Introduction To Three Dimensional Geometry

(i) (1, – 1, 3) :- 4^th octant,

(ii) (– 1, 2, 4) :- 2^nd octant,

(iii) (– 2, – 4, –7) :- 7^th octant,

(iv) (– 4, 2, – 5) :- 6^th octant.

(i) (1, 2, 3) :- 1^st Octant,

(ii) (4, – 2, 3) :- 4^th Octant,

(iii) (4, –2, –5) :- 8^th Octant,

(iv) (4, 2, –5) :- 5^th Octant,

(v) (– 4, 2, 5) :- 2^nd Octant,

(vi) (–3, –1, 6) :- 3^rd Octant,

(vii) (2, – 4, – 7) :- 8^th Octant,

(viii) (– 4, 2, – 5) :- 6^th Octant.

(i) (3, 4, 2) :- A(3,0,0), B(0,4,0), C(0,0,2)

(ii) (–5, 3, 7) :- A(−5,0,0), B(0,3,0), C(0,0,7)

(iii) (4, – 3, – 5) :- A(4,0,0), B(0,−3,0), C(0,0,−5)

(i) (3, 4, 5) :- A(3,4,0), B(0,4,5), C(3,0,5)

(ii) (–5, 3, 7) :- A(−5,3,0), B(0,3,7), C(−5,0,7)

(iii) (4, – 3, – 5) :- A(4,−3,0), B(0,−3,−5), C(4,0,−5)

The points (2, 0, 0) and (–3, 0, 0) are at a distance of:-

|2 − (−3)| = 5 units.

The distance from the origin to (6, 6, 7) is :-

=11 units.

Given; x² + y² = 1 ⇒ 1 – x²– y² = 0

Distance of the point from origin is :-

=√1

= 1 unit.

When x = 1 the distance of that point from origin will be 1 unit.

Given; The points A (1, – 1, 3), B (2, – 4, 5) and (5, – 13, 11).

=4√14

=AC

∴ Points A, B and C are collinear.

Given; Three consecutive vertices of a parallelogram ABCD are A (6, – 2, 4), B (2, 4, – 8), C (–2, 2, 4).

Let the forth vertex be D(x,y,z).

Midpoint of diagonal AC

Midpoint of diagonal BD

∴ D (2, −4, 16) is the forth vertex.

Given; The vertices A (0, 4, 1), B (2, 3, – 1) and C (4, 5, 0).

⇒ AC² = AB² + BC²

∴ Triangle ABC a right angled.

Given; The centroid is origin and two vertices are (2, 4, 6) and (0, –2, –5).

Let the third vertex be (x,y,z);

For a triangle the coordinates of the centroid is given by the average of the coordinates of its vertices.

∴ the third vertex is (−2, −2, −1).

Given; The mid-point of whose sides are D (1, 2, – 3), E (3, 0, 1) and F (– 1, 1, – 4).

For a triangle the coordinates of the centroid is also given by the average of the coordinates of midpoints of its sides.

∴ (1, 1, −2) is the required centroid.

Given; The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, – 1).

Let he vertices be A(x₁, y₁, z₁), B(x₂, y₂, z₂) and A(x₃, y₃, z₃) respectively.

∴ x₂ = 3, y₂ = 12, z₂ = 17.

∴ x₁ = 10 – x₂ = 7, y₁ = 14 – y₂ = 2, z₁ = 22 – z₂ = 5.

∴ x₃ = – x₂ = −3, y₃ = 16 – y₂ = 4, z₃ = 10 – z₂ = −7.

∴ A (7, 2, 5), B(3, 12, 17), C(−3, 4, −7) are the required vertices.

Given; Three consecutive vertices of a parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2)

Let the forth vertex be D(x,y,z).

Midpoint of diagonal AC

Midpoint of diagonal BD

⇒ x=4

⇒ y=7

⇒ z=6

∴ D(4, 7, 6) is the forth vertex.

Given; The line segment joining the points A (2, 1, – 3) and B (5, – 8, 3).

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be the points which trisects the line segment.

⇒ P divides AB in the ratio 2:1;

⇒ Q divides AP in the ratio 1:1;

∴ (4, −5, 1) and (3, −2, −1) are the coordinate of the points which trisect the line segment joining the points A (2, 1, – 3) and B (5, – 8, 3).

Given; Triangle ABC having vertices A (a, 1, 3), B (– 2, b, – 5) and C (4, 7, c) and origin is the centriod.

For a triangle the coordinates of the centroid is given by the average of the coordinates of its vertices.

Given; A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) are the vertices of a triangle, The internal bisector of the angle A meets BC at the point D.

⇒ ABC is an isosceles triangle and thus the internal bisector of the angle A meets BC at its midpoint.

