Let P be a plane containing the line (x - 1)/3 = (y + 6)/4 = (z + 5)/2 and parallel to the line (x - 1)/4 = (y - 2)/(- 3) = (z + 5)/7. If the point (1, -1, α) lies on the plane P, then the value of |5α| is equal to ____________.

Solution

Maths sir · 4 steps

The plane's normal is perpendicular to both direction vectors, so take their cross product.

1. What we know. The plane contains the first line, so it passes through (1, -6, -5) and contains the direction (3, 4, 2). It is parallel to the second line, so it also contains the direction (4, -3, 7).

   n⃗ = |î, ĵ, k̂; 3, 4, 2; 4, -3, 7| = 34î - 13ĵ - 25k̂

2. Cross product. The î part is 28 + 6 = 34. The ĵ part is -(21 - 8) = -13. The k̂ part is -9 - 16 = -25.
3. Plane equation. Through (1, -6, -5) the plane is 34(x - 1) - 13(y + 6) - 25(z + 5) = 0. Put in (1, -1, α): 0 - 13(5) - 25(α + 5) = 0.
4. So. -65 - 25α - 125 = 0 gives α = -190/25 = -7.6. Then 5α = -38, so |5α| = 38.

The common slip: 12. Dropping the minus sign on the ĵ component gives the normal (34, 13, -25), so α = -2.4 and |5α| = 12, not 38.

The idea to keep. A plane that contains one line and is parallel to another has a normal along the cross product of the two directions. Then use any point of the line it contains.

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