JEE Mathematics Introduction to Three Dimensional Geometry MCQs Set 02

Answer: (c) \( z=0 \)

Explanation:
1. Every point of the xy-plane lies at height zero, so its z-coordinate is 0.

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

Answer: (c) \( y=0,z=0,x\ne0 \)

Explanation:
1. On the x-axis both y and z are 0, and the point is not the origin, so \( x\ne0 \).

Teacher's Note:
1. The x-axis has points of the form (x, 0, 0).

Answer: (a) \( \sqrt{b^2+c^2} \)

Explanation:
1. The foot of the perpendicular on the x-axis is \( (a,0,0) \).
2. Distance \( =\sqrt{0+b^2+c^2} \).

Teacher's Note:
1. Drop the perpendicular to the axis and find the distance.

Answer: (b) octant II

Explanation:
1. The signs are \( (-,+,+) \).
2. Octant I is \( (+,+,+) \) and octant II is \( (-,+,+) \).

Teacher's Note:
1. Octants I to IV have z positive, and II has x negative with y positive.

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

Explanation:
1. For vertices A, B, C in order the fourth vertex is \( D=A+C-B \).
2. \( D=(-1+2-3,\ 0+2-1)=(-2,1) \).

Teacher's Note:
1. The diagonals of a parallelogram bisect each other.

Answer: (c) \( \left(\dfrac34,\dfrac54,\dfrac74\right) \)

Explanation:
1. The point equidistant from the origin and \( (3/2,0,0) \) has \( x=3/4 \).
2. Similarly \( y=5/4 \) and \( z=7/4 \), the circumcentre of the box corner.

Teacher's Note:
1. Equidistance from the origin and a point on an axis fixes that coordinate at half.

Answer: (a) \( 5:4 \)

Explanation:
1. Let the ratio be \( m:n \). The z-coordinate is \( \frac{-4m+5n}{m+n}=0 \).
2. So \( 4m=5n \), \( m:n=5:4 \) (internally).

Teacher's Note:
1. On the xy-plane z = 0.

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

Explanation:
1. The perpendicular to the xy-plane changes only the z-coordinate, which becomes 0.

Teacher's Note:
1. Set z = 0.

Answer: (b) \( 5:7 \)

Explanation:
1. Take the point \( (2+t,\ 1+3t,\ 5-2t) \). Then \( x+y-z=(2+t)+(1+3t)-(5-2t)=-2+6t=\frac12 \).
2. So \( t=\frac{5}{12} \) and the ratio is \( t:(1-t)=5:7 \).

Teacher's Note:
1. Parametrise the line and substitute into the plane.

Answer: (d) second, sixth

Explanation:
1. \( (-,+,+) \) is the second octant.
2. \( (-,+,-) \) is the sixth octant.

Teacher's Note:
1. Octants V to VIII are the mirror images of I to IV below the xy-plane, in the same order.

Answer: (a) \( (7,0,0),(0,9,0),(0,0,4) \)

Explanation:
1. The foot on the x-axis keeps only the x-coordinate: \( (7,0,0) \).
2. Similarly M = \( (0,9,0) \) and N = \( (0,0,4) \).

Teacher's Note:
1. Each foot keeps one coordinate.

Answer: (a) 7 units

Explanation:
1. The edges are \( |5-2|=3 \), \( |9-3|=6 \) and \( |7-5|=2 \).
2. Diagonal \( =\sqrt{9+36+4}=7 \).

Teacher's Note:
1. The two points are opposite corners of the box.

Answer: (b) parallelogram

Explanation:
1. \( A+C=(6,-7,11)=B+D \), so the diagonals bisect each other: a parallelogram.
2. \( AB=7 \) but \( BC=\sqrt{35} \), so it is neither a rhombus nor a square.

Teacher's Note:
1. Compare the mid-points of AC and BD, then the adjacent sides.

Answer: (b) seventh octant

Explanation:
1. The signs are \( (-,-,-) \), which is the seventh octant.

Teacher's Note:
1. All negative coordinates give octant VII.

Answer: (d) all of the above

Explanation:
1. \( L=(3,4,0) \) so \( PL=5 \); \( M=(0,4,5) \) so \( PM=3 \); \( N=(3,0,5) \) so \( PN=4 \).
2. All three statements are true.

Teacher's Note:
1. Each perpendicular distance is the absolute value of one coordinate.

Answer: (d) \( \left(\dfrac{49}{4},0,0\right) \)

Explanation:
1. Let \( P=(a,0,0) \). \( PA^2=(a-3)^2+8 \) and \( PB^2=(a-5)^2+41 \).
2. Equating: \( a^2-6a+17=a^2-10a+66 \), so \( 4a=49 \) and \( a=\frac{49}{4} \).

Teacher's Note:
1. A point on the x-axis has the form (a, 0, 0).

Answer: (a) a right-angled isosceles triangle

Explanation:
1. The squared sides are \( AB^2=1+1+16=18 \), \( BC^2=9+9+0=18 \) and \( CA^2=16+4+16=36 \).
2. \( AB=BC \) and \( AB^2+BC^2=CA^2 \), so the triangle is right-angled and isosceles.

Teacher's Note:
1. Compute all three squared sides.

Answer: (b) \( (0,1,3) \)

Explanation:
1. Let \( P=(0,y,z) \). From \( PA=PC \): \( 4+(z-3)^2=(z-1)^2 \), which gives \( z=3 \).
2. From \( PB=PC \): \( (y-3)^2+(z-2)^2=y^2+(z-1)^2 \), which with \( z=3 \) gives \( y=1 \).

Teacher's Note:
1. A point in the YZ-plane has x = 0.

Answer: (d) none of these

Explanation:
1. The foot on the y-axis is \( (0,5,0) \).
2. Distance \( =\sqrt{9+0+36}=\sqrt{45}=3\sqrt5 \), which is not among the options.

Teacher's Note:
1. The distance from the y-axis uses the x and z coordinates.

Answer: (a) \( 2:3 \) externally

Explanation:
1. On the YZ-plane x = 0. With ratio \( m:n \): \( \frac{6m+4n}{m+n}=0 \), so \( m:n=-2:3 \).
2. A negative ratio means external division in the ratio \( 2:3 \).

Teacher's Note:
1. Set x = 0 in the section formula.

Answer: (b) \( 2:3 \) internally

Explanation:
1. \( \frac{3m-2n}{m+n}=0 \) gives \( m:n=2:3 \).
2. The ratio is positive, so the division is internal.

Teacher's Note:
1. A positive ratio means internal division.

Answer: (a) \( a=-2 \)

Explanation:
1. The sums of the coordinates must be zero: \( a-2+4=0 \), \( 1+b+7=0 \), \( 3-5+c=0 \).
2. So \( a=-2 \), \( b=-8 \), \( c=2 \); only (a) is true.

Teacher's Note:
1. The centroid is the average of the vertices.

Answer: (d) II and III

Explanation:
1. \( PQ=(2,2,-2) \) and \( PR=(6,6,-6) \), so \( PR=3\,PQ \) and R lies beyond Q.
2. \( PR:RQ=3:2 \) with R outside the segment PQ, so II is true and I is false. Q lies between P and R with \( PQ:QR=2:4=1:2 \), so III is true.

Teacher's Note:
1. Compare the displacement vectors from P.

Answer: (c) Both are true

Explanation:
1. Two intersecting lines determine a plane, so I is true.
2. Points of the xy-plane have z = 0, so II is true.

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

Answer: (d) All are true

Explanation:
1. On each axis the other two coordinates vanish, so all three statements are true.

Teacher's Note:
1. Each axis fixes two coordinates at zero.