JEE Mathematics Introduction to Three Dimensional Geometry MCQs Set 04

Answer: (b) \( x^2+a^2=c^2-y^2-z^2 \)

Explanation:
1. Sum of squares: \( (x - a)^2 + y^2 + z^2 + (x + a)^2 + y^2 + z^2 = 2x^2 + 2a^2 + 2y^2 + 2z^2 \).
2. Set this equal to \( 2c^2 \) and divide by 2: \( x^2 + a^2 + y^2 + z^2 = c^2 \).
3. So \( x^2 + a^2 = c^2 - y^2 - z^2 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (b) (5, 4)

Explanation:
1. Direction from (3,2,-4) to (9,8,-10) is (6, 6, -6), proportional to (1, 1, -1).
2. Direction from \( (\lambda, \mu, -6) \) to (3, 2, -4) is \( (3 - \lambda, 2 - \mu, 2) \), which must be a multiple of (1, 1, -1); the z-part 2 means the multiple is \( -2 \).
3. So \( 3 - \lambda = -2 \) gives \( \lambda = 5 \), and \( 2 - \mu = -2 \) gives \( \mu = 4 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

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

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (b) 3 : 2 externally

Explanation:
1. PR = (6, 6, -6) and RQ = (-4, -4, 4), so PR = -(3/2) RQ.
2. The ratio PR : RQ is 3 : 2 with opposite directions, so \( R \) lies outside segment PQ.
3. \( R \) divides PQ externally in the ratio 3 : 2.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (a) collinear

Explanation:
1. Distances: between (-2,4,7) and (3,-6,-8): \( \sqrt{25 + 100 + 225} = 5 sqrt14 \).
2. Between (3,-6,-8) and (1,-2,-2): \( \sqrt{4 + 16 + 36} = 2 sqrt14 \). Between (-2,4,7) and (1,-2,-2): \( \sqrt{9 + 36 + 81} = 3 sqrt14 \).
3. \( 2 sqrt14 + 3 sqrt14 = 5 sqrt14 \), so the points are collinear.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

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

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (c) \( y^2-2y-4x+2z+6=0 \)

Explanation:
1. Distance squared from (2,1,-1): \( (x - 2)^2 + (y - 1)^2 + (z + 1)^2 \).
2. Distance squared from the y-axis: \( x^2 + z^2 \).
3. Equate: \( x^2 - 4x + 4 + y^2 - 2y + 1 + z^2 + 2z + 1 = x^2 + z^2 \), so \( y^2 - 2y - 4x + 2z + 6 = 0 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (b) 3 : 1

Explanation:
1. The XOY plane is \( z = 0 \). \( A \) point \( (m\cdot (-1) + n\cdot 3)/(m + n) = 0 \) gives \( m = 3n \).
2. So the ratio \( m \) : \( n = 3 \) : 1 (internal, since both \( z \) values have opposite signs).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (b) square

Explanation:
1. Take \( A(1,1,1), B(-2,4,1), C(-1,5,5), D(2,2,5) \). AB = (-3,3,0) = DC and BC = (1,1,4) = AD, so it is a parallelogram.
2. AB . AD = \( -3 + 3 + 0 = 0 \), so the angle is a right angle, and |AB| = |AD| = \( sqrt18 \).
3. Equal adjacent sides with a right angle: a square.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (b) -1

Explanation:
1. Points on QR: \( (2 + 3t, 2, 1 - 3t) \).
2. \( x = 4 \) gives \( t = 2/3 \).
3. \( z = 1 - 2 = -1 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (c) \( -8x+4y-16z-16=k \)

Explanation:
1. \( PA^2 = (x - 2)^2 + (y - 3)^2 + (z - 4)^2 \) and \( PB^2 = (x + 2)^2 + (y - 5)^2 + (z + 4)^2 \).
2. Difference: \( -8x + 4y - 16z + (4 + 9 + 16 - 4 - 25 - 16) = -8x + 4y - 16z - 16 \).
3. So the locus is \( -8x + 4y - 16z - 16 = k \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (a) 8

