NDA Three-dimensional Geometry previous year questions with solutions

Answer: (b) 84

Method. sum of squares of medians = 34 × sum of squares of sides

AB² = 4 + 36 = 40, BC² = 36 + 16 = 52, CA² = 4 + 16 = 20; sum = 112.
AD² + BE² + CF² = 34 × 112 = 84.
(Directly: D = (0, 3, 2), AD² = 4 + 9 + 4 = 17; E = (1, 0, 2), BE² = 1 + 36 + 4 = 41; F = (1, 3, 0), CF² = 1 + 9 + 16 = 26; 17 + 41 + 26 = 84.)

Answer: (a) 9 square units

Method. area = 12|AB × AC|

A(0, 7, 10): AB = ⟨-1, -1, -4⟩, AC = ⟨-4, 2, -4⟩.
AB × AC = ⟨(-1)(-4) − (-4)(2), (-4)(-4) − (-1)(-4), (-1)(2) − (-1)(-4)⟩ = ⟨12, 12, -6⟩.
|AB × AC| = ⁠√(144⁠ + 144 + 36) = ⁠√324⁠ = 18.
Area = 9 square units.

Answer: (b) (2, 0, -1)

Method. diagonals of a parallelogram bisect each other: D = A + C − B

Midpoint of AC = midpoint of BD, so D = A + C − B.
D = (1 − 1 + 2, -1 + 2 − 1, 0 + 7 − 8) = (2, 0, -1).

Answer: (b) 27/⁠77

Method. angle between vectors CB and CD by the dot product

D = (2, 0, -1) from the previous item.
CB = B − C = ⟨-1, -1, 1⟩, CD = D − C = ⟨3, -2, -8⟩.
CB · CD = -3 + 2 − 8 = -9; |CB|² = 3, |CD|² = 9 + 4 + 64 = 77.
cos² θ = 813 × 77 = 2777.

Answer: (c) ⁠√66⁠ square units

Method. adjacent sides are perpendicular; area = product of their lengths

AB = B − A = ⟨1, 3, -1⟩, BC = C − B = ⟨2, -1, -1⟩.
AB · BC = 2 − 3 + 1 = 0, so the right angle is at B and AB, BC are adjacent sides.
|AB| = ⁠√(1⁠ + 9 + 1) = ⁠√11,⁠ |BC| = ⁠√(4⁠ + 1 + 1) = ⁠√6.⁠
Area = ⁠√11⁠ × ⁠√6⁠ = ⁠√66⁠ square units.

Answer: (d) 3

Method. right angle at B: BA · BC = 0

BA = A − B = ⟨-k, 1, -3⟩, BC = C − B = ⟨2, k, -1⟩.
BA · BC = -2k + k + 3 = 3 − k = 0.
k = 3.

Answer: (a) -15

Method. parametric point on the line AB

P = A + t(B − A) = (1 + t, 2 + 8t, 3 − 2t).
z = 7: 3 − 2t = 7, so t = -2.
P = (-1, -14, 7); x + y = -15.

Answer: (b) 2 only

Method. common midpoint of diagonals gives a parallelogram; a rectangle also needs a right angle (equal diagonals)

Mid-point of AC = (3, 5, 5); mid-point of BD = (3, 5, 5). Statement 2 is true, so ABCD is a parallelogram.
AB⃗ = (-4, -8, -8), BC⃗ = (6, 10, 6); AB⃗·BC⃗ = -24 − 80 − 48 = -152 ≠ 0.
Angle B is not 90° (equivalently AC = ⁠√12⁠ ≠ BD = ⁠√620),⁠ so it is not a rectangle. Statement 1 is false.

Shortcut. Diagonal lengths ⁠√12⁠ and ⁠√620⁠ differ, so no rectangle.

Answer: (d) plane parallel to xy-plane

Method. z = constant is a plane parallel to the xy-plane

The condition fixes only z; x and y are free.
All points with z = 7 form the plane at height 7, parallel to the xy-plane (z = 0).

Answer: (c) externally in the ratio 2 : 3

Method. the xy-plane (z = 0) divides the join in the ratio -z₁ : z₂

Let z = 0 divide the join in ratio λ : 1. Then 6λ + 4λ + 1 = 0, so λ = -23.
A negative ratio means external division, in the ratio 2 : 3.
(Both points have z > 0, so they lie on the same side of the plane; division must be external.)

Answer: (b) 6449

Method. angle between vectors by dot product, then cos 2θ = 2cos²θ − 1

AB⃗ = (12, 0, 0), AC⃗ = (12, 6, 0), AG⃗ = (12, 6, 4), |AG| = ⁠√(144⁠ + 36 + 16) = 14.
cos α = 14412 · 14 = 67, so cos²α = 3649.
cos β = 144 + 36⁠√180⁠ · 14 = ⁠√18014,⁠ so cos²β = 180196 = 4549.
cos 2α + cos 2β = 2(cos²α + cos²β) − 2 = 2 · 8149 − 2 = 6449.

Answer: (a) Zero

Method. collinearity: the two joining vectors must be proportional

Let P(k, 1, 3), Q(1, -2, k + 1), R(15, 2, -4).
PQ⃗ = (1 − k, -3, k − 2), PR⃗ = (15 − k, 1, -7).
Collinear needs 1 − k15 − k = -31 = k − 2-7.
From the y and z parts: k − 2 = 21, so k = 23.
Then 1 − k15 − k = -22-8 = 114 ≠ -3, so the x part fails.
No value of k works: zero values.

Answer: (c) 8

Method. octants formed by the three coordinate planes

The planes x = 0, y = 0, z = 0 split space by the sign of each coordinate.
Each of x, y, z can be positive or negative: 2 × 2 × 2 = 8 sign patterns.
So space is divided into 8 octants.

