Question

Let \(\pi\) be the plane passing through the point \((3,-3,1)\) and perpendicular to the line joining the points \[ (3,4,-1) \] and \[ (2,-1,5). \] If the equation of the plane containing the points \[ (3,4,-1),\quad (-1,2,5) \] and perpendicular to the plane \(\pi\) is \[ ax+y+cz-d=0, \] then \[ 3(a+c)= \]

• A. \(-d\)
• B. \(2d\)
• C. \(d\)
• D. \(-2d\)

Hint:

If a plane contains a line and is perpendicular to another plane, then its normal vector is obtained by taking the cross product of the given line direction vector and the normal vector of the other plane.

Answer 1

The Correct Option is C

Step 1: Find the normal vector of plane \(\pi\).
The plane \(\pi\) is perpendicular to the line joining \[ (3,4,-1) \] and \[ (2,-1,5). \] Hence its normal vector is the direction vector of the line: \[ \vec n_1=(2-3,\,-1-4,\;5-(-1)). \] \[ \vec n_1=(-1,-5,6). \]

Step 2: Find a vector lying in the required plane.
The required plane contains the points \[ (3,4,-1) \] and \[ (-1,2,5). \] Therefore, a direction vector in this plane is \[ \vec v=(-1-3,\;2-4,\;5-(-1)). \] \[ \vec v=(-4,-2,6). \] \[ \vec v=(-2,-1,3). \]

Step 3: Find the normal vector of the required plane.
The required plane is perpendicular to \(\pi\).
Therefore its normal vector must be perpendicular to both \[ \vec n_1=(-1,-5,6) \] and \[ \vec v=(-2,-1,3). \] Hence \[ \vec n=\vec n_1\times \vec v. \] \[ \vec n= \begin{vmatrix} \hat i & \hat j & \hat k -1 & -5 & 6 -2 & -1 & 3 \end{vmatrix}. \] \[ = \hat i(-15+6) -\hat j(-3+12) +\hat k(1-10). \] \[ =(-9,-9,-9). \] Thus a proportional normal vector is \[ (1,1,1). \]

Step 4: Write the equation of the plane.
Using the point \[ (3,4,-1), \] the plane is \[ (x-3)+(y-4)+(z+1)=0. \] \[ x+y+z-6=0. \] Comparing with \[ ax+y+cz-d=0, \] we get \[ a=1,\qquad c=1,\qquad d=6. \]

Step 5: Evaluate \(3(a+c)\).
\[ 3(a+c)=3(1+1). \] \[ =6. \] Since \[ d=6, \] we obtain \[ 3(a+c)=d. \]

Step 6: Final conclusion.
Therefore, \[ \boxed{d} \]