∴ The coordinates of D is

Given; Three points A (2, 3, 4), B (–1, 2, – 3) and C (– 4, 1, – 10)

=√59

=√59

=√236

=2√59

⇒ AB + BC = AC; Points A, B and C are collinear.

AC: BC = 2√59:√59 = 2:1

∴ from the lengths of AB, BC and AC we can say that C divides AB in the ratio 2:1 externally.

Given; The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4).

Let he vertices be A(x₁, y₁, z₁), B(x₂, y₂, z₂) and A(x₃, y₃, z₃) respectively.

⇒ x₁ =2-x₂, y₁ =10-y₂, z₁ =-2- z₂

⇒ x₃ =-x₂ , y₃ =8-y₂ , z₃ =-4-z₂

∴ x₂ = −1,

y₂ = 6,

z₂ = −7.

∴ x₁ = 2 − x₂ = 3,

y₁ = 10 − y₂ = 4,

z₁ = −4 – z₂ = 5.

∴ x₃ = – x₂ = 1,

y₃ = 8 – y₂ = 2,

z₃ = −4 − z₂ = 3.

∴ A(−1, 6, −7), B(3, 4, 5), C(1, 2, 3) are the required vertices.

Centroid of a triangle is given by the average of the coordinates of its vertices or midpoint of sides.

Centroid is

Given; Three points A (0, – 1, – 7), B (2, 1, – 9) and C (6, 5, – 13)

⇒ AB + BC = AC; ∴ Points A, B and C are collinear.

AB:AC = 2√3:6√3 = 1:3

∴ from the lengths of AB, BC and AC we can say that the first point divides the join of the other two in the ratio 1:3 externally.

Given; Cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin.

The Coordinates of the vertices are;

(0, 0, 0),

(2, 0, 0),

(0, 2, 0),

(0, 0, 2),

(2, 2, 0),

(0, 2, 2),

(2, 0, 2),

(2, 2, 2).

From basic ides of 3-D geometry, we know that-

x-coordinate of a point is its distance from yz plane.

∴ Distance of Point P(3, 4, 5) from yz plane is given by its x coordinate.

∵ x-coordinate of point P = 3

∴ Distance of (3, 4, 5) from yz plane is 3 units

Hence, option(A) is the only correct choice.

As y-axis lies on xy plane and yz.

So, its distance from xy and yz plane is 0.

∴ By basic definition of 3-D coordinate we can say that x-coordinate and z–coordinate are 0.

As, perpendicular is drawn from point P to y-axis, so distance of point of intersection of this line from xz plane remains the same.

∴ y-coordinate of the new point say Q = 4

Or we can say that corresponding point on y-axis is (0, 4, 0)

∴ Length of perpendicular = distance between P and Q

From distance formula-

PQ =

∴ Length of foot of perpendicular drawn from the point P (3, 4, 5) on y-axis is √34 units.

Hence, option(B) is the only correct choice.

Let P be the point whose coordinate is (3, 4, 5) and Q represents the origin.

From distance formula-

PQ =

=√50

∴ Distance of the point (3, 4, 5) from the origin (0, 0, 0) is √50 units.

Hence, option(A) is the only correct choice.

Let P be the point whose coordinate is (a, 0, 1) and Q represents the point (0, 1, 2).

Given, PQ = √27

From distance formula-

PQ =

⇒

Squaring both sides-

a² + 2 = 27

⇒ a² = 25

⇒ a = ±√25 = ±5

∴ a = 5 or a = -5

Option (A) and (B) both have one correct answer but only option (C) contains both the answers

Hence, option(C) is the most suitable choice.

As xy and xz both planes contains x– axis and x-axis is the line of intersection of these planes

No other option contains 2 planes and both of which contains x -axis.

Hence, option(A) is the only correct choice.

As y-axis lies on xy plane and yz.

So its distance from xy and yz plane is 0.

∴ By basic definition of 3-D coordinate we can say that x-coordinate and z–coordinate are 0.

∴ any point on Y-axis has x and z coordinate = 0

∴ x = 0 and z = 0 can be considered as the equation of y-axis.

Hence, option(C) is the only correct choice.

∵ All the coordinates of the given point (-2,-3,-4) are negative.

∴ point belongs to seventh quadrant.

Hence, option(B) is the only correct choice.

From basic definition of 3D geometry we know that x-axis is perpendicular to yz plane.

As plane is parallel to yz plane, so x-axis must be perpendicular to that plane.

Hence, option(A) is the only correct choice.

As x-axis lies on xy plane and xz.

So its distance from xy and xz plane is 0.

∴ By basic definition of 3-D coordinate we can say that y-coordinate and z–coordinate are 0.