Answer: (b) z = 5

Method. a plane parallel to the xy-plane has equation z = constant

A plane parallel to the xy-plane has every point at the same height: z = c.
It meets the z-axis at (0, 0, c), so the intercept 5 gives c = 5.
Equation: z = 5.

Answer: (a) 1, -1

Method. collinearity: the third point is on the line through the other two

Direction from (2, 0, -1) to (4, 2, 3): (2, 2, 4). General point (2 + 2t, 2t, -1 + 4t).
z = -3 ⇒ -1 + 4t = -3 ⇒ t = -12.
x = 2 − 1 = 1, y = -1.

Answer: (b) (3, -3, -1)

Method. centroid = average of the vertices

x = 2 + 5 + 23 = 3, y = -3 − 3 − 33 = -3, z = 3 − 4 − 23 = -1.
Centroid (3, -3, -1).

Answer: (c) 3, 1

Method. a parallelogram inscribed in the skew quadrilateral OPQR has sides parallel to the diagonals PR and OQ

LM lies in plane OPR and KN in plane PQR; parallel lines in these two planes are both parallel to their common line PR. Likewise LK ∥ MN ∥ OQ.
If OL = s·OR, then OM = s·OP, LM = s·PR, and LK = (1 − s)·OQ.
So ⁠√2⁠ = s·PR and ⁠√2⁠ = (1 − s)·OQ ⇒ ⁠√2PR⁠ + ⁠√2OQ⁠ = 1, with PR² = 8 + (6 − 2m)², OQ² = 4 + 4n².
(c) m = 3, n = 1: PR = ⁠√8,⁠ OQ = ⁠√8,⁠ and 12 + 12 = 1 (holds) (s = 12, the midpoints).
(a) m = 6, n = 2: PR = ⁠√44,⁠ OQ = ⁠√20,⁠ sum ≈ 0.53 (fails). (b) m = 1, n = 3: PR = ⁠√24,⁠ OQ = ⁠√40,⁠ sum ≈ 0.51 (fails).
So m = 3, n = 1.

Shortcut. Taking L, M, N, K as midpoints, LM = 12PR and LK = 12OQ, so PR = OQ = 2⁠√2:⁠ 8 + (6 − 2m)² = 8 ⇒ m = 3 and 4 + 4n² = 8 ⇒ n = 1 (taking n > 0).

Answer: (d) 90° - θ

Method. Pythagoras with side lengths: the triangle is right-angled at P

PQ² = 2² + (-1)² + 1² = 6.
PR² = (-1)² + 3² + 5² = 35.
QR² = (-3)² + 4² + 4² = 41 = PQ² + PR², so ∠QPR = 90°.
Angles of the triangle add to 180°: ∠PRQ = 180° - 90° - θ = 90° - θ.

Answer: (a) (73, -83, 0)

Method. parametric form of the line through two points; set z = 0

Direction: (5 + 3, -6 − 4, 4 + 8) = (8, -10, 12). Point: (-3 + 8t, 4 − 10t, -8 + 12t).
On the XY-plane z = 0: -8 + 12t = 0 ⇒ t = 23.
x = -3 + 163 = 73, y = 4 − 203 = -83.
Point: (73, -83, 0).

Shortcut. The XY-plane divides the join in the ratio 8 : 4 = 2 : 1 (from the z-coordinates -8 and 4), so x = 2.5 + 1·(-3)3 = 73, y = 2·(-6) + 43 = -83.

Answer: (c) (a2, b2, c2)

Method. equidistant point from four points of a sphere (centre of the sphere through them)

Let the point be (x, y, z). Equal distance from (0, 0, 0) and (a, 0, 0): x² = (x − a)² ⇒ x = a2.
Similarly, from (0, 0, 0) and (0, b, 0): y = b2; from (0, 0, 0) and (0, 0, c): z = c2.
The point is (a2, b2, c2).

Answer: (c) collinear

Method. collinearity by comparing displacement vectors

PQ⃗ = (4 − 3, 5 − 2, 2 − 4) = (1, 3, -2).
QR⃗ = (5 − 4, 8 − 5, 0 − 2) = (1, 3, -2) = PQ⃗, so P, Q, R are collinear.
PS⃗ = (2 − 3, -1 − 2, 6 − 4) = (-1, -3, 2) = -PQ⃗, so S is on the same line.
All four points are collinear.

No correct option. This question is defective as printed in UPSC's paper.

Method. parametric form of the line through two points; set x = 0

Direction: (3 − 1, -1 − 2, 2 + 1) = (2, -3, 3). Point: (1 + 2t, 2 − 3t, -1 + 3t).
On the yz-plane x = 0: 1 + 2t = 0 ⇒ t = -12.
y = 2 + 32 = 72, z = -1 − 32 = -52.
The line meets the yz-plane at (0, 72, -52), which is none of the options.

No valid answer as printed: the line meets the yz-plane at (0, 72, −52), but the 2017 page, retyped by another publisher (no UPSC original survives), prints options (a) and (d) identically as (0, −72, 52). Option (d) was most likely (0, 72, −52) with its signs lost in retyping.

Answer: (c) Both I and II

Method. cos²α + cos²β + cos²γ = 1 with cos β = sin α

α + β = 90° gives cos β = sin α, so cos² α + sin² α + cos² γ = 1, hence cos γ = 0.
With α, β ≥ 0 and α + β = 90°, α ∈ [0°, 90°].
cos α + cos β = cos α + sin α = ⁠√2⁠ sin(α + 45°), maximum ⁠√2⁠ at α = 45°: I is correct.
cos α + cos β + cos γ = cos α + sin α ≥ 1 on [0°, 90°], with 1 attained at α = 0° (the x-axis itself): II is correct.

