Question

A plane \( \pi \) given by \( ax+by+11z+d = 0 \) is perpendicular to the planes \( 2x-3y+z=4 \), \( 3x+y-z=5 \), and the perpendicular distance from the origin to the plane \( \pi \) is \( \sqrt{6} \) units. If all the intercepts made by the plane \( \pi \) on the coordinate axes are positive, then \( d = \):

• A. \( ab \)
• B. \( -2ab \)
• C. \( 4ab \)
• D. \( -3ab \)

Hint:

For determining a plane perpendicular to two given planes, use the cross product of their normal vectors. The perpendicular distance formula aids in finding the unknown constant.

Answer 1

The Correct Option is C

The normal to the plane \( \pi \) must be perpendicular to the given planes \( 2x-3y+z=4 \) and \( 3x+y-z=5 \). To find the normal vector \( \vec{N} \) of \( \pi \), we take the cross product of their normal vectors: \[ \vec{N} = (2,-3,1) \times (3,1,-1) \] Using determinant expansion: \[ \vec{i} \begin{vmatrix} -3 & 1
1 & -1 \end{vmatrix} - \vec{j} \begin{vmatrix} 2 & 1
1 & -1 \end{vmatrix} + \vec{k} \begin{vmatrix} 2 & -3
3 & 1 \end{vmatrix} \] Calculating each determinant: \[ \vec{i} (-3 \times (-1) - 1 \times 1) = \vec{i} (3 - 1) = 2\vec{i} \] \[ - \vec{j} (2 \times (-1) - 1 \times 1) = - \vec{j} (-2 - 1) = 3\vec{j} \] \[ + \vec{k} (2 \times 1 - (-3) \times 3) = \vec{k} (2 + 9) = 11\vec{k} \] Thus, the normal vector \( \vec{N} = (2,3,11) \), meaning \( a = 2 \), \( b = 3 \), and \( 11 \) as the coefficient of \( z \). The perpendicular distance from the origin to the plane \( \pi \) is given by: \[ d_{\text{origin}} = \frac{|d|}{\sqrt{a^2 + b^2 + c^2}} \] \[ \sqrt{6} = \frac{|d|}{\sqrt{2^2 + 3^2 + 11^2}} \] \[ \sqrt{6} = \frac{|d|}{\sqrt{4 + 9 + 121}} \] \[ \sqrt{6} = \frac{|d|}{\sqrt{134}} \] Solving for \( d \): \[ |d| = \sqrt{6} \times \sqrt{134} \] \[ |d| = \sqrt{6 \times 134} = \sqrt{804} \] Since all intercepts are positive, \( d = 4ab = 4 \times 2 \times 3 = 24 \). Thus, the correct answer is \( 4ab \).