Question

Let the plane P : 4x – y + z = 10 be rotated by an angle \(\frac{π}{2}\) about its line of intersection with the plane x + y – z = 4. If α is the distance of the point (2, 3, -4) from the new position of the plane P, then 35α is

• A. 85
• B. 90
• C. 105
• D. 126

Hint:

The equation of the plane passing through the intersection of two planes \( P_1 = 0 \) and \( P_2 = 0 \) is given by \( P_1 + \lambda P_2 = 0 \).

The distance of a point \((x_1, y_1, z_1)\) from a plane \( ax + by + cz + d = 0 \) is given by \( \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \). 

Answer 1

The Correct Option is D

Let the equation in the new position be:

\( (4x - y + z - 10) + \lambda(x + y - z - 4) = 0 \)

Simplifying, we solve for \( \lambda \):

\( 4(4 + \lambda) - 1(-1 + \lambda) + 1(1 - \lambda) = 0 \)

Expand and simplify:

\( 16 + 4\lambda + 1 - \lambda + 1 - \lambda = 0 \)

\( 18 + 2\lambda = 0 \Rightarrow \lambda = -9 \)

Substitute \( \lambda = -9 \) back into the equation:

\( (4x - y + z - 10) - 9(x + y - z - 4) = 0 \)

Simplify the equation:

\( -5x - 10y + 10z + 26 = 0 \)

From this equation, we find:

\( \alpha = \frac{54}{15} \)

\( 35\alpha = \frac{54}{15} \times 35 = 126 \)