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 π.
Plane π is perpendicular to the line joining (3, 4, -1) and (2, -1, 5). Its normal vector is n⃗₁ = (2-3, -1-4, 5-(-1)) = (-1, -5, 6).

Step 2: Find a vector lying in the required plane.
The required plane contains (3, 4, -1) and (-1, 2, 5). A direction vector is v⃗ = (-1-3, 2-4, 5-(-1)) = (-4, -2, 6) ∼ (-2, -1, 3).

Step 3: Determine the normal vector of the required plane.
The required plane is perpendicular to π, so its normal n⃗ must be perpendicular to both n⃗₁ and v⃗. n⃗ = n⃗₁ × v⃗ = |î ĵ k̂; -1 -5 6; -2 -1 3| = (-15+6)î - (-3+12)ĵ + (1-10)k̂ = (-9, -9, -9) ∼ (1, 1, 1).

Step 4: Write the plane equation and extract constants.
Using point (3, 4, -1): (x-3) + (y-4) + (z+1) = 0 → x + y + z - 6 = 0. Comparing with ax + y + cz - d = 0 gives a = 1, c = 1, d = 6.

Step 5: Evaluate 3(a + c).
3(1 + 1) = 6 = d.

Step 6: Final conclusion.
The value equals d.