Question

Let \( O(0) \), \( A(\hat{i} + \hat{j} + \hat{k}) \), \( B(-2\hat{i} + 3\hat{k}) \), \( C(2\hat{i} + \hat{j}) \), and \( D(4\hat{k}) \) be the position vectors of the points \( O, A, B, C, \) and \( D \). If a line passing through \( A \) and \( B \) intersects the plane passing through \( O, C, \) and \( D \) at the point \( R \), then the position vector of \( R \) is:

• A. \(4\hat{i} + 2\hat{j} - \hat{k}. \)
• B. \( 2\hat{i} + \hat{j} + \hat{k} \)
• C. \( -7\hat{i} - 6\hat{j} - 5\hat{k} \)
• D. \( 3\hat{i} + 2\hat{j} - 5\hat{k} \)

Hint:

For finding the intersection of a line with a plane, use the parametric equations of the line and substitute them into the plane equation to solve for the parameter.

Answer 1

The Correct Option is A

Step 1: Finding the Equation of the Line Passing Through \( A \) and \( B \) The parametric equation of the line passing through points \( A(1,1,1) \) and \( B(-2,0,3) \) is given by: \[ \overrightarrow{r} = \overrightarrow{A} + \lambda (\overrightarrow{B} - \overrightarrow{A}) \] \[ \overrightarrow{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) + \lambda (-2\hat{i} + 3\hat{k} - (1\hat{i} + 1\hat{j} + 1\hat{k})) \] \[ \overrightarrow{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) + \lambda (-3\hat{i} - 1\hat{j} + 2\hat{k}). \] Expanding the terms: \[ x = 1 - 3\lambda, \quad y = 1 - \lambda, \quad z = 1 + 2\lambda. \] 
Step 2: Finding the Equation of the Plane Passing Through \( O, C, \) and \( D \) The normal vector to the plane is found by taking the cross-product of vectors \( \overrightarrow{OC} \) and \( \overrightarrow{OD} \): \[ \overrightarrow{OC} = (2\hat{i} + \hat{j} - 0\hat{k}), \] \[ \overrightarrow{OD} = (0\hat{i} + 0\hat{j} + 4\hat{k}). \] Taking the cross-product: \[ \overrightarrow{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 0 & 0 & 4 \end{vmatrix} \] \[ = \hat{i} (1 \times 4 - 0 \times 0) - \hat{j} (2 \times 4 - 0 \times 0) + \hat{k} (2 \times 0 - 1 \times 0). \] \[ = 4\hat{i} - 8\hat{j} + 0\hat{k}. \] Thus, the equation of the plane is: \[ 4x - 8y = 0. \] 
Step 3: Finding the Intersection of the Line and the Plane Substituting \( x = 1 - 3\lambda \), \( y = 1 - \lambda \) into the plane equation: \[ 4(1 - 3\lambda) - 8(1 - \lambda) = 0. \] \[ 4 - 12\lambda - 8 + 8\lambda = 0. \] \[ -4 - 4\lambda = 0. \] \[ 4\lambda = -4 \quad \Rightarrow \quad \lambda = -1. \] Substituting \( \lambda = -1 \) into the parametric equations: \[ x = 1 - 3(-1) = 4, \quad y = 1 - (-1) = 2, \quad z = 1 + 2(-1) = -1. \] Thus, the position vector of \( R \) is: \[ \overrightarrow{R} = 4\hat{i} + 2\hat{j} - \hat{k}. \] 

Answer 2

Step 1: Finding the Parametric Equation of the Line Through Points \( A \) and \( B \)
The line passing through \( A(1,1,1) \) and \( B(-2,0,3) \) can be expressed as:
\[ \overrightarrow{r} = \overrightarrow{A} + \lambda (\overrightarrow{B} - \overrightarrow{A}) \]
First, find the direction vector:
\[ \overrightarrow{B} - \overrightarrow{A} = (-2 - 1)\hat{i} + (0 - 1)\hat{j} + (3 - 1)\hat{k} = -3\hat{i} - \hat{j} + 2\hat{k} \]
So, the parametric form is:
\[ \overrightarrow{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) + \lambda (-3\hat{i} - \hat{j} + 2\hat{k}) \]
Splitting into components:
\[ x = 1 - 3\lambda, \quad y = 1 - \lambda, \quad z = 1 + 2\lambda. \]

Step 2: Finding the Equation of the Plane Through Points \( O, C, \) and \( D \)
Consider vectors from the origin \( O(0,0,0) \) to points \( C(2,1,0) \) and \( D(0,0,4) \):
\[ \overrightarrow{OC} = 2\hat{i} + 1\hat{j} + 0\hat{k}, \quad \overrightarrow{OD} = 0\hat{i} + 0\hat{j} + 4\hat{k} \]
The normal vector \( \overrightarrow{N} \) to the plane is the cross product:
\[ \overrightarrow{N} = \overrightarrow{OC} \times \overrightarrow{OD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 0 & 0 & 4 \end{vmatrix} \]
Calculate determinant:
\[ \overrightarrow{N} = \hat{i}(1 \times 4 - 0 \times 0) - \hat{j}(2 \times 4 - 0 \times 0) + \hat{k}(2 \times 0 - 1 \times 0) = 4\hat{i} - 8\hat{j} + 0\hat{k}. \]
Hence, the plane equation is:
\[ 4x - 8y = 0. \]

Step 3: Finding the Intersection Point of the Line and the Plane
Substitute \( x = 1 - 3\lambda \) and \( y = 1 - \lambda \) into the plane equation:
\[ 4(1 - 3\lambda) - 8(1 - \lambda) = 0 \]
Expand and simplify:
\[ 4 - 12\lambda - 8 + 8\lambda = 0 \quad \Rightarrow \quad -4 - 4\lambda = 0 \]
Solve for \( \lambda \):
\[ 4\lambda = -4 \quad \Rightarrow \quad \lambda = -1 \]
Substitute \( \lambda = -1 \) back into the parametric equations:
\[ x = 1 - 3(-1) = 4, \quad y = 1 - (-1) = 2, \quad z = 1 + 2(-1) = -1. \]
Therefore, the position vector of the intersection point \( R \) is:
\[ \overrightarrow{R} = 4\hat{i} + 2\hat{j} - \hat{k}. \]