Answer: (d) Neither I nor II

Method. parallel: proportional direction ratios; perpendicular: zero dot product

PQ = Q − P = ⟨6, 2, 0⟩, RS = S − R = ⟨-4, -2, 2⟩: not proportional (third components 0 and 2), so I is false.
PR = R − P = ⟨8, -1, 3⟩, QS = S − Q = ⟨-2, -5, 5⟩.
PR · QS = -16 + 5 + 15 = 4 ≠ 0, so II is false.
Neither statement is correct.

Answer: (b) 0

Method. squared distances from coordinate differences

PQ² = 36 + 4 + 0 = 40.
QS² = 4 + 25 + 25 = 54.
PR² = 64 + 1 + 9 = 74.
PQ² + 2QS² - 2PR² = 40 + 108 − 148 = 0.

Answer: (b) -cos² γ

Method. cos(A + B)cos(A − B) = cos² A − sin² B with cos²α + cos²β + cos²γ = 1

cos(α + β) cos(α − β) = cos² α − sin² β.
= cos² α − (1 − cos² β) = cos² α + cos² β − 1.
Since cos² α + cos² β + cos² γ = 1, this equals -cos² γ.

Answer: (a) -31/⁠49

Method. direction cosine with the y-axis, then cos 2θ = 2cos²θ − 1

p = 2q = 3r = 6λ gives p: q: r = 6 : 3 : 2, magnitude ⁠√(36⁠ + 9 + 4) = 7.
cos θ = ±37 (sign depends on λ; it does not matter below).
cos 2θ = 2 × 949 − 1 = -3149.

Answer: (b) 29101

Method. direction perpendicular to two given vectors via the cross product

l + 2m + n = 0 and 2l − 2m + 3n = 0 say ⟨l, m, n⟩ is perpendicular to ⟨1, 2, 1⟩ and ⟨2, -2, 3⟩.
⟨1, 2, 1⟩ × ⟨2, -2, 3⟩ = ⟨2.3 − 1·(-2), 1.2 − 1.3, 1·(-2) − 2.2⟩ = ⟨8, -1, -6⟩.
Magnitude² = 64 + 1 + 36 = 101, so l² = 64101, m² = 1101, n² = 36101.
l² + m² - n² = 64 + 1 − 36101 = 29101.

Answer: (d) 90°

Method. read direction ratios by setting each equal expression to t

2x = 3y = -z = t: (x, y, z) = t(12, 13, -1), direction ratios (3, 2, -6).
6x = -y = -4z = t: (x, y, z) = t(16, -1, -14), direction ratios (2, -12, -3).
3.2 + 2·(-12) + (-6)(-3) = 6 − 24 + 18 = 0.
The lines are perpendicular: 90°.

Answer: (a) 4

Method. cos θ from direction ratios, set θ = π/⁠4 and solve

cos π4 = 2k − 3 + 103⁠√(k²⁠ + 34) = 1⁠√2.⁠
⁠√2(2k⁠ + 7) = 3⁠√(k²⁠ + 34); squaring: 8k² + 56k + 98 = 9k² + 306.
k² - 56k + 208 = 0, (k − 4)(k − 52) = 0, so k = 4 or 52 (both keep 2k + 7 > 0).
Only 4 is among the options.

Answer: (d) (11, 2, -10)

Method. a line perpendicular to both has direction along the cross product

With k = 4 the direction ratios are (2, -1, 2) and (4, 3, 5).
(2, -1, 2) × (4, 3, 5) = ((-1)(5) − 2.3, 2.4 − 2.5, 2.3 − (-1)·4) = (-11, -2, 10).
This is proportional to (11, 2, -10).
Check: 22 − 2 − 20 = 0 and 44 + 6 − 50 = 0.

Answer: (c) Both 1 and 2

Method. direction ratios are any non-zero multiple; perpendicular to z-axis means third ratio 0

1. y-axis has direction cosines ⟨0, 1, 0⟩; ⟨0, 4, 0⟩ is 4 times this. True.
2. Perpendicular to z-axis ⟨0, 0, 1⟩ needs 5.0 + 6.0 + 0.1 = 0, which holds. True.
Both correct.

Answer: (c) Both 1 and 2

Method. direction ratios are any non-zero multiples of the direction cosines; l² + m² + n² = 1

1. If a, b, c are direction ratios, so are ka, kb, kc for every k ≠ 0: infinitely many. True.
2. Always l² + m² + n² = 1. For the x-axis, (l, m, n) = (1, 0, 0) and l + m + n = 1 too. True.
Both statements are correct.

Answer: (b) π4

Method. angle between lines from direction ratios: cos θ = (a₁a₂ + b₁b₂ + c₁c₂)/⁠(|d₁||d₂|)

a₁a₂ + b₁b₂ + c₁c₂ = 6.3 + 3.3 + 6.0 = 27.
⁠√(6²⁠ + 3² + 6²) = ⁠√81⁠ = 9; ⁠√(3²⁠ + 3² + 0²) = 3⁠√2.⁠
cos θ = 279 · 3⁠√2⁠ = 1⁠√2,⁠ so θ = π4.

Answer: (b) 1127

Method. rewrite the line in symmetric form to read its direction ratios

x − 1 = 2(y + 3) = 1 − z is x − 11 = y + 31/⁠2 = z − 1-1.
Direction ratios 1, 12, -1, i.e. 2, 1, -2; ⁠√(4⁠ + 1 + 4) = 3.
l, m, n = 23, 13, -23.
l⁴ + m⁴ + n⁴ = 16 + 1 + 1681 = 3381 = 1127.

