Question

If \(\mathbf{i} + \mathbf{j},\; \mathbf{j} + \mathbf{k},\; \mathbf{k} + \mathbf{i},\; \mathbf{i} - \mathbf{j},\; \mathbf{j} - \mathbf{k}\) are the position vectors of the points \(A, B, C, D, E\) respectively, then the point of intersection of the line \(AB\) and the plane passing through \(C, D, E\) is:

• A. \(\mathbf{i} + \mathbf{j} + \mathbf{k}\)
• B. \(\tfrac{1}{2}\,\mathbf{i} \;+\;\,\mathbf{j} \;+\;\tfrac{1}{2}\,\mathbf{k}\)
• C. \(\tfrac{1}{2}\bigl(\mathbf{i} + \mathbf{j} + \mathbf{k}\bigr)\)
• D. \(\tfrac{1}{2}\,\mathbf{i} \;-\;\mathbf{j} \;+\;\tfrac{1}{2}\,\mathbf{k}\)

Hint:

For line-plane intersection: parametrize the line, form the plane equation (via normal), then solve simultaneously.
- Check the parameter value carefully to avoid sign errors.

Answer 1

The Correct Option is B

Step 1: Label each point by its position vector.
\[ A = \mathbf{i} + \mathbf{j},\quad B = \mathbf{j} + \mathbf{k},\quad C = \mathbf{k} + \mathbf{i},\quad D = \mathbf{i} - \mathbf{j},\quad E = \mathbf{j} - \mathbf{k}. \] Step 2: Parametric form of line \(AB\).
\[ \overrightarrow{AB} = B - A = (\mathbf{j}+\mathbf{k}) \;-\; (\mathbf{i}+\mathbf{j}) = -\mathbf{i} + \mathbf{k}. \] Any point on line \(AB\) is \[ P(t) = A + t\,\overrightarrow{AB} = (\mathbf{i}+\mathbf{j}) + t(-\mathbf{i} + \mathbf{k}) = (1 - t)\,\mathbf{i} + \mathbf{j} + t\,\mathbf{k}. \] Step 3: Plane through \(C, D, E\).
- Two direction vectors are \(\overrightarrow{CD} = D - C\) and \(\overrightarrow{CE} = E - C\).
- A normal to that plane is \(\overrightarrow{CD}\times \overrightarrow{CE}\).
- Then use point \(C\) to find the plane equation and solve for the intersection with line \(AB\). Step 4: Substituting yields \(t = \tfrac{1}{2}\).
Hence the intersection point is \[ P\!\Bigl(\tfrac{1}{2}\Bigr) = \Bigl(1 - \tfrac12\Bigr)\mathbf{i} + \mathbf{j} + \tfrac12\,\mathbf{k} = \tfrac12\,\mathbf{i} + \mathbf{j} + \tfrac12\,\mathbf{k}. \] By the given labeling, this corresponds to \(\tfrac12\,\mathbf{i} + \mathbf{j} + \tfrac12\,\mathbf{k}\).