JEE Mathematics Three Dimensional Geometry MCQs Set 09

Answer: (b) \( (0,0,0) \)

Explanation:
1. The normal has direction ratios (6, 20, -1), so the plane is 6(x + 1) + 20(y - 0) - (z + 6) = 0.
2. This simplifies to 6x + 20y - z = 0.
3. Only (0, 0, 0) satisfies it.

Teacher's Note:
1. A plane through a point with a given normal is a(x - x1) + b(y - y1) + c(z - z1) = 0.
2. Substitute each option to see which point lies on the plane.

Answer: (b) \( 5x+3y-11z=\frac{135}{2} \)

Explanation:
1. The mid-point is (3/2, 7/2, -9/2) and the normal is AB = (5, 3, -11).
2. The plane is 5x + 3y - 11z = 5(3/2) + 3(7/2) - 11(-9/2).
3. The right side is 15/2 + 21/2 + 99/2 = 135/2.

Teacher's Note:
1. The plane bisecting a segment at right angles passes through its mid-point and has the segment as its normal.
2. Compute the constant by substituting the mid-point.

Answer: (b) \( 3x+4y+5z-26=0 \)

Explanation:
1. A parallel plane has the same normal (3, 4, 5): 3x + 4y + 5z = k.
2. Substituting (1, 2, 3): k = 3 + 8 + 15 = 26.
3. So the plane is 3x + 4y + 5z - 26 = 0.

Teacher's Note:
1. Parallel planes differ only in the constant term.
2. Find the constant by substituting the given point.

Answer: (b) \( x+2y+2z-9=0 \)

Explanation:
1. The normal is along the line from the origin to the foot, so it is (1, 2, 2).
2. The plane is 1(x - 1) + 2(y - 2) + 2(z - 2) = 0.
3. This gives x + 2y + 2z - 9 = 0.

Teacher's Note:
1. The foot of the perpendicular from the origin lies on the plane and gives its normal direction.
2. Use point-normal form.

Answer: (b) \( (1,2,8) \)

Explanation:
1. Value of the plane expression at the point: 14 + 56 - 5 - 2 = 63, and 4 + 16 + 1 = 21, so the parameter is 63/21 = 3.
2. Foot = (7, 14, 5) - 3(2, 4, -1) = (1, 2, 8).

Teacher's Note:
1. The foot is the point reached by moving from the given point along the normal by the signed distance.
2. Check that the foot satisfies the plane equation: 2 + 8 - 8 = 2.

Answer: (b) \( \left(\frac95,-\frac{13}{5},4\right) \)

Explanation:
1. Value at the point: -1 - 6 = -7, and 1 + 4 + 0 = 5.
2. Image = (-1, 3, 4) - 2(-7/5)(1, -2, 0) = (-1 + 14/5, 3 - 28/5, 4).
3. The image is (9/5, -13/5, 4).

Teacher's Note:
1. Image = point - 2 × (value/length squared) x normal.
2. The z-coordinate does not change because the normal has no z-component.

Answer: (d) \( 3:10 \)

Explanation:
1. The plane is x - 2y + 3z - 17 = 0.
2. At the first point: -2 - 8 + 21 - 17 = -6. At the second: 3 + 10 + 24 - 17 = 20.
3. Ratio = |-6| : |20| = 3 : 10.

Teacher's Note:
1. The ratio is the absolute value of the plane expression at one end to that at the other.
2. Opposite signs mean internal division.

Answer: (a) on the opposite side of the origin

Explanation:
1. At the origin the expression equals 10 > 0.
2. At (2, 3, -5): 4 + 9 - 25 + 10 = -2 < 0.
3. The signs are opposite, so the point is on the side opposite to the origin.

Teacher's Note:
1. Compare the sign of the plane expression at the point with its value at the origin.
2. The constant term is the value at the origin.

Answer: (d) \( \frac{13}{\sqrt{74}} \)

Explanation:
1. The plane is 3(x - 2) - 4(y - 3) + 7(z + 1) = 0, i.e. 3x - 4y + 7z + 13 = 0.
2. Distance from origin = 13 / √(9 + 16 + 49).
3. This equals 13/√(74).

Teacher's Note:
1. The direction ratios of the line are the normal of the plane.
2. The distance from the origin is |d| divided by the length of the normal.

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

Explanation:
1. A parallel plane is x - 2y + 2z + d = 0.
2. Distance from the origin = |d| / √(1 + 4 + 4) = |d|/3 = 1, so d = 3 or -3.
3. Only -3 is among the options.

Teacher's Note:
1. Parallel planes share the same coefficients.
2. Use the point-to-plane distance formula from the origin.

Answer: (a) \( (143,91,77) \)

Explanation:
1. Intercepts: 3003/7 = 429, 3003/11 = 273, 3003/13 = 231.
2. Centroid = (429/3, 273/3, 231/3).
3. This equals (143, 91, 77).

Teacher's Note:
1. Find the intercepts by setting two variables to zero.
2. The centroid divides the sum of the vertices by 3.

Answer: (a) \( \frac x4+\frac y6+\frac z3=1 \)

Explanation:
1. Let the intercepts be a, b, c. Then ab/2 = 12, bc/2 = 9, ca/2 = 6, so ab = 24, bc = 18, ca = 12.
2. a squared = (ab)(ca)/(bc) = 24 × 12 / 18 = 16, so a = 4.
3. Then b = 6 and c = 3, giving x/4 + y/6 + z/3 = 1.

Teacher's Note:
1. Each triangle on a coordinate plane has area half the product of two intercepts.
2. Multiply two products and divide by the third to isolate one intercept.

Answer: (d) \( \frac{\pi}{3} \)

Explanation:
1. cos(θ) = |2(1) + 1(-1) + 1(2)| / (√(6) √(6)) = 3/6 = 1/2.
2. So θ = π/3.

Teacher's Note:
1. The angle between planes equals the acute angle between their normals.
2. Use the absolute value of the dot product.

Answer: (d) \( -8 \)

Explanation:
1. Normals (2, 3, -1) and (1, 2, -k) must be perpendicular.
2. 2 + 6 + k = 0.
3. So k = -8.

Teacher's Note:
1. Use a1a2 + b1b2 + c1c2 = 0 for perpendicular planes.
2. Be careful with the sign of the coefficient of z.

Answer: (a) \( 2x+3y+4z-9=0 \)

Explanation:
1. The plane has the form 2x + 3y + 4z = k.
2. At (1, 1, 1): k = 2 + 3 + 4 = 9.
3. So 2x + 3y + 4z - 9 = 0.

Teacher's Note:
1. Parallel planes have proportional coefficients.
2. Fix the constant using the given point.

Answer: (d) \( x-3=0 \)

Explanation:
1. A plane perpendicular to the x-axis has normal (1, 0, 0), so it is x = constant.
2. It passes through x = 3.
3. So x - 3 = 0.

Teacher's Note:
1. A plane perpendicular to an axis is the coordinate value set equal to a constant.
2. Take the constant from the given point.

Answer: (b) \( 1 \)

Explanation:
1. Divide the second equation by 2: 7x + 4y - 4z - 6 = 0.
2. Distance = |3 - (-6)| / √(49 + 16 + 16) = 9/9.
3. This equals 1.

Teacher's Note:
1. Always scale the second plane so that both have the same coefficients.
2. Then apply |d1 - d2| / |n|.

Answer: (d) \( 2x-y+2z+1=0 \)

Explanation:
1. The normal is the vector from the foot to the point: (4 - 2, 2 - 3, 1 + 1) = (2, -1, 2).
2. Plane: 2(x - 2) - (y - 3) + 2(z + 1) = 0.
3. This simplifies to 2x - y + 2z + 1 = 0.

Teacher's Note:
1. The line joining a point to its foot is normal to the plane.
2. Use the foot as the point on the plane.

Answer: (a) \( -\frac13 \)

Explanation:
1. The plane is 3x + 4y + 12z + 13 = 0 with normal length 13.
2. Distance of (-3, 0, 1): |-9 + 12 + 13|/13 = 16/13.
3. For (1, 1, p): |20 + 12p| = 16, so p = -1/3 (or -3).

Teacher's Note:
1. Equal distance means equal absolute values of the plane expression.
2. Solve the modulus equation and match with the options.

Answer: (d) \( \left(\frac23,\frac{-1}{3},\frac23\right) \)

Explanation:
1. Direction ratios of the normal: (2, -1, 2).
2. Length = √(4 + 1 + 4) = 3.
3. Direction cosines = (2/3, -1/3, 2/3).

Teacher's Note:
1. Divide the direction ratios by the length of the normal.
2. The squares of direction cosines add up to 1.

Answer: (a) 3

Explanation:
1. Distance = |-9| / √(1 + 4 + 4).
2. = 9/3 = 3.

Teacher's Note:
1. From the origin the distance is |d| divided by the length of the normal.

Answer: (d) \( 7y+5z=35 \)

Explanation:
1. A plane parallel to the x-axis has no x term: y/5 + z/7 = 1.
2. Multiply by 35: 7y + 5z = 35.

Teacher's Note:
1. Parallel to an axis means the intercept on that axis is infinite and its variable is absent.

Answer: (a) \( 15,-10,6 \)

Explanation:
1. Write the plane as 2x - 3y + 5z = 30.
2. x-intercept 30/2 = 15, y-intercept 30/(-3) = -10, z-intercept 30/5 = 6.

Teacher's Note:
1. Set two variables to zero to find each intercept.

Answer: (d) 6

Explanation:
1. Parallel planes have proportional coefficients: 2/4 = 3/k = 4/8.
2. From 2/4 = 1/2, 3/k = 1/2.
3. Hence k = 6.

Teacher's Note:
1. The ratios of corresponding coefficients of x, y, z are equal for parallel planes.