Answer: (c) 3

Method. projection of a segment on the y-axis is |y₂ - y₁|

The y-axis has direction cosines 0, 1, 0.
Projection of AB = |(x₂ - x₁)·0 + (y₂ - y₁)·1 + (z₂ - z₁)·0| = |4 − 7| = 3.

Answer: (d) 1

Method. l² + m² + n² = 1 for any direction cosines

Whatever the direction ratios, the direction cosines are obtained by dividing them by ⁠√((a⁠ + b)² + (b + c)² + (c + a)²).
Hence the sum of the squares of the direction cosines is always 1.

Answer: (a) 12, 12, 1⁠√2⁠

Method. direction ratios from the coefficients of the parameter, then normalise

(p + 1, p − 3, ⁠√2⁠ p) = (1, -3, 0) + p(1, 1, ⁠√2),⁠ so the direction ratios are 1, 1, ⁠√2.⁠
⁠√(1⁠ + 1 + 2) = 2.
Direction cosines: 12, 12, ⁠√22⁠ = 12, 12, 1⁠√2.⁠

Answer: (b) 60°

Method. l² + m² + n² = 1 gives the two possible lines; angle from the dot product of direction cosines

n = cos 60° = 12, m = ⁠√3⁠ l.
l² + 3l² + 14 = 1 ⇒ l² = 316 ⇒ l = ±⁠√34,⁠ m = ±34 (same sign).
Lines: (⁠√34,⁠ 34, 12) and (-⁠√34,⁠ -34, 12).
cos θ = -316 − 916 + 14 = -12 ⇒ θ = 120°.
The acute angle between the lines is 180° - 120° = 60°.

Answer: (d) < 0, 0, 1 >

Method. the z-axis makes angles 90°, 90°, 0° with the axes

Angles with x-, y-, z-axes: 90°, 90°, 0°.
Direction cosines: cos 90°, cos 90°, cos 0° = < 0, 0, 1 >.

Answer: (c) 1 and 3 only

Method. cos² α + cos² β + cos² γ = 1

(1) cos² α + cos² β = 1 − cos² γ = sin² γ: correct.
(2) sin² α + sin² β = 2 − (cos² α + cos² β) = 2 − sin² γ = 1 + cos² γ ≠ cos² γ: wrong.
(3) sin² α + sin² β + sin² γ = 3 − (cos² α + cos² β + cos² γ) = 3 − 1 = 2: correct.
So 1 and 3 only.

Answer: (b) 4

Method. angle between lines: cos θ = a₁a₂ + b₁b₂ + c₁c₂⁠√(a₁²⁠ + b₁² + c₁²) ⁠√(a₂²⁠ + b₂² + c₂²)

cos π4 = 2x − 3 + 10⁠√9⁠ · ⁠√(x²⁠ + 34) ⇒ 1⁠√2⁠ = 2x + 73⁠√(x²⁠ + 34).
Squaring: 2(2x + 7)² = 9(x² + 34) ⇒ 8x² + 56x + 98 = 9x² + 306.
x² - 56x + 208 = 0 ⇒ (x − 4)(x − 52) = 0 ⇒ x = 4 or 52.
Both give 2x + 7 > 0, so both are valid; the smaller value is 4.

Answer: (b) parallel to y-axis

Method. direction cosines are the cosines of the angles with the axes

Direction cosines (0, 1, 0) give cos α = 0, cos β = 1, cos γ = 0.
So α = 90°, β = 0°, γ = 90°: the line makes zero angle with the y-axis.
Hence the line is parallel to the y-axis.

Answer: (c) ae + cg + 1 = 0

Method. perpendicular lines: dot product of direction ratios is zero

x = ay + b, z = cy + d: as y increases by 1, x increases by a and z by c, so direction ratios are (a, 1, c).
Likewise the second line has direction ratios (e, 1, g).
Perpendicular ⇔ a·e + 1.1 + c·g = 0, i.e. ae + cg + 1 = 0.

Answer: (b) x − 2y + 3z − 28 = 0

Method. point-normal form with normal OP

Normal OP = ⟨2, -4, 6⟩ ∝ ⟨1, -2, 3⟩.
1(x − 2) − 2(y + 4) + 3(z − 6) = 0.
x − 2y + 3z − 28 = 0.

Answer: (b) 3

Method. mirror plane: normal along PP′, passing through the midpoint

P(1, 3, 4), image Q(-3, 5, 2): PQ = ⟨-4, 2, -2⟩ ∝ ⟨2, -1, 1⟩, so (A, B, C) = t(2, -1, 1).
A + B + C = 2t = 2, so t = 1: A = 2, B = -1, C = 1.
Midpoint (-1, 4, 3) lies on the plane: 2(-1) − 4 + 3 + D = 0.
D = 3.

Answer: (b) 3x + 2y + z = 6

Method. point-normal form of a plane

The line's direction ⟨3, 2, 1⟩ is the plane's normal.
3(x − 1) + 2(y − 1) + 1(z − 1) = 0.
3x + 2y + z = 6.

Answer: (b) 1

Method. direction cosines of the normal from the plane's coefficients

Normal to 2x − 3y + 6z + 4 = 0 has direction ratios ⟨2, -3, 6⟩, magnitude ⁠√(4⁠ + 9 + 36) = 7.
So l = 27, m = -37, n = 67 (the opposite signs give the same squares).
7l² + m² - n² = 28 + 9 − 3649 = 149.
49(7l² + m² - n²) = 1.

