JEE Mathematics Planes MCQs Set 01

Answer: (a) \( \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3 \)

Explanation:
1. Let the plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) with A(a, 0, 0), B(0, b, 0), C(0, 0, c).
2. The centroid is \( \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \), so \( a = 3\alpha,\ b = 3\beta,\ c = 3\gamma \).
3. The plane is \( \frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1 \), i.e. \( \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3 \).

Teacher's Note:
1. Use the intercept form of a plane.
2. The centroid divides each coordinate by 3.

Answer: (b) \( 4x + 3y + 5z = 50 \)

Explanation:
1. Equate the two parametric points: \( (3r + 1, r + 2, 2r + 3) \) and \( (s + 3, 2s + 1, 3s + 2) \). Solving \( 3r + 1 = s + 3 \), \( r + 2 = 2s + 1 \) gives \( r = 1,\ s = 1 \), and the third coordinate agrees: the point is \( (4, 3, 5) \).
2. The plane through a point P at the greatest distance from the origin is perpendicular to OP, so its normal is \( (4, 3, 5) \).
3. \( 4(x - 4) + 3(y - 3) + 5(z - 5) = 0 \), i.e. \( 4x + 3y + 5z = 50 \).

Teacher's Note:
1. Find the common point of the two lines first.
2. The farthest plane through a point is perpendicular to the line joining it to the origin.

Answer: (a) \( \frac{35\sqrt{2}}{12} \)

Explanation:
1. The planes are parallel; their distance apart is \( \frac{|-12 - 2|}{\sqrt{4 + 9 + 36}} = \frac{14}{7} = 2 \).
2. The angle \( \theta \) between the line and the planes satisfies \( \sin\theta = \frac{|3\cdot 2 - 4\cdot 3 + 5\cdot 6|}{\sqrt{50}\cdot 7} = \frac{24}{35\sqrt{2}} \).
3. \( PQ = \frac{2}{\sin\theta} = \frac{35\sqrt{2}}{12} \).

Teacher's Note:
1. The perpendicular distance between the planes equals \( PQ\sin\theta \).
2. Use the formula for the angle between a line and a plane.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The midpoint of AB is \( (2, 2, 5) \), and \( 2 - 2 + 5 = 5 \), so the plane passes through it: R is true.
2. \( \vec{BA} = (2, -2, 2) \) is parallel to the plane's normal \( (1, -1, 1) \), so AB is perpendicular to the plane and A is the mirror image of B: A is true.
3. R alone (the plane passes through the midpoint) does not guarantee perpendicularity, so R does not explain A.

Teacher's Note:
1. A mirror image needs the midpoint on the plane and a join parallel to the normal.
2. R covers only the first of these two conditions.

Answer: (a) \( 10\sqrt{3} \)

Explanation:
1. The line through the point parallel to \( x = y = z \) is \( (1 + \lambda,\ -5 + \lambda,\ 9 + \lambda) \).
2. On the plane: \( (1 + \lambda) - (-5 + \lambda) + (9 + \lambda) = 5 \), so \( 15 + \lambda = 5 \) and \( \lambda = -10 \).
3. The foot is \( (-9, -15, -1) \) and the distance is \( |\lambda|\sqrt{3} = 10\sqrt{3} \).

Teacher's Note:
1. Distance along a direction equals \( |\lambda| \) times the length of the direction vector.
2. Substitute the parametric point in the plane.

Answer: (a) \( x - 2y + 2z - 3 = 0 \)

Explanation:
1. A parallel plane is \( x - 2y + 2z + d = 0 \).
2. Distance from the origin: \( \frac{|d|}{\sqrt{1 + 4 + 4}} = \frac{|d|}{3} = 1 \), so \( d = \pm 3 \).
3. The listed plane is \( x - 2y + 2z - 3 = 0 \) (\( d = -3 \)).

Teacher's Note:
1. Parallel planes share the same normal coefficients.
2. Only the constant term changes.

Answer: (c) \( \frac{7}{2} \)

Explanation:
1. Write the second plane as \( 2x + y + 2z + \frac{5}{2} = 0 \) and the first as \( 2x + y + 2z - 8 = 0 \).
2. Distance \( = \frac{|-8 - \frac{5}{2}|}{\sqrt{4 + 1 + 4}} = \frac{21/2}{3} \).
3. So the distance is \( \frac{7}{2} \).

Teacher's Note:
1. Make the coefficients of x, y, z identical in both planes.
2. Use \( \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \).

Answer: (c) \( x + 3y + 6z = 7 \)

Explanation:
1. Planes through the line: \( (2x - 5y + z - 3) + \lambda(x + y + 4z - 5) = 0 \), with normal \( (2 + \lambda,\ \lambda - 5,\ 4\lambda + 1) \).
2. Parallel to \( x + 3y + 6z = 1 \): \( \frac{2 + \lambda}{1} = \frac{\lambda - 5}{3} = \frac{4\lambda + 1}{6} \). The first equality gives \( \lambda = -\frac{11}{2} \), and the third ratio then equals \( \frac{-21}{6} = -\frac{7}{2} \), consistent with the others.
3. The plane becomes \( -\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0 \), i.e. \( x + 3y + 6z = 7 \).

