Question

The vector equation of any plane passing through the line of intersection of the planes $ \bar{r} \cdot \bar{n}_1 = q_1 $ and $ \bar{r} \cdot \bar{n}_2 = q_2 $ is given by $ \bar{r} \cdot (\bar{n}_1 + \lambda \bar{n}_2) = q_1 + \lambda q_2 $ for $ \lambda \in \bar{R} $. The vector equation of a plane passing through the point $ 2\bar{i} - 3\bar{j} + \bar{k} $ and the line of intersection of the planes $ \bar{r} \cdot (\bar{i} - 2\bar{j} + 3\bar{k}) = 5 $, $ \bar{r} \cdot (3\bar{i} + \bar{j} - 2\bar{k}) = 7 $ is

• A. $ \bar{r} \cdot (-2\bar{i} - 3\bar{j} + 5\bar{k}) = -2 $
• B. $ \bar{r} \cdot (7\bar{i} - \bar{k}) = 19 $
• C. $ \bar{r} \cdot (4\bar{i} - \bar{j} + \bar{k}) = 12 $
• D. $ \bar{r} \cdot (8\bar{i} + 5\bar{j} - 9\bar{k}) = 16 $

Hint:

Any plane passing through the line of intersection of two planes $ \bar{r} \cdot \bar{n}_1 = q_1 $ and $ \bar{r} \cdot \bar{n}_2 = q_2 $ has the equation $ \bar{r} \cdot (\bar{n}_1 + \lambda \bar{n}_2) = q_1 + \lambda q_2 $. Use the given point to find the value of $ \lambda $.

Answer 1

The Correct Option is C

Step 1: Identify $ \bar{n_1, q_1, \bar{n}_2, q_2 $ from the given plane equations.}
Plane 1: $ \bar{r} \cdot (\bar{i} - 2\bar{j} + 3\bar{k}) = 5 $ So, $ \bar{n}_1 = \bar{i} - 2\bar{j} + 3\bar{k} $ and $ q_1 = 5 $. Plane 2: $ \bar{r} \cdot (3\bar{i} + \bar{j} - 2\bar{k}) = 7 $ So, $ \bar{n}_2 = 3\bar{i} + \bar{j} - 2\bar{k} $ and $ q_2 = 7 $. 
Step 2: Write the equation of any plane passing through the line of intersection of the two planes.
The equation is $ \bar{r} \cdot (\bar{n}_1 + \lambda \bar{n}_2) = q_1 + \lambda q_2 $. Substituting the values of $ \bar{n}_1, q_1, \bar{n}_2, q_2 $: $ \bar{r} \cdot ((\bar{i} - 2\bar{j} + 3\bar{k}) + \lambda (3\bar{i} + \bar{j} - 2\bar{k})) = 5 + 7\lambda $ $ \bar{r} \cdot ((1 + 3\lambda)\bar{i} + (-2 + \lambda)\bar{j} + (3 - 2\lambda)\bar{k}) = 5 + 7\lambda $ 
Step 3: Use the fact that the plane passes through the point $ 2\bar{i - 3\bar{j} + \bar{k} $.}
Let the position vector of this point be $ \bar{r}_0 = 2\bar{i} - 3\bar{j} + \bar{k} $. This point must satisfy the equation of the plane. $ (2\bar{i} - 3\bar{j} + \bar{k}) \cdot ((1 + 3\lambda)\bar{i} + (-2 + \lambda)\bar{j} + (3 - 2\lambda)\bar{k}) = 5 + 7\lambda $ $ 2(1 + 3\lambda) + (-3)(-2 + \lambda) + 1(3 - 2\lambda) = 5 + 7\lambda $ $ 2 + 6\lambda + 6 - 3\lambda + 3 - 2\lambda = 5 + 7\lambda $ $ 11 + \lambda = 5 + 7\lambda $
Step 4: Solve for $ \lambda $.
$ 11 - 5 = 7\lambda - \lambda $ $ 6 = 6\lambda $ $ \lambda = 1 $ 
Step 5: Substitute the value of $ \lambda $ back into the equation of the plane.
$ \bar{r} \cdot ((1 + 3(1))\bar{i} + (-2 + 1)\bar{j} + (3 - 2(1))\bar{k}) = 5 + 7(1) $ $ \bar{r} \cdot ((1 + 3)\bar{i} + (-1)\bar{j} + (3 - 2)\bar{k}) = 5 + 7 $ $ \bar{r} \cdot (4\bar{i} - \bar{j} + \bar{k}) = 12 $ 
Step 6: Compare the resulting equation with the given options.
The equation $ \bar{r} \cdot (4\bar{i} - \bar{j} + \bar{k}) = 12 $ matches option (3). 
Step 7: Conclusion.
The vector equation of the required plane is $ \bar{r} \cdot (4\bar{i} - \bar{j} + \bar{k}) = 12 $.