Answer: (d) 1

Method. intercept form of a plane; distance from origin

Intercepts 1, 1, 1 give the plane x + y + z = 1.
p = |0 + 0 + 0 − 1|⁠√(1⁠ + 1 + 1) = 1⁠√3.⁠
3p² = 3 × 13 = 1.

Answer: (c) ⟨1⁠√6,⁠ 1⁠√6,⁠ 2⁠√6⟩⁠

Method. intercept form: normal of x/⁠a + y/⁠b + z/⁠c = 1 is ⟨1/⁠a, 1/⁠b, 1/⁠c⟩

Plane: x2 + y2 + z1 = 1, normal ⟨12, 12, 1⟩ ∝ ⟨1, 1, 2⟩.
⁠√(1⁠ + 1 + 4) = ⁠√6.⁠
Direction cosines ⟨1⁠√6,⁠ 1⁠√6,⁠ 2⁠√6⟩.⁠

Answer: (a) <3, 2, 1>

Method. substitute the point to find k, then read the normal from the coefficients

(2, 3, -6) on the plane: 4k + 2 − 2 = 2, so k = 2.
Plane: x + 2y3 + z3 = 2, i.e. 3x + 2y + z = 6.
Normal direction ratios ⟨3, 2, 1⟩.

Answer: (b) 11

Method. intercept form x/⁠p + y/⁠q + z/⁠r = 1

With k = 2 the plane is 3x + 2y + z = 6, i.e. x2 + y3 + z6 = 1.
p = 2, q = 3, r = 6; p + q + r = 11.

Answer: (c) (1, 1, 1)

Method. foot of perpendicular from origin to ax + by + cz = d is da² + b² + c²(a, b, c)

The perpendicular from O runs along the normal (1, 1, 1): point (t, t, t).
On the plane: t + t + t = 3, so t = 1.
Foot = (1, 1, 1).

Answer: (a) 10x − 2y + 2z = 28

Method. family of planes through a line of intersection: P₁ + λP₂ = 0

Plane: (2x + y + 2z − 9) + λ(4x − 5y − 4z − 1) = 0.
At (3, 2, 1): (6 + 2 + 2 − 9) + λ(12 − 10 − 4 − 1) = 1 − 3λ = 0 ⇒ λ = 13.
Multiply by 3: 3(2x + y + 2z − 9) + (4x − 5y − 4z − 1) = 0 ⇒ 10x − 2y + 2z − 28 = 0.
That is 10x − 2y + 2z = 28.

Answer: (a) 14

Method. distance between parallel planes |d₁ - d₂|⁠√(a²⁠ + b² + c²) with equal coefficients

Write 4x − 2y + 4z + 9 = 0 as 8x − 4y + 8z + 18 = 0.
Distance = |21 − 18|⁠√(64⁠ + 16 + 64) = 312 = 14.

Answer: (a) 1 unit

Method. distance of a point from a plane |ax₁ + by₁ + cz₁ + d|/⁠√(a²⁠ + b² + c²)

Distance = |3.2 − 6.3 + 2.4 + 11|⁠√(9⁠ + 36 + 4) = |6 − 18 + 8 + 11|7 = 77 = 1 unit.

Answer: (c) Both 1 and 2

Method. angle between planes via normals; distance between parallel planes after equalising coefficients

1. Normals (2, -1, 1) and (1, 1, 2): cos θ = 2 − 1 + 2⁠√6⁠ · ⁠√6⁠ = 12 ⇒ θ = π3. Correct.
2. Write 2x − y + 2z + 4 = 0 as 6x − 3y + 6z + 12 = 0.
Distance = |12 − 2|⁠√(36⁠ + 9 + 36) = 109. Correct.
Both statements are correct.

Answer: (a) 2x − y − 2z = 2

Method. a plane through three points: test each option against all three points

Test 2x − y − 2z = 2:
(-2, 6, -6): -4 − 6 + 12 = 2 (holds); (-3, 10, -9): -6 − 10 + 18 = 2 (holds); (-5, 0, -6): -10 − 0 + 12 = 2 (holds).
The other options fail at (-2, 6, -6): (b) gives -16 ≠ 3, (c) gives -2 ≠ 6, (d) gives -2 ≠ 3.
So the plane is 2x − y − 2z = 2.

Answer: (b) 3 units

Method. distance of the origin from the plane ax + by + cz = d is |d|/⁠√(a²⁠ + b² + c²)

Length of normal = |9|⁠√(1²⁠ + 2² + (-2)²) = 9⁠√9⁠ = 93 = 3 units.

Answer: (a) x + 2y + 3z − 6 = 0

Method. family of planes through a line of intersection: P₁ + λP₂ = 0

Plane: (x + y + z − 1) + λ(2x + 3y + 4z − 7) = 0, normal (1 + 2λ, 1 + 3λ, 1 + 4λ).
Perpendicular to x − 5y + 3z = 5: (1 + 2λ) − 5(1 + 3λ) + 3(1 + 4λ) = 0 ⇒ -1 − λ = 0 ⇒ λ = -1.
Plane: (x + y + z − 1) − (2x + 3y + 4z − 7) = 0 ⇒ -x − 2y − 3z + 6 = 0.
That is x + 2y + 3z − 6 = 0.

Answer: (c) 3x + 3 = 2y − 4 = 6z + 18

Method. line along the normal of the plane, in symmetric form

Direction of L = normal of P = ⟨2, 3, 1⟩.
L: x + 12 = y − 23 = z + 31.
Multiply each member by 6: 3(x + 1) = 2(y − 2) = 6(z + 3), i.e. 3x + 3 = 2y − 4 = 6z + 18.

