Question

The point of intersection of the line joining the points \( \bar{i} + 2\bar{j} + \bar{k} \), \( 2\bar{i} - \bar{j} - \bar{k} \) and the plane passing through the points \( \bar{i}, 2\bar{j}, 3\bar{k} \) is

• A. \( \bar{i} + 2\bar{j} + 3\bar{k} \)
• B. \( \frac{1}{7}(3\bar{i} - \bar{j} + \bar{k}) \)
• C. \( \bar{i} - 3\bar{j} - 2\bar{k} \)
• D. \( \frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k}) \)

Hint:

Using the intercept form for planes passing through axes points is much faster than finding the normal vector using cross products.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

We need to find the equation of the plane and the parametric equation of the line, then solve for the intersection point.
Step 2: Key Formula or Approach:

Plane intercept form: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). Line parametric form: \( \vec{r} = \vec{a} + \lambda(\vec{b}-\vec{a}) \).
Step 3: Detailed Explanation:

Plane Equation: Points are \( (1,0,0), (0,2,0), (0,0,3) \). \[ \frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1 \implies 6x + 3y + 2z = 6 \] Line Equation: Points \( A(1, 2, 1) \) and \( B(2, -1, -1) \). Vector \( \vec{AB} = (1, -3, -2) \). Line: \( x = 1+\lambda, \ y = 2-3\lambda, \ z = 1-2\lambda \). Intersection: Substitute line into plane eq: \[ 6(1+\lambda) + 3(2-3\lambda) + 2(1-2\lambda) = 6 \] \[ 6 + 6\lambda + 6 - 9\lambda + 2 - 4\lambda = 6 \] \[ 14 - 7\lambda = 6 \implies 7\lambda = 8 \implies \lambda = \frac{8}{7} \] Find Point: \[ x = 1 + \frac{8}{7} = \frac{15}{7} \] \[ y = 2 - \frac{24}{7} = -\frac{10}{7} \] \[ z = 1 - \frac{16}{7} = -\frac{9}{7} \] Vector: \( \frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k}) \).
Step 4: Final Answer:

The point is \( \frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k}) \).