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

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (d) 2 and 3

Explanation:
1. PR = \( (6,6,-6) = 3 \) PQ, so \( R \) is beyond \( Q \) on the line PQ; PR : RQ = 6 : \( 4 = 3 \) : 2 externally, so statement 2 is correct and 1 is false.
2. PQ = (2,2,-2) and QR = (4,4,-4), so \( Q \) lies between \( P \) and \( R \) with PQ : QR = 1 : 2, so statement 3 is correct.
3. Statements 2 and 3 are correct.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. Set \( (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2 \).
2. \( -2x + 1 - 6z + 9 = -6x + 9 + 2z + 1 \), so \( 4x - 8z = 0 \).
3. The locus is \( x - 2z = 0 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

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

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (a) (-1, 2, 8)

Explanation:
1. New coordinates = old coordinates minus the new origin.
2. \( (0 - 1, 4 - 2, 5 - (-3)) = (-1, 2, 8) \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (c) -2 : 3

Explanation:
1. The yz-plane is \( x = 0 \). For ratio \( m \) : \( n, x = (3m + 2n)/(m + n) = 0 \) gives \( 3m = -2n \).
2. \( m \) : \( n = -2 \) : 3, an external division.
3. The ratio is \( -2 \) : 3.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (a) -2

Explanation:
1. Direction from the first to the second point: (1, -3, -2).
2. Direction from the first to the third: \( (3, -9, k - 4) \), which must be 3 times (1, -3, -2).
3. \( k - 4 = -6 \), so \( k = -2 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. Equidistant from (0,0,0) and \( (a,0,0) \): \( x = a/2 \). Similarly \( y = a/2 \) and \( z = a/2 \).
2. The point is \( (a/2, a/2, a/2) \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. The xy-plane is \( z = 0 \). For ratio \( m \) : \( n, z = (6m + 4n)/(m + n) = 0 \) gives \( 6m = -4n \).
2. \( m \) : \( n = -2 \) : 3, so the division is external in the ratio 2 : 3.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. AB = \( \sqrt{4 + 16 + 16} = 6 \) and AC = \( \sqrt{4 + 4 + 1} = 3 \), so the bisector meets BC at \( D \) with BD : DC = 2 : 1.
2. \( D = (2C + B)/3 = (2, 13/3, 6) \).
3. AD = \( \sqrt{4 + (7 - 13/3)^2 + 4} = \sqrt{8 + 64/9} = \sqrt{136/9} = (2/3) sqrt34 \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. The bisector of angle \( A \) meets BC at \( D \) with BD : DC = AB : AC.
2. AB = \( \sqrt{4 + 1 + 4} = 3 \) and AC = \( \sqrt{144 + 16 + 9} = 13 \), so BD : DC = 3 : 13.
3. \( D = (13B + 3C)/16 = ((65 - 27)/16, (39 + 18)/16, (26 - 9)/16) = (19/8, 57/16, 17/16) \).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

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

Explanation:
1. Let the point be \( (x, y, 0) \). Then \( PA^2 = PC^2 \) gives \( (x - 2)^2 + y^2 + 9 = x^2 + y^2 + 1 \), so \( x = 3 \).
2. \( PB^2 = PC^2 \) gives \( x^2 + (y - 3)^2 + 4 = x^2 + y^2 + 1 \), so \( -6y + 13 = 1 \) and \( y = 2 \).
3. The point is (3, 2, 0).

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (a) parallelogram

Explanation:
1. Take \( A(4,7,8), B(2,3,4), C(-1,-2,1), D(1,2,5) \). AB = (-2,-4,-4) and DC = (-2,-4,-4), so AB is parallel and equal to DC.
2. AD = (-3,-5,-3), and |AB| = 6 while |AD| = \( sqrt43 \), so adjacent sides differ.
3. AB . AD = \( 6 + 20 + 12 = 38 \), not zero, so no right angle. It is a parallelogram only.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.

Answer: (c) 8

Explanation:
1. The sides of the triangle are twice the sides of the medial triangle, so \( AB^2 + BC^2 + CA^2 = 4 \) (sum of squares of the sides of the medial triangle).
2. Medial triangle sides: \( \sqrt{l^2 + m^2}, \sqrt{m^2 + n^2}, \sqrt{n^2 + l^2} \), with sum of squares \( 2(l^2 + m^2 + n^2) \).
3. So the total is \( 4 x 2 (l^2 + m^2 + n^2) = 8 (l^2 + m^2 + n^2) \), and the ratio is 8.

Teacher's Note:
1. Use the distance formula, section formula and the idea that a plane like xy has \( z = 0 \) to set up the equation.
2. Cross-check by substituting the final point back into the condition.