Answer: (d) <2, 2, -10>

Method. a line parallel to a plane is perpendicular to its normal

Need ⟨a, b, c⟩ · ⟨2, 3, 1⟩ = 0.
(a) -6 + 6 + 1 = 1; (b) 6 + 6 − 6 = 6; (c) 2 + 9 + 2 = 13; (d) 4 + 6 − 10 = 0.
Only ⟨2, 2, -10⟩ works.

Answer: (b) ⟨2, -23, -16⟩

Method. line of intersection is along the cross product of the normals

Normals ⟨6, 4, -5⟩ and ⟨1, -2, 3⟩.
⟨6, 4, -5⟩ × ⟨1, -2, 3⟩ = ⟨4.3 − (-5)(-2), (-5)·1 − 6.3, 6·(-2) − 4.1⟩ = ⟨2, -23, -16⟩.
Check: 12 − 92 + 80 = 0 and 2 + 46 − 48 = 0. Option (a) ⟨2, 23, 16⟩ fails: 12 + 92 − 80 ≠ 0.

Answer: (a) 2x − 20y + 29z + 2 = 0

Method. plane through the intersection: P₁ + λP₂ = 0, normal perpendicular to the given line

P: (6x + 4y − 5z − 2) + λ(x − 2y + 3z) = 0, normal ⟨6 + λ, 4 − 2λ, -5 + 3λ⟩.
Parallel to ⟨1, 3, 2⟩: (6 + λ) + 3(4 − 2λ) + 2(-5 + 3λ) = 8 + λ = 0, so λ = -8.
P: -2x + 20y − 29z − 2 = 0, i.e. 2x − 20y + 29z + 2 = 0.
Check: ⟨2, -20, 29⟩ · ⟨1, 3, 2⟩ = 2 − 60 + 58 = 0.

Answer: (b) ⟨-7, 6, 1⟩

Method. line of intersection is along the cross product of the normals

Normals ⟨1, 1, 1⟩ and ⟨2, 3, -4⟩.
⟨1, 1, 1⟩ × ⟨2, 3, -4⟩ = ⟨1·(-4) − 1.3, 1.2 − 1·(-4), 1.3 − 1.2⟩ = ⟨-7, 6, 1⟩.
Check: -7 + 6 + 1 = 0 and -14 + 18 − 4 = 0.

Answer: (a) 6

Method. direction cosines from the direction ratios ⟨-7, 6, 1⟩

From the previous item, S has direction ratios ⟨-7, 6, 1⟩; magnitude² = 49 + 36 + 1 = 86.
l² = 4986, m² = 3686, n² = 186.
l² - m² - n² = 49 − 36 − 186 = 1286 = 643.
43(l² - m² - n²) = 6.

Answer: (c) ⟨0, 1, -1⟩

Method. line of intersection is along the cross product of the normals

Normals ⟨1, 1, 1⟩ and ⟨2, -1, -1⟩.
⟨1, 1, 1⟩ × ⟨2, -1, -1⟩ = ⟨1·(-1) − 1·(-1), 1.2 − 1·(-1), 1·(-1) − 1.2⟩ = ⟨0, 3, -3⟩.
Dividing by 3: ⟨0, 1, -1⟩.

Answer: (d) (-4, -3, 3)

Method. parametrise the line, then substitute into the plane

Add the two plane equations of L: 3x + 12 = 0, so x = -4 everywhere on L; then -4 + y + z + 4 = 0 gives z = -y.
L: (-4, t, -t).
In P: -4 + 2t − 3t + 1 = 0, so t = -3.
Point (-4, -3, 3). Check: -4 − 3 + 3 + 4 = 0, -8 + 3 − 3 + 8 = 0, -4 − 6 + 9 + 1 = 0.

Shortcut. Test the options: only (-4, -3, 3) satisfies all three equations.

Answer: (c) (4, 1, 4)

Method. parametric point of the line substituted into the plane

Line: (x, y, z) = (1 + 3t, -1 + 2t, 2 + 2t).
Substitute in x + 2y + 3z = 18: (1 + 3t) + 2(-1 + 2t) + 3(2 + 2t) = 5 + 13t = 18, so t = 1.
Point: (1 + 3, -1 + 2, 2 + 2) = (4, 1, 4).

Shortcut. Only (4, 1, 4) satisfies 4 + 2 + 12 = 18 and has x − 1 : y + 1 : z − 2 = 3 : 2 : 2.

Answer: (d) (137, -117, 197) and (17, -37, -57)

Method. point at distance r along a line: (x₁, y₁, z₁) ± r(l, m, n)

⁠√(9⁠ + 4 + 36) = 7, so the direction cosines are (37, -27, 67).
Points: (1, -1, 1) ± 2(37, -27, 67) = (1, -1, 1) ± (67, -47, 127).
+ gives (137, -117, 197); - gives (17, -37, -57).

Answer: (d) x − 2y + z − 1 = 0

Method. a line is parallel to a plane when its direction is perpendicular to the plane's normal

Direction of the line: (3, 4, 5). Test normal · direction:
(a) 6 + 8 + 5 = 19; (b) 6 − 4 − 10 = -8; (c) 6 + 8 − 10 = 4; (d) 3 − 8 + 5 = 0.
Only (d) gives 0. The point (2, 3, 4) gives 2 − 6 + 4 − 1 = -1 ≠ 0, so the line is parallel to the plane and not in it.

Answer: (b) 2 and 3 only

Method. parametric form of the line AB

A = (1, -1, 0), B = (0, 1, 1); AB⃗ = (-1, 2, 1).
Points of the line: (1 − t, -1 + 2t, t).
1. x = -1 gives t = 2, point (-1, 3, 2) ≠ (-1, -3, 1): not on the line.
2. (-1, 3, 2) is t = 2: on the line.
3. (-2, 5, 3) is t = 3: on the line.
So 2 and 3 only.

