Exams › JEE Advanced › Maths

A point P moves in the plane pi: 2x - 3y + 6z - 4 = 0 such that the area of triangle PAB, where A = (2, 2, 1) and B = (-1, -4, -1), is always equal to 14 square units. The locus of P is a pair of lines in the plane pi. The two planes containing these lines and perpendicular to pi are 6x + ay + bz + d1 = 0 and 6x + ay + bz + d2 = 0 with d1 > d2. Find the value of (d1 - d2 + a + b) / 17.

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

AB = (-3, -6, -2), |AB| = 7. Area of PAB = (1/2)*base*height = (1/2)*7*h = 14 -> h = 4. The locus of P is two lines in plane pi parallel to AB at distance 4 from the line AB. The line AB passes through A(2,2,1) with direction (-3,-6,-2). The plane through AB perpendicular to pi will contain the direction of AB and the normal to pi (2,-3,6). Normal to pi: n = (2,-3,6). AB direction: d = (-3,-6,-2). The perpendicular to AB within plane pi: cross product of n and d (this gives the direction perpendicular to AB within pi). n x d = |i j k; 2 -3 6; -3 -6 -2| = i((-3)(-2)-6(-6)) - j(2(-2)-6(-3)) + k(2(-6)-(-3)(-3)) = i(6+36) - j(-4+18) + k(-12-9) = (42, -14, -21). Simplify: divide by 7: (6, -2, -3). So the direction perpendicular to AB in plane pi is (6,-2,-3). The planes containing the locus lines and perpendicular to pi have normal direction = n x d = (42,-14,-21) or simplified (6,-2,-3). The planes are of form 6x - 2y - 3z + d = 0. But the problem states 6x + ay + bz + d = 0, so a = -2, b = -3. The two planes are at distance 4 from the line AB on either side. Foot of perpendicular from A to one such plane: line AB passes through A(2,2,1). The midline plane passes through A projected onto the plane pi perpendicularly to AB. The plane through AB in direction (6,-2,-3): 6(2) + (-2)(2) + (-3)(1) + d_mid = 0 -> 12 - 4 - 3 + d_mid = 0 -> d_mid = -5. The two planes are at distance 4 from this: distance between plane 6x-2y-3z+d=0 and a point (x0,y0,z0) = |6x0-2y0-3z0+d|/sqrt(36+4+9) = |6x0-2y0-3z0+d|/7. Shift along direction (6,-2,-3) by 4 units: the two planes correspond to d = -5 + 4*7 = -5+28 = 23 and d = -5 - 28 = -33. So d1 = 23, d2 = -33 (d1 > d2). a = -2, b = -3. (d1 - d2 + a + b)/17 = (23 - (-33) + (-2) + (-3))/17 = (56 - 5)/17 = 51/17 = 3. Answer: 3.

Related JEE Advanced Maths questions

• Perpendiculars are drawn from points on the line (x+2)/(2) = (y+1)/(-1) = (z)/(3) to the plane x + y + z = 3. The feet of perpendiculars lie on the line -
• A line ℓ passing through the origin is perpendicular to the lines ℓ₁: (3 + t) i + (-1 + 2t) j + (4 + 2t) k, -∞ < t < ∞ ℓ₂: (3 + 2s) i + (3 + 2s) j + (2 + s) k, -∞ < s < ∞ Then the coordinates(s) of the point(s) on ℓ₂ at a distance of √17 from the point of intersection of ℓ and ℓ₁ is(are)
• In three-dimensional space, consider the planes P1: y = 0 and P2: x + z = 1. Let P3 represent a plane distinct from both P1 and P2, passing through the line formed by the intersection of P1 and P2. If the point (0, 1, 0) lies at a distance of 1 unit from P3, and the point (α, β, γ) lies at a distance of 2 units from P3, which of the following equations is/are valid?
• If the point (3, 1, 7) is reflected across the plane x − y + z = 3, and the resulting image point is P, what is the equation of the plane that passes through P and includes the line given by x/1 = y/2 = z/1?
• What is the equation of a plane that passes through the point (1, 1, 1) and is orthogonal to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7?
• Consider a cube Q defined by the vertices {(x1, x2, x3) ∈ R³: x1, x2, x3 ∈ {0, 1}}. Let F represent the collection of all twelve lines formed by the diagonals of the six faces of the cube. Additionally, let S denote the group of four lines that pass through the main diagonals of the cube, such as the line connecting (0,0,0) and (1,1,1). If d(ℓ1, ℓ2) represents the shortest distance between two lines ℓ1 and ℓ2, determine the maximum value of d(ℓ1, ℓ2) when ℓ1 is chosen from F and ℓ2 is chosen from S.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →