JEE Mathematics Introduction to Three Dimensional Geometry MCQs Set 01

Answer: (b) x² + a² = c² − y² − z²

Explanation:
1. (x − a)² + y² + z² + (x + a)² + y² + z² = 2c².
2. 2x² + 2a² + 2y² + 2z² = 2c², so x² + a² = c² − y² − z².

Teacher's Note:
1. Expand both squared distances and add.

Answer: (b) (5, 4)

Explanation:
1. The direction from (3, 2, −4) to (9, 8, −10) is (6, 6, −6), i.e. (1, 1, −1).
2. (λ − 3, μ − 2, −2) must be a multiple of (1, 1, −1): the multiple is 2, so λ = 5 and μ = 4.

Teacher's Note:
1. Collinear points give proportional direction vectors.

Answer: (a) \( \left(0,\frac{13}{5},2\right) \)

Explanation:
1. Points on the line are (3 − 5t, 5 − 4t, −7 + 15t). On the yz-plane x = 0, so t = 3/5.
2. Then y = 5 − 12/5 = 13/5 and z = −7 + 9 = 2.

Teacher's Note:
1. On the yz-plane the x-coordinate is zero.

Answer: (b) 3 : 2 externally

Explanation:
1. R − P = (6, 6, −6) = 3(Q − P), so R lies beyond Q on ray PQ.
2. PR : RQ = 6 : 4 = 3 : 2, with R outside the segment: externally 3 : 2.

Teacher's Note:
1. Compare R − P with Q − P.

Answer: (a) Collinear

Explanation:
1. The vector from the first to the second point is (5, −10, −15) and from the first to the third is (3, −6, −9).
2. (5, −10, −15) = (5/3)(3, −6, −9), so the points are collinear.

Teacher's Note:
1. Check whether the two direction vectors are proportional.

Answer: (c) The points A(−2, −1), B(1, 0), C(4, 3), D(1, 2) are vertices of a parallelogram

Explanation:
1. In (c), AB = (3, 1) = DC and AD = (3, 3) = BC, so ABCD is a parallelogram.
2. In (a), all four sides equal √8 and both diagonals equal 4, so ABCD is a square, not a non-square rectangle.
3. In (b), AB = √13 and BC = √52 are unequal, so it is not a square.

Teacher's Note:
1. Compute side vectors and diagonals.

Answer: (b) \( \left(-\frac12,-2,\frac{11}{2}\right) \)

Explanation:
1. The centroid G divides the median AD in the ratio 2 : 1, so D = (3G − A)/2.
2. 3G − A = (3 − 4, 3 − 7, 3 + 8) = (−1, −4, 11), so D = (−1/2, −2, 11/2).

Teacher's Note:
1. Use G = (A + 2D)/3.

Answer: (c) y² − 2y − 4x + 2z + 6 = 0

Explanation:
1. The distance from the y-axis is √(x² + z²).
2. x² + z² = (x − 2)² + (y − 1)² + (z + 1)² simplifies to 0 = −4x + y² − 2y + 2z + 6.

Teacher's Note:
1. Square the equal distances and cancel x² and z².

Answer: (b) 3 : 1

Explanation:
1. On the XOY plane z = 0. If the ratio is λ : 1, z = (−λ + 3)/(λ + 1) = 0 gives λ = 3.
2. The ratio is 3 : 1 (internal).

Teacher's Note:
1. Set the z-coordinate of the dividing point to zero.

Answer: (b) Square

Explanation:
1. With A, B, C, D in order: AB = (−3, 3, 0) = DC, so ABCD is a parallelogram.
2. AD = (1, 1, 4) and AB · AD = −3 + 3 + 0 = 0, so it is a rectangle; |AB| = |AD| = 3√2, so all sides are equal.
3. The most specific description is a square.

Teacher's Note:
1. Check side vectors, then perpendicularity, then equal adjacent sides.

Answer: (b) −1

Explanation:
1. A point of QR is (2 + 3t, 2, 1 − 3t). x = 4 gives t = 2/3.
2. Then z = 1 − 2 = −1.

Teacher's Note:
1. Parametrise the line QR.

Answer: (c) \( \sqrt{38} \)

Explanation:
1. The edge lengths are the differences of coordinates: |5 − 2| = 3, |9 − 4| = 5, |7 − 5| = 2.
2. The diagonal joins the two given points: √(9 + 25 + 4) = √38.

Teacher's Note:
1. The diagonal is the distance between the two given points.

Answer: (c) −8x + 4y − 16z − 16 = k

Explanation:
1. PA² − PB² = [(x − 2)² − (x + 2)²] + [(y − 3)² − (y − 5)²] + [(z − 4)² − (z + 4)²] = −8x + (4y − 16) − 16z.
2. So the locus is −8x + 4y − 16z − 16 = k, a plane.

Teacher's Note:
1. The squared terms cancel, leaving a plane.

Answer: (a) 8

Explanation:
1. The distance of a point from the xy-plane is the absolute value of its z-coordinate, which is 8.

Teacher's Note:
1. The xy-plane is z = 0.

Answer: (d) 2 and 3

Explanation:
1. PQ = (2, 2, −2) and PR = (6, 6, −6) = 3 PQ, so R is beyond Q, with PR : RQ = 3 : 2, externally; statement 1 is false and 2 is true.
2. Q lies between P and R with PQ : QR = 2 : 4 = 1 : 2, so statement 3 is true.

Teacher's Note:
1. Compare PR with PQ as vectors.

Answer: (d) x − 2z = 0

Explanation:
1. (x − 1)² + (z − 3)² = (x − 3)² + (z + 1)² (the y terms cancel).
2. −2x − 6z + 10 = −6x + 2z + 10, so 4x = 8z, i.e. x − 2z = 0.

Teacher's Note:
1. Equate squared distances.

Answer: (b) (6, 7, 0)

Explanation:
1. The foot of the perpendicular on the xy-plane keeps x and y and sets z = 0, giving (6, 7, 0).

Teacher's Note:
1. Only the z-coordinate changes.

Answer: (a) (−1, 2, 8)

Explanation:
1. New coordinates = old − shift = (0 − 1, 4 − 2, 5 + 3) = (−1, 2, 8).

Teacher's Note:
1. Subtract the coordinates of the new origin.

Answer: (c) −2 : 3

Explanation:
1. If the ratio is m : n, x = (3m + 2n)/(m + n) = 0 on the yz-plane, so 3m = −2n.
2. m : n = −2 : 3, i.e. the division is external.

Teacher's Note:
1. The yz-plane is x = 0.

Answer: (a) −2

Explanation:
1. The direction of the first two points is (1, −3, −2).
2. (8, −7, k) − (5, 2, 4) = (3, −9, k − 4) must equal 3(1, −3, −2), so k − 4 = −6 and k = −2.

Teacher's Note:
1. Use proportional direction ratios.

Answer: (b) \( \left(\frac a2,\frac a2,\frac a2\right) \)

Explanation:
1. Equal distance from (0, 0, 0) and (a, 0, 0) gives x = a/2; similarly y = a/2 and z = a/2.
2. The point is (a/2, a/2, a/2).

Teacher's Note:
1. Equate the distance from the origin with each of the other three.

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

Explanation:
1. On the xy-plane z = 0. With ratio λ : 1, (6λ + 4)/(λ + 1) = 0 gives λ = −2/3.
2. A negative ratio of magnitude 2/3 means external division in the ratio 2 : 3.

Teacher's Note:
1. The two z-coordinates 4 and 6 are both positive, so the plane lies outside the segment.

Answer: (c) \( \frac23\sqrt{34} \)

Explanation:
1. AB = √(4 + 16 + 16) = 6 and AC = √(4 + 4 + 1) = 3, so the bisector meets BC at D with BD : DC = 2 : 1.
2. D = (B + 2C)/3 = (2, 13/3, 6).
3. AD = √(4 + 64/9 + 4) = √(136/9) = (2/3)√34.

Teacher's Note:
1. The angle bisector divides the opposite side in the ratio of the adjacent sides.

Answer: (a) \( \left(\frac{19}{8},\frac{57}{16},\frac{17}{16}\right) \)

Explanation:
1. AB = √(4 + 1 + 4) = 3 and AC = √(144 + 16 + 9) = 13, so BD : DC = 3 : 13.
2. D = (13B + 3C)/16 = ((65 − 27)/16, (39 + 18)/16, (26 − 9)/16) = (19/8, 57/16, 17/16).

Teacher's Note:
1. BD : DC = AB : AC.

Answer: (a) (3, 2, 0)

Explanation:
1. Let the point be (x, y, 0). PA² = PC² gives (x − 2)² + y² + 9 = x² + y² + 1, so x = 3.
2. PB² = PC² gives x² + (y − 3)² + 4 = x² + y² + 1, so y = 2.

Teacher's Note:
1. Set z = 0 and equate squared distances.

Answer: (a) parallelogram

Explanation:
1. Taking them in this order, AB = (−2, −4, −4) and DC = (−2, −4, −4), so ABCD is a parallelogram.
2. |AB| = 6 but |AD| = √43, so it is not a rhombus, and AB · AD = 38 ≠ 0, so it is not a rectangle.

Teacher's Note:
1. Test the side vectors for equality, length and perpendicularity.

Answer: (c) 8

Explanation:
1. The medial triangle has sides half of the original, so AB² + BC² + CA² = 4 (DE² + EF² + FD²) where D, E, F are the mid-points.
2. DE² + EF² + FD² = (l² + m²) + (m² + n²) + (n² + l²) = 2(l² + m² + n²), so the ratio is 8.

Teacher's Note:
1. The mid-point triangle has sides parallel and half in length.