Answer: (c) ⁠√6⁠

Method. |AB⃗| = |position vector of B − position vector of A|

AB⃗ = (ĵ + k̂) − (î − ĵ) = -î + 2ĵ + k̂.
|AB⃗| = ⁠√(1⁠ + 4 + 1) = ⁠√6.⁠

Answer: (b) 5 units

Method. the line with direction (1, 0, 0) through the origin is the x-axis

The line passes through the origin with direction ratios (1, 0, 0), so it is the x-axis.
Distance of (x, y, z) from the x-axis = ⁠√(y²⁠ + z²) = ⁠√(9⁠ + 16) = 5 units.

Answer: (b) (2, 5, 5)

Method. general point of the line (1 + t, 3 + 2t, -2 + 7t)

(b) (2, 5, 5): x = 2 ⇒ t = 1; then y = 3 + 2 = 5 (holds), z = -2 + 7 = 5 (holds).
(a) x = 3 ⇒ t = 2, y = 7 ≠ 5. (c) x = -1 ⇒ t = -2, z = -16 ≠ 5. (d) x = 2 ⇒ t = 1, y = 5 ≠ -1.
So (2, 5, 5) lies on the line.

Answer: (d) 7

Method. a line lies in a plane if its point lies in the plane and its direction is ⊥ the normal

Direction (1, 1, 2)·normal (2, -4, 1) = 2 − 4 + 2 = 0, so the line is parallel to the plane.
The point (4, 2, k) must lie in the plane: 2.4 − 4.2 + k = 7 ⇒ k = 7.

Answer: (d) 3x + 2y − 3z = -2, 3x − 6y + 3z = 0

Method. the line lies in both planes: each plane contains (1, 2, 3) and has normal ⊥ (2, 3, 4)

The line passes through (1, 2, 3) with direction (2, 3, 4).
(d) 3x + 2y − 3z: 3 + 4 − 9 = -2 (holds); normal (3, 2, -3)·(2, 3, 4) = 6 + 6 − 12 = 0 (holds).
3x − 6y + 3z: 3 − 12 + 9 = 0 (holds); normal (3, -6, 3)·(2, 3, 4) = 6 − 18 + 12 = 0 (holds). Normals not parallel, so the planes meet in exactly this line.
Others fail: (a) normal (1, 1, 1)·(2, 3, 4) = 9 ≠ 0; (b) 4 + 8 − 15 − 3 = -6 ≠ 0; (c) 3 + 4 − 9 = -2 ≠ 0.

Answer: (c) (4, 5, -2)

Method. foot of perpendicular: parametric point on BC with AD · BC = 0

BC⃗ = (2, 8, -3). A point on BC: D = (2t, -11 + 8t, 4 − 3t).
AD⃗ = (2t − 1, 8t − 19, -3t).
AD⃗ · BC⃗ = 2(2t − 1) + 8(8t − 19) − 3(-3t) = 77t − 154 = 0 ⇒ t = 2.
D = (4, 5, -2).

Answer: (c) 7

Method. concentric sphere has the same centre; radius = distance from centre to origin

Centre of S = (2, 3, 6), whatever k is.
The concentric sphere through the origin has radius ⁠√(2²⁠ + 3² + 6²) = ⁠√49⁠ = 7.

Answer: (a) -15

Method. r² = g² + f² + h² - k

Centre (2, 3, 6), so r² = 4 + 9 + 36 − k = 49 − k.
49 − k = 64, so k = -15.

Answer: (b) ⁠√(23)⁠

Method. smallest sphere through three points has their circumcircle as a great circle

AB = BC = CA = ⁠√2,⁠ so ABC is equilateral.
The smallest sphere has centre at the circumcentre of ABC and radius equal to its circumradius.
Circumradius of an equilateral triangle of side a = a⁠√3⁠ = ⁠√2√3⁠ = ⁠√(23)⁠.

Answer: (a) x + y + z − 1 = 0

Method. centre of S = circumcentre (= centroid) of the equilateral triangle ABC

Centre of S = centroid of ABC = (13, 13, 13).
x + y + z − 1 = 13 + 13 + 13 − 1 = 0, so it lies on x + y + z − 1 = 0.
The other left-hand sides give (b) 2, (c) 2, (d) 4, none zero.

Answer: (b) 5⁠√32⁠

Method. centre and radius of a sphere in general form

Divide by 2: x² + y² + z² + 32x + 32y + 32z − 3 = 0.
Centre (-34, -34, -34); r² = 3 × 916 + 3 = 27 + 4816 = 7516.
r = 5⁠√34,⁠ so diameter = 5⁠√32.⁠

Answer: (d) 4x + 8y + 8z + 15 = 0

Method. substitute the centre into each plane

Centre (from the previous item) = (-34, -34, -34).
(a) 2(-94) − 3 = -152 ≠ 0; (b) 4(-94) − 3 = -12 ≠ 0.
(c) 4(-34) + 8(-34) + 8(-34) − 15 = -3 − 6 − 6 − 15 = -30 ≠ 0.
(d) -3 − 6 − 6 + 15 = 0. The centre lies on 4x + 8y + 8z + 15 = 0.

Answer: (a) -2

Method. centre of the sphere = mid-point of the diameter = (-u, -v, -w)

Mid-point of the diameter = (32, 0, 12) = (-u, -v, -w).
u + v + w = -(32 + 0 + 12) = -2.
Check: (x − 1)(x − 2) + (y + 1)(y − 1) + (z − 2)(z + 1) = 0 gives x² + y² + z² - 3x − z − 1 = 0, constant -1 as given.

