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The correct ANSWER is (d) parallelogram

The best explanation: We know, DISTANCE between two POINTS (x1, y1, z1) and (x2, Y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Distance AB = \(\sqrt{(1-5)^2+(2+2)^2+(-1-1)^2}\) = 6

Distance BC = \(\sqrt{(5-8)^2+(-2+7)^2+(1-4)^2} = \sqrt{43}\)

Distance CD = \(\sqrt{(8-4)^2+(-7+3)^2+(4-2)^2}\) = 6

Distance AD = \(\sqrt{(1-4)^2+(2+3)^2+(-1-2)^2} = \sqrt{43}\)

Since AB=CD and BC=AD i.e. opposite two sides are equal so, it is a parallelogram.

The correct answer is (a) True

For explanation I would say: Yes, we have distance formula in 3-D GEOMETRY also. Distance between two points (x1, Y1, z1) and (x2, Y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Correct OPTION is (d) 7^th

For explanation I WOULD say: Since the point with all negative coordinates i.e. of the form (-, -, -) LIE in 7^th octant.

So, (-1, – 5, -7) LIES in 7^th octant.

Right option is (a) equilateral TRIANGLE

Easy explanation: We KNOW, distance between two points (x1, y1, z1) and (X2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Distance AB = \(\sqrt{(1-3)^2+(2-1)^2+(3-2)^2} = \sqrt{6}\)

Distance BC = \(\sqrt{(3-2)^2+(1-3)^2+(2-1)^2} = \sqrt{6}\)

Distance CA = \(\sqrt{(2-1)^2+(3-2)^2+(1-3)^2} = \sqrt{6}\)

Since AB=BC=CA so, it forms equilateral triangle.

The correct answer is (d) 8

The EXPLANATION is: 3-D geometry can have 3 COORDINATES each with positive or NEGATIVE value.

Total OCTANTS = 2*2*2 = 8.

Correct choice is (b) Y-Z PLANE

To explain I WOULD say: In 3D coordinate system, a point with x-coordinate zero and y & z coordinate having non-zero VALUE MUST lie on Y-Z plane. So, (0, 2, 4) lie on Y-Z plane.

Right option is (b) (7, 15, 4)

To EXPLAIN I would say: POINTS which trisect the line divides it into 2:1 and 1:2.

The coordinates of a point dividing the line segment joining (x1, y1, z1) and (X2, y2, z2) internally in the ratio m : N is \((\FRAC{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})\).

For 1:2, coordinates of point are \((\frac{1*13+2*4}{1+2},\frac{1*27+2*9}{1+2},\frac{-4+2*8}{1+2})\) = (7, 15, 4)

For 2:1, coordinates of point are \((\frac{2*13+1*4}{1+2},\frac{2*27+1*9}{1+2},\frac{-8+1*8}{1+2})\) = (10, 21, 0)

Right CHOICE is (d) -1/2

The explanation: The coordinates of a point dividing the line segment joining (X1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is \((\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})\).

Let ratio be K : 1.

So, z-coordinate of the point will be (k*6+1*3)/(k+1).

We know, for X-Y plane, z coordinate is zero.

(6k+1*3)/(k+1) = 0 => k=-1/2

Right answer is (a) X-Y PLANE

Easiest explanation: In 3d coordinate system, a point with z-coordinate ZERO and x & y coordinate having non-zero value must LIE on X-Y plane. So, (4, 2, 0) lie on X-Y plane.

Correct choice is (d) right angled isosceles TRIANGLE

Explanation: We know, distance between two points (X1, y1, z1) and (x2, Y2, z2) is \(\SQRT{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Distance AB = \(\sqrt{(7-6)^2+(0+1)^2+(10-6)^2} = \sqrt{18}\)

Distance BC = \(\sqrt{(6-9)^2+(-1+4)^2+(6-6)^2} = \sqrt{18}\)

Distance AC = \(\sqrt{(7-9)^2+(0+4)^2+(10-6)^2} = \sqrt{36}\)

Since AB = BC and AB^2+BC^2 = AC^2 so, it forms right angled isosceles triangle.

Right choice is (c) (7, 8, 9)

For explanation: The coordinates of a point DIVIDING the line segment joining (x1, y1, z1) and (x2, y2, z2) externally in the ratio m : n is \((\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n},\frac{mz_2-nz_1}{m-n})\).

So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1 is \((\frac{2*4-1*1}{2-1},\frac{2*5-1*2}{2-1},\frac{2*6-1*3}{2-1})\) = (7, 8, 9).

The CORRECT choice is (a) 2 units

To explain: We know, DISTANCE of a POINT from Y-Z plane is equal to the value of its x-coordinate.

So, distance of point (2, 3, 5) from Y-Z plane is 2 units.

Right answer is (c) 6^th

To elaborate: SINCE the POINT with only y coordinate positive i.e. of the form (-, +, -) lie in 6^th octant.

