Question

If \( L \) is the line of intersection of two planes \[ x + 2y + 2z = 15 \quad \text{and} \quad x - y + z = 4 \] and the direction ratios of the line \( L \) are \( (a, b, c) \), then the value of \[ \frac{a^2 + b^2 + c^2}{b^2} \] is:

• A. \( 14 \)
• B. \( 10 \)
• C. \( 22 \)
• D. \( 26 \)

Hint:

For the intersection of two planes, find the direction ratios using the cross product of their normal vectors. Then compute the required expression step by step.

Answer 1

The Correct Option is D

Step 1: Find the Direction Ratios of the Line

The direction ratios of the line of intersection of two planes are given by:

\[ \mathbf{n_1} \times \mathbf{n_2} \]

where \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are the normal vectors of the given planes.

For the first plane:

\[ x + 2y + 2z = 15 \Rightarrow \mathbf{n_1} = (1, 2, 2). \]

For the second plane:

\[ x - y + z = 4 \Rightarrow \mathbf{n_2} = (1, -1, 1). \]

Computing the Cross Product:

\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 1 & -1 & 1 \end{vmatrix} \]

Expanding along the first row:

\[ = \mathbf{i} \begin{vmatrix} 2 & 2 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix}. \]

Evaluating the determinants:

\[ = \mathbf{i} (2 \times 1 - 2 \times (-1)) - \mathbf{j} (1 \times 1 - 2 \times 1) + \mathbf{k} (1 \times (-1) - 2 \times 1). \]

\[ = \mathbf{i} (2 + 2) - \mathbf{j} (1 - 2) + \mathbf{k} (-1 - 2). \]

\[ = 4\mathbf{i} + \mathbf{j} - 3\mathbf{k}. \]

Thus, the direction ratios are:

\[ (a, b, c) = (4, 1, -3). \]

Step 2: Compute \( \frac{a^2 + b^2 + c^2}{b^2} \)

\[ a^2 + b^2 + c^2 = 4^2 + 1^2 + (-3)^2. \]

\[ = 16 + 1 + 9 = 26. \]

\[ b^2 = 1^2 = 1. \]

\[ \frac{a^2 + b^2 + c^2}{b^2} = \frac{26}{1} = 26. \]

Step 3: Conclusion

Thus, the correct answer is:

\[ \mathbf{26}. \]