Answer: (d) x² + y² + z² - 2x − 6y − 8z = 0

Method. concentric spheres differ only in the constant term

Concentric sphere: x² + y² + z² - 2x − 6y − 8z + d = 0.
Through the origin: d = 0.
Equation: x² + y² + z² - 2x − 6y − 8z = 0.

Answer: (a) -2

Method. centre of the sphere = mid-point of the diameter = (-u, -v, -w)

Mid-point of AB = (32, 0, 12) = (-u, -v, -w).
u + v + w = -(32 + 0 + 12) = -2.
Check: (x − 1)(x − 2) + (y + 1)(y − 1) + (z − 2)(z + 1) = 0 expands to x² + y² + z² - 3x − z − 1 = 0, constant -1 as given.

Answer: (b) 14

Method. angle in a semicircle: P on the sphere sees the diameter AB at 90°

For P on the sphere, ∠APB = 90°, so PA² + PB² = AB² (also true at P = A or B).
AB² = (2 − 1)² + (1 + 1)² + (-1 − 2)² = 1 + 4 + 9 = 14.

Answer: (b) 2 only

Method. tangency: distance of centre from the line equals the radius

Centre (2, 3, 4); r² = 4 + 9 + 16 + 16 = 45, r = ⁠√45.⁠
Distance of the centre from the z-axis = ⁠√(2²⁠ + 3²) = ⁠√13⁠ ≠ ⁠√45:⁠ z-axis cuts the sphere, not tangent. Statement 1 false.
2 + 3 + 4 − 9 = 0: centre lies on the plane. Statement 2 true.

Answer: (b) 4

Method. sphere through O and the axis intercepts (a, 0, 0), (0, b, 0), (0, 0, c) is x² + y² + z² - ax − by − cz = 0

The plane meets the axes at A(2, 0, 0), B(0, 12k, 0), C(0, 0, 4).
Sphere through O, A, B, C: x² + y² + z² - 2x − 12ky − 4z = 0.
Comparing with the given -3y: 12k = 3, so k = 4.

Answer: (b) 0.5 < p < 1

Method. distance of a point from a plane: |ax₁ + by₁ + cz₁ + d|/⁠√(a²⁠ + b² + c²)

Centre of x² + y² + z² - 2x − 3y − 4z = 0 is (1, 32, 2); the plane is 6x + 4y + 3z − 12 = 0 (k = 4).
p = |6 + 6 + 6 − 12|⁠√(36⁠ + 16 + 9) = 6⁠√61.⁠
⁠√61⁠ ≈ 7.81, so p ≈ 0.77: 0.5 < p < 1.

Answer: (d) 6x = 4y = 3z

Method. line through origin with direction along the centre: x/⁠a = y/⁠b = z/⁠c

Centre (1, 32, 2); the line is x1 = y3/⁠2 = z2, i.e. x = 2y3 = z2.
Multiply each part by 6: 6x = 4y = 3z.

Answer: (c) 8

Method. distance of the centre from each axis equals r

Let the centre be (a, b, c). Distances from the x-, y-, z-axes are ⁠√(b²⁠ + c²), ⁠√(c²⁠ + a²), ⁠√(a²⁠ + b²).
Setting each equal to r: b² + c² = c² + a² = a² + b² = r², so a² = b² = c² = r²2.
a, b, c = ±r⁠√2,⁠ independently: 2 × 2 × 2 = 8 centres.
So 8 spheres.

Answer: (d) 4 units

Method. radius of a sphere touching a plane = distance of the centre from the plane

r = |6.1 − 3·(-2) + 2.3 − 4|⁠√(36⁠ + 9 + 4) = |6 + 6 + 6 − 4|7 = 147 = 2.
Diameter = 2r = 4 units.

Answer: (c) 7

Method. sphere x² + y² + z² + 2ux + 2vy + 2wz + d = 0 has radius ⁠√(u²⁠ + v² + w² - d)

Centre (3, -4, 5), d = 1.
r = ⁠√(9⁠ + 16 + 25 −⁠ 1) = ⁠√49⁠ = 7.

Answer: (c) 9(x² + y² + z²) = 4r²

Method. sphere through the origin cuts the axes at twice the centre's coordinates

Sphere through O: x² + y² + z² - 2ux − 2vy − 2wz = 0, centre (u, v, w), radius² = u² + v² + w² = r².
Putting y = z = 0: x = 2u, so A(2u, 0, 0); similarly B(0, 2v, 0), C(0, 0, 2w).
Centroid (x, y, z) = (2u3, 2v3, 2w3) ⇒ u = 3x2, v = 3y2, w = 3z2.
94(x² + y² + z²) = r² ⇒ 9(x² + y² + z²) = 4r².

Answer: (a) x² + y² + z² + 4x − 6y − 8z = 7

Method. sphere (x − a)² + (y − b)² + (z − c)² = r²

(x + 2)² + (y − 3)² + (z − 4)² = 36.
x² + y² + z² + 4x − 6y − 8z + 4 + 9 + 16 = 36.
x² + y² + z² + 4x − 6y − 8z = 7.

Answer: (c) ax + by + cz = 2

Method. intercept form of the plane; centre of the sphere through O and the three intercepts

Let the plane be xα + yβ + zγ = 1, so A(α, 0, 0), B(0, β, 0), C(0, 0, γ).
It passes through (a, b, c): aα + bβ + cγ = 1.
The sphere OABC is x² + y² + z² - αx − βy − γz = 0, with centre (α2, β2, γ2).
Let the centre be (x, y, z): α = 2x, β = 2y, γ = 2z.
Then a2x + b2y + c2z = 1, i.e. ax + by + cz = 2.