JEE Mathematics Three Dimensional Geometry MCQs Set 19

Answer: (c) \( d\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \)

Explanation:
1. Putting y = z = 0 gives x-intercept d/a; similarly d/b and d/c.
2. Sum = d(1/a + 1/b + 1/c).

Teacher's Note:
1. The intercept on an axis is found by setting the other two variables to zero.
2. The plane is written in intercept form as x/(d/a) + y/(d/b) + z/(d/c) = 1.

Answer: (b) (1, 1, 1)

Explanation:
1. The intercepts are 1/a, 1/b, 1/c, so their reciprocals are a, b, c.
2. The condition gives a + b + c = 1, which says that (1, 1, 1) satisfies ax + by + cz = 1.

Teacher's Note:
1. Look for the point that turns the condition into the plane equation.
2. Other options would require different relations between a, b, c.

Answer: (c) \( \frac{3}{\sqrt{26}},\frac{-1}{\sqrt{26}},\frac{4}{\sqrt{26}} \)

Explanation:
1. The normal has direction ratios 3, −1, 4 with length √26.
2. Since the constant 5 is positive, the perpendicular from the origin points along the normal itself.
3. Direction cosines = (3, −1, 4)/√26.

Teacher's Note:
1. Direction cosines must have sum of squares equal to 1, so options (a) and (b) are ruled out.
2. Make the constant term positive to fix the sign.

Answer: (a) 2√3

Explanation:
1. Distance = 1/√(4 + 9 + λ²) = 1/5.
2. So 13 + λ² = 25, λ² = 12, and λ = 2√3 since λ > 0.

Teacher's Note:
1. The distance of ax + by + cz = d from the origin is |d|/√(a² + b² + c²).
2. Use the condition λ > 0 to select the root.

Answer: (b) \( \frac{5}{\sqrt{43}},\frac{-3}{\sqrt{43}},\frac{3}{\sqrt{43}} \)

Explanation:
1. Expanding: 5x − 3y + 3z − 10 = 0, so the normal has ratios 5, −3, 3.
2. Length = √(25 + 9 + 9) = √43, so the direction cosines are (5, −3, 3)/√43.

Teacher's Note:
1. Expand brackets before reading off coefficients.
2. Divide by the length of the normal vector to get direction cosines.

Answer: (c) 2x − 21y − 13z = 0

Explanation:
1. The plane contains O, A and B, so its normal is OA × OB.
2. (1, −3, 5) × (4, 1, −1) = (3 − 5, 20 + 1, 1 + 12) = (−2, 21, 13).
3. So the plane is 2x − 21y − 13z = 0.

Teacher's Note:
1. A plane through three points has normal equal to the cross product of two edge vectors.
2. Check with A: 2 + 63 − 65 = 0, and with B: 8 − 21 − (−13) = 0.

Answer: (b) \( -\frac{7}{5} \)

Explanation:
1. The point (1, −1, −1) lies on the plane: 3 + 2 − 5 = 0.
2. The direction must be perpendicular to the normal: 3(1) − 2(−2) + 5λ = 0, so 7 + 5λ = 0 and λ = −7/5.

Teacher's Note:
1. For a line to lie in a plane, one point must satisfy the plane and the direction must be parallel to it.
2. Both conditions are needed.

Answer: (c) \( \left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right) \)

Explanation:
1. The line is perpendicular to both normals, so its direction is (3, 4, 1) × (1, −2, −3) = (−10, 10, −10).
2. That is proportional to (−1, 1, −1), length √3.
3. Direction cosines = (−1/√3, 1/√3, −1/√3).

Teacher's Note:
1. Option (a) lists ratios, not direction cosines.
2. The cross product of the normals gives the common parallel direction.

Answer: (b) \( \lambda=-\frac{1}{3},\ \mu=-\frac{8}{3} \)

Explanation:
1. Substituting x = 3 in both planes: λ + μ = −3 and −2λ + μ = −2.
2. Subtracting: 3λ = −1, so λ = −1/3 and μ = −8/3.

Teacher's Note:
1. A point on a line lies on both planes, giving two equations in the two unknowns.
2. Check by putting the values back into both equations.

Answer: (a) \( \frac{10}{13} \)

Explanation:
1. Distance = |5(2) − 12(0)|/√(25 + 144) = 10/13.

Teacher's Note:
1. The z-coordinate does not matter because the plane has no z term.
2. Use |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²).

Answer: (b) α² + β² + γ² = α′² + β′² + γ′²

Explanation:
1. The plane passes through the origin, and reflection in it preserves distances from points of the plane, in particular from the origin.
2. Hence OP = OQ, so α² + β² + γ² = α′² + β′² + γ′².

Teacher's Note:
1. Reflection in a plane through the origin preserves the distance from the origin.
2. PQ is parallel to the normal, so option (d) fails in general.

Answer: (b) \( \sin^{-1}\frac{\sqrt{6}}{5} \)

Explanation:
1. Direction (1, 1, 1), normal (4, −3, 5).
2. sin θ = |4 − 3 + 5|/(√3 · √50) = 6/(5√6) = √6/5.
3. So θ = sin⁻¹(√6/5).

Teacher's Note:
1. The angle between a line and a plane uses sine with the dot product of direction and normal.
2. The angle with the normal would use cosine.

Answer: (c) 7(x − z) = 6

Explanation:
1. The line is the intersection of 3x + y − 5z − 2 = 0 and x − 2y + 3z − 2 = 0.
2. Family: (3 + λ)x + (1 − 2λ)y + (−5 + 3λ)z − 2 − 2λ = 0.
3. Perpendicular to x − y + z = 3: (3 + λ) − (1 − 2λ) + (−5 + 3λ) = 0, so 6λ = 3 and λ = 1/2.
4. This gives 7x − 7z − 6 = 0, i.e. 7(x − z) = 6.