So, (-1, 5, -7) LIES in 6^th octant.

The CORRECT CHOICE is (c) X-Z plane

For explanation: In 3d coordinate system, a point with y-coordinate zero and x & z coordinate having non-zero value must LIE on X-Z plane. So, (3, 0, 4) lie on X-Z plane.

Correct choice is (c) \(\sqrt{11}\)

To explain I would say: We KNOW, midpoint of (x1, y1, z1) and (x2, y2, z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).

Midpoint of LINE PR is (5, 4, 6).

LENGTH of median through Q is distance between midpoint of PR and Q i.e. \(\sqrt{(5-2)^2+(4-5)^2+(6-5)^2} = \sqrt{(3)^2+(-1)^2+(1)^2} = \sqrt{11}\).

Correct choice is (B) 2^nd

Explanation: Since the point with only x COORDINATE negative i.e. of the form (-, +, +) lie in 2^nd octant.

So, (-1, 5, 7) LIES in 2^nd octant.

The correct option is (a) True

For EXPLANATION I would say: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Distance AB = \(\sqrt{(3-5)^2+(2-3)^2+(1+2)^2} = \sqrt{14}\)

Distance BC = \(\sqrt{(5+1)^2+(3-0)^2+(-2-7)^2} = 3\sqrt{14}\)

Distance AC = \(\sqrt{(3+1)^2+(2-0)^2+(1-7)^2} = 2\sqrt{14}\)

SINCE AB+AC = BC so, the three points are COLLINEAR.

Right OPTION is (a) 1^st

The EXPLANATION: Since the POINT with all POSITIVE COORDINATES i.e. of the form (+, +, +) so lie in 1^st octant.

So, (1, 5, 7) lies in 1^st octant.

Correct choice is (c) 5 units

To elaborate: We KNOW, distance of a point from X-Y plane is equal to the VALUE of its z-coordinate.

So, distance of point (2, 3, 5) from X-Y plane is 5 units.

Right option is (b) (3, 6, 8)

The best explanation: We know, midpoint of (x1, y1, Z1) and (x2, Y2, z2) is (x1+x2) /2, (y1+y2) /2, (z1+z2)/2).

So, midpoint of (1, 4, 6) and (5, 8, 10) is ((1+5)/ 2, (4+8)/ 2, (6+10)/2) is (3, 6, 8).

Right choice is (C) isosceles triangle

Explanation: We know, distance between two points (x1, y1, z1) and (x2, Y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Distance AB = \(\sqrt{(3-5)^2+(0-3)^2+(3-2)^2} = \sqrt{14}\)

Distance BC =\(\sqrt{(5-6)^2+(3-5)^2+(2-5)^2} = \sqrt{14}\)

Distance AC =\(\sqrt{(3-6)^2+(0-5)^2+(3-5)^2} = \sqrt{38}\)

Since AB =BC so, it FORMS isosceles triangle.

Correct option is (B) 3 units

The best I can EXPLAIN: We KNOW, distance of a POINT from X-Z plane is equal to the value of its y-coordinate.

So, distance of point (2, 3, 5) from X-Z plane is 3 units.

The correct ANSWER is (d) 5 UNITS

To EXPLAIN: We know, DISTANCE between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

So, distance between two points (5, 6, 7) and (2, 6, 3) will be \(\sqrt{(5-2)^2+(6-6)^2+(7-3)^2}\) = \(\sqrt{(3)^2+(4)^2}\) = 5 units.

Correct choice is (c) Z-axis

For EXPLANATION I would SAY: In 3d COORDINATE system, a POINT with x and y coordinate zero and z-coordinate having non-zero value must lie on z-axis. So, (0, 0, 3) lie on z-axis.

Right option is (c) 4

The BEST I can explain: 2-D GEOMETRY can have 2 COORDINATES each with positive or negative value.

Total QUADRANTS = 2*2 = 4.

The correct ANSWER is (a) (7, 5, 3)

Explanation: If coordinates of vertices of a triangle are (x1, Y1, z1), (X2, y2, z2), (x3, Y3, Z3) the coordinates of centroid of the triangle are ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)

So, coordinates of centroid of the given triangle are ((7+5+9)/3, (6+4+5)/3, (4+6+8)/3) = (7, 5, 3).

The correct choice is (a) (3, 4, 5)

For explanation I would say: The COORDINATES of a point dividing the line SEGMENT joining (x1, Y1, z1) and (x2, y2, z2) INTERNALLY in the ratio m : n is \((\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})\).

So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio 2:1 is \((\frac{2*4+1*1}{2+1},\frac{2*5+1*2}{2+1},\frac{2*6+1*3}{2+1})\) = (3, 4, 5).

Right choice is (a) X-axis

Easy EXPLANATION: In 3d coordinate system, a POINT with y and z coordinate zero and x-coordinate having non-zero VALUE must lie on x-axis. So, (4, 0, 0) lie on x-axis.