Let λ₁, λ₂ be the values of λ for which the points (5/2, 1, λ) and (-2,0,1) are at equal distance from the plane 2 x+3 y-6 z+7=0. If λ₁ > λ₂, then the distance of the point (λ₁-λ₂, λ₂, λ₁) from the line (x-5)/1=(y-1)/2=(z+7)/2 is ____________.

Solution

Maths sir · 4 steps

Equal distance from a plane means two cases, plus and minus; then drop a perpendicular to the line.

1. Equal distances. Both points are measured from the same plane, so the denominator √(2²+3²+6²) = 7 is common, and only |2x+3y-6z+7| has to match. For (-2, 0, 1) it is |-4+0-6+7| = 3. For (5/2, 1, λ) it is |5+3-6λ+7| = |15-6λ|.
2. Find both values. 15-6λ = 3 gives λ = 2, and 15-6λ = -3 gives λ = 3. So λ₁ = 3, λ₂ = 2, and the point is P(1, 2, 3).
3. Set up the line. The line passes through A(5, 1, -7) with direction d⃗ = (1, 2, 2), so |d⃗| = 3. Then AP⃗ = (-4, 1, 10), with |AP⃗|² = 16+1+100 = 117 and AP⃗·d⃗ = -4+2+20 = 18.

   distance² = |AP⃗|² - ((AP⃗·d⃗)/|d⃗|)² = 117 - (18/3)² = 117 - 36 = 81

4. So. The distance of (1, 2, 3) from the line is √81 = 9.

The idea to keep. Distance from a point to a line is a right triangle: hypotenuse |AP⃗|, one side the projection on d⃗. And equal distances from a plane always means two cases, plus and minus.

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