Teacher's Note:
1. Perpendicular planes have perpendicular normals, so the dot product is zero.
2. Here the line was read as two planes with the same right side 2.

Answer: (d) \( \cos^{-1}\left(\frac{2}{5}\sqrt{\frac{2}{3}}\right) \)

Explanation:
1. cos θ = |1·3 + 1·(−4) + 1·5|/(√3 · √50) = 4/(5√6).
2. 4/(5√6) = (2/5)·(2/√6) = (2/5)√(2/3).
3. So θ = cos⁻¹((2/5)√(2/3)).

Teacher's Note:
1. The angle between planes is the angle between their normals.
2. Simplify surds to compare with the options.

Answer: (c) \( \frac{7}{5\sqrt{2}} \)

Explanation:
1. Write them as 4x − 5y + 3z − 5 = 0 and 4x − 5y + 3z + 2 = 0.
2. Distance = |−5 − 2|/√(16 + 25 + 9) = 7/√50 = 7/(5√2).

Teacher's Note:
1. For parallel planes, distance = |d₁ − d₂|/|n|.
2. Make sure the coefficients of x, y, z are identical first.

Answer: (a) 11x − 3y = 0

Explanation:
1. Normalised planes: (4x + 3y − 5z)/(5√2) and (5x − 12y + 13z)/(13√2).
2. Adding: 13(4x + 3y − 5z) + 5(5x − 12y + 13z) = 77x − 21y = 0, i.e. 11x − 3y = 0.
3. The other bisector, 27x + 99y − 130z = 0, is not among the options.

Teacher's Note:
1. Planes equally inclined to two planes are the angle bisector planes.
2. Use both + and − signs and compare.

Answer: (c) \( \frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}} \)

Explanation:
1. The direction (−2, 1, −1) satisfies 2(−2) + 1 − 3(−1) = 0, so the line is parallel to the plane.
2. Its projection on the plane is a parallel line with the same direction, length √6.
3. Direction cosines = (−2, 1, −1)/√6.

Teacher's Note:
1. If the line is parallel to the plane, its projection is parallel to the line itself.
2. Check d·n = 0 before doing any projection work.

Answer: (c) a⁻² + b⁻² + c⁻² − a′⁻² − b′⁻² − c′⁻² = 0

Explanation:
1. For intercepts a, b, c, the distance p of the plane from the origin satisfies 1/p² = 1/a² + 1/b² + 1/c².
2. The distance from the origin does not depend on the axes, so the same holds for a′, b′, c′ with the same p.
3. Hence a⁻² + b⁻² + c⁻² − a′⁻² − b′⁻² − c′⁻² = 0.

Teacher's Note:
1. The distance from a fixed point to a plane is independent of the coordinate system.
2. Equate the two expressions for 1/p².

Answer: (c) aa′ + cc′ + 1 = 0

Explanation:
1. Taking y as parameter, the direction ratios are (a, 1, c) and (a′, 1, c′).
2. Perpendicular: aa′ + 1 + cc′ = 0.

Teacher's Note:
1. Use y as the parameter to read direction ratios.
2. The constants b, d, b′, d′ do not affect direction.

Answer: (a) 11x + 10y − 5z = 0

Explanation:
1. Family: (x − y + 2z − 3) + λ(4x + 3y − z − 1) = 0.
2. Matching ratios (1 + 4λ) : (−1 + 3λ) : (2 − λ) with 11 : 10 : −5 gives λ = −3.
3. Then the constant −3 − λ = 0, so the plane is 11x + 10y − 5z = 0.

Teacher's Note:
1. A plane through the same line belongs to the pencil of the two planes.
2. Check that the constant term also matches.

Answer: (d) 9

Explanation:
1. With intercepts a, b, c: 1/a² + 1/b² + 1/c² = 1/1² = 1.
2. The centroid is (a/3, b/3, c/3), so a = 3x, b = 3y, c = 3z.
3. (1/9)(x⁻² + y⁻² + z⁻²) = 1, giving k = 9.

Teacher's Note:
1. Use the intercept form to connect the distance from the origin to the intercepts.
2. Express the intercepts through the centroid.

Answer: (a) 1 and 3

Explanation:
1. The plane must contain the point (1, −1, 0) and have a normal perpendicular to the direction (1, −2, 1).
2. Statement 1: normal (1, 1, 1) gives 1 − 2 + 1 = 0, and the point gives 0 — valid.
3. Statement 2: normal (3, 2, −1) gives 3 − 4 − 1 ≠ 0 — invalid.
4. Statement 3: 4 − 2 − 2 = 0, and 4 − 1 − 0 = 3 — valid.
5. Statement 4: the normal works (3 − 4 + 1 = 0) but the point gives 3 − 2 = 1 ≠ 0 — invalid.

Teacher's Note:
1. Test two conditions for each plane: the direction is parallel to the plane and one point of the line lies on it.
2. Failing either one rules out the plane.

Answer: (b) 2 and 3

Explanation:
1. Values of 2x − y − 3z − 5: A = −7, B = 7, C = 5, D = −18.
2. A segment is cut by the plane when its end points have opposite signs: AB and AC.
3. AD (both negative) and BC (both positive) are not cut.

Teacher's Note:
1. Substitute the end points in the plane expression and compare signs.
2. Opposite signs mean the points are on opposite sides.