Teacher's Note:
1. Use the family of planes through the intersection of two planes.
2. Choose the parameter that makes the normals proportional.

Answer: (b) \( 2\sqrt{42} \)

Explanation:
1. The line through P parallel to the given direction is \( (1 + \lambda,\ -2 + 4\lambda,\ 3 + 5\lambda) \). It meets the plane where \( 2 + 2\lambda - 6 + 12\lambda - 12 + (-20\lambda) + 22 = 0 \), i.e. \( 6 - 6\lambda = 0 \), so \( \lambda = 1 \).
2. The meeting point is \( R(2, 2, 8) \) and \( PR = \sqrt{1 + 16 + 25} = \sqrt{42} \).
3. R is the midpoint of PQ, so \( PQ = 2\sqrt{42} \).

Teacher's Note:
1. The foot along the given direction is the midpoint of the point and its image.
2. Compute \( PR \) and double it.

Answer: (b) \( \frac{10}{\sqrt{83}} \)

Explanation:
1. The normal is the cross product of the directions \( (1, -2, 3) \) and \( (2, -1, -1) \): \( (5, 7, 3) \).
2. Plane: \( 5(x - 1) + 7(y + 1) + 3(z + 1) = 0 \), i.e. \( 5x + 7y + 3z + 5 = 0 \).
3. Distance of \( (1, 3, -7) \): \( \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}} = \frac{10}{\sqrt{83}} \).

Teacher's Note:
1. The normal is perpendicular to both direction vectors, so use their cross product.
2. Then apply the point-to-plane distance formula.

Answer: (d) \( \frac{1}{3\sqrt{2}} \)

Explanation:
1. Direction of \( L_1 \): \( (2, -2, 3)\times(1, -1, 1) = (1, 1, 0) \); direction of \( L_2 \): \( (1, 2, -1)\times(3, -1, 2) = (3, -5, -7) \).
2. The normal to the plane containing both is \( (1, 1, 0)\times(3, -5, -7) = (-7, 7, -8) \). A point of \( L_1 \) is \( (0, 5, 4) \) (check: \( -10 + 12 - 2 = 0 \) and \( -5 + 4 + 1 = 0 \)).
3. Plane: \( -7x + 7(y - 5) - 8(z - 4) = 0 \), i.e. \( 7x - 7y + 8z + 3 = 0 \). Distance from the origin \( = \frac{3}{\sqrt{162}} = \frac{1}{3\sqrt{2}} \).

Teacher's Note:
1. Two lines lie in a common plane, so the normal is the cross product of their directions.
2. Use one point of a line to fix the constant.

Answer: (b) \( \sqrt{\frac{2}{3}} \)

Explanation:
1. The segment has direction \( (1, 0, 1) \) and length \( \sqrt{2} \).
2. The angle \( \theta \) with the plane: \( \sin\theta = \frac{|1 + 0 + 1|}{\sqrt{2}\sqrt{3}} = \frac{2}{\sqrt{6}} \), so \( \cos\theta = \frac{1}{\sqrt{3}} \).
3. Projection length \( = \sqrt{2}\cdot\frac{1}{\sqrt{3}} = \sqrt{\frac{2}{3}} \).

Teacher's Note:
1. The projection on a plane has length \( \text{length}\times\cos\theta \), where \( \theta \) is the angle with the plane.
2. \( \sin\theta \) uses the dot product with the normal.

Answer: (a) \( \cos^{-1}\left(\frac{19}{35}\right) \)

Explanation:
1. Normal to OPQ: \( \vec{OP}\times\vec{OQ} = (1, 2, 1)\times(2, 1, 3) = (5, -1, -3) \).
2. Normal to PQR: \( \vec{PQ}\times\vec{PR} = (1, -1, 2)\times(-2, -1, 1) = (1, -5, -3) \).
3. \( \cos\theta = \frac{5 + 5 + 9}{\sqrt{35}\sqrt{35}} = \frac{19}{35} \).

Teacher's Note:
1. The angle between two planes equals the angle between their normals (or its supplement).
2. Use cross products of edge vectors for the normals.

Answer: (c) \( \{\sqrt{3}, -\sqrt{3}\} \)

Explanation:
1. Four points are coplanar when \( \vec{AB}, \vec{AC}, \vec{AD} \) have zero triple product. With \( A(-\lambda^2, 1, 1) \): \( \vec{AB} = (1 + \lambda^2, -\lambda^2 - 1, 0) \), \( \vec{AC} = (1 + \lambda^2, 0, -\lambda^2 - 1) \) and \( \vec{AD} = (\lambda^2 - 1, -2, 0) \).
2. The determinant of these three vectors equals \( (\lambda^2 - 3)(\lambda^2 + 1)^2 \).
3. Setting it to zero, and since \( \lambda^2 + 1 > 0 \), we get \( \lambda^2 = 3 \), so \( \lambda = \pm\sqrt{3} \).

Teacher's Note:
1. Use the coplanarity determinant for four points.
2. The factor \( (\lambda^2 + 1)^2 \) never vanishes for real \( \lambda \).