For a, b ∈ ℤ and |a-b| ≤ 10, let the angle between the plane P: ax+y-z=b and the line l: x-1=a-y=z+1 be cos ⁻¹(1/3). If the distance of the point (6,-6,4) from the plane P is 3 √6, then a⁴+b² is equal to :

Solution

Maths sir · 4 steps

The normal with the line's direction gives a sine here, so turn the given cosine into a sine first.

1. Read off vectors. Write the line as (x-1)/1=(y-a)/-1=(z+1)/1, so its direction is d⃗=(1,-1,1). The normal to P is n⃗=(a,1,-1).

   sin θ = (|n⃗·d⃗|)/(|n⃗| |d⃗|) = (|a-2|)/(√3 √(a²+2))

2. Find a. From cos θ=1/3 we get sin θ=2√2/3. Squaring gives 9(a-2)²=24(a²+2), which simplifies to 5a²+12a+4=0, that is (5a+2)(a+2)=0. Since a is an integer, a=-2.
3. Use the distance. Now P is -2x+y-z=b with |n⃗|=√6. The distance from (6,-6,4) is (|-12-6-4-b|)/√6=(|22+b|)/√6=3√6, so |22+b|=18 and b=-4 or b=-40. Only b=-4 satisfies |a-b|≤ 10, because |-2-(-40)|=38.
4. So. a⁴+b²=16+16= 32, which is option (c).

The idea to keep. The normal is at 90° to the plane, so (|n⃗·d⃗|)/|n⃗||d⃗| is the sine of the line-plane angle. Convert the given angle to its sine before you equate.

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