Let the equation of plane passing through the line of intersection of the planes x+2 y+a z=2 and x-y+z=3 be 5 x-11 y+b z=6 a-1. For c ∈ ℤ, if the distance of this plane from the point (a,-c, c) is 2/√a, then (a+b)/c is equal to :

Solution

Maths sir · 5 steps

Write the family of planes, match it term by term with the given plane, and a, b drop out together.

1. Family of planes. Every plane through the line of intersection has the form (x+2y+az-2)+λ(x-y+z-3)=0. Grouping terms gives (1+λ)x+(2-λ)y+(a+λ)z=2+3λ.

   (1+λ)/5=(2-λ)/-11=(a+λ)/b=(2+3λ)/(6a-1)

2. Find lambda, a, b. The first two ratios give -11-11λ=10-5λ, so λ=-7/2, and every ratio equals -1/2. Then 2+3λ=-17/2 gives 6a-1=17, so a=3. Also a+λ=-1/2 gives b=1.
3. Use the distance. The plane is 5x-11y+z=17, and √(25+121+1)=√147=7√3. From (3,-c,c) the distance is (|15+11c+c-17|)/7√3=(|12c-2|)/7√3. Setting this equal to 2/√3 gives |12c-2|=14.
4. Pick the integer. 12c-2=14 gives c=4/3, which is not an integer. 12c-2=-14 gives c=-1, so c=-1.
5. So. (a+b)/c=(3+1)/-1=-4, which is option (d).

The idea to keep. For a plane through the intersection of two planes, write P₁+λ P₂=0 and compare coefficients as equal ratios. The two known coefficients fix λ first, and the rest follow.

Answer from the NTA final answer key. Solution by PrepWiser.