 6: Final Answer \[ Correct Answer: (3) \ (3, 1, 4) \]