∴ any point on X-axis has y and z coordinate = 0

∴ y = 0 and z = 0 can be considered as the equation of x-axis.

Hence, option(A) is the only correct choice.

∵ x coordinate gives the distance of a point from yz plane.

∴ x = 0 implies that the point lies on yz plane.

Thus, locus of a point for which x = 0 is the yz plane

Hence, option(B) is the only correct choice.

According to question, if we draw the parallelepiped in 3D space.

Points P(5, 8,10) and Q(3, 6, 8) appear to be opposite vertices of parallelepiped.

∴ Length of PQ gives the length of diagonal of parallelepiped.

Using distance formula we can find PQ as-

PQ =

∴ Length of diagonal of the parallelepiped is 2√3 units.

Hence, option(A) is the only correct choice.

As L lies on xy plane, so its distance from xy plane is zero or we can say that z-coordinate is 0.

∵ L is perpendicular from P(3, 4, 5) to xy plane so its distance from yz and xz plane remains the same as that of point P

∴ Coordinates of L are (3, 4, 0)

∵ None of the option(A,B or C) matches with our answer.

Hence, Option(D) is the most suitable choice.

As x-axis lies on xy plane and xz.

So its distance from xy and xz plane is 0.

∴ By basic definition of 3-D coordinate we can say that y-coordinate and z–coordinate are 0.

As, perpendicular is drawn from point P(3, 4, 5) to x-axis , so distance of point of intersection of this line from yz plane remains the same.

∴ x-coordinate of the new point(L) = 3

Or we can say that coordinates of L are (3, 0, 0)

Hence, option(A) is the only correct choice.

First Octant

Explanation:

If O represents origin then OX represents the positive X – axis , OY represents positive Y axis and OZ represents positive Z-axis.

∴ any point lying in this region will have all coordinates positive.

∴ It determines 1^st octant

Six

Explanation:

A rectangular parallelepiped has always three pairs of rectangular faces.

It can be verified by counting the number of rectangles in the adjoining figure showing a rectangular parallelepiped.

Given Point.

Explanation: By definition of x,y and z coordinates we can conclude about the above answer.

8

Explanation:

The 3 coordinate planes divide the space into 8 different parts and each part is called an octant.

(0, y, z)

Explanation:

∵ P lies in yz plane and x – coordinate gives the perpendicular distance of a given point from yz plane.

∴ x coordinate of P is 0

But still point P has some perpendicular distances from xy and xz planes. So we define this point by (0, y, z) where y is the perpendicular distance from point P to xz plane and z is the ⊥ distance from point P to xy plane.

x = 0

Explanation:

∵ x coordinate gives the distance of a point from yz plane.

∴ x = 0 implies that the point lies on yz plane.

Thus, locus of a point for which x = 0 is the yz plane or we can say that equation of yz plane is x = 0.

(0,0,z)

Explanation:

∵ P lies on z-axis and this axis is common to xz and yz planes or we can say that it lies on xz and yz planes.

∴ its y and x coordinates are going to be 0 but still it may have any distance form xy axis as it is ⊥ to it.

∴ Co-ordinates of P are of the form (0, 0, z)

x = 0, y = 0.

Explanation:

Let P be an arbitrary point on z-axis. Its locus will give the equation of z - axis

∵ P lies on z-axis and this axis is common to xz and yz planes or we can say that it lies on xz and yz planes.

∴ it’s y and x coordinates are going to be 0 but still it may have any distance form xy axis as it is ⊥ to it.

∴ Co-ordinates of P are of the form (0, 0, z)

Clearly x = 0 and y = 0 together defines any point on z – axis.

∴ locus of P is x = 0 & y = 0 and also the equation of z-axis.

z-coordinate

Explanation:

If a line is parallel to xy plane, perpendicular distance of each and every point on line from xy plane remains.

∵ z-coordinate of a point gives its perpendicular distance from xy plane.

This implies that z-coordinate of every point must remain the same.

y and z coordinates.

Explanation:

If a line is parallel to x-axis then every point on it will be at a constant distance from xy and xz plane.

∵ z-coordinate of a point gives its perpendicular distance from xy plane and y-coordinate of a point gives its perpendicular distance from xz plane

∴ every point will have same y and z coordinates.

yz plane

Explanation:

∵ x = a ⇒ x-coordinate is constant.

∵ x-coordinate of a point gives its perpendicular distance from yz plane.

So, every point on the plane x = a lies at a fixed distance from yz plane.

This means x = a is a plane parallel to yz plane.

x-axis

Explanation:

∵ x-axis is perpendicular to yz plane, so it is going to be perpendicular to any plane parallel to yz.

√333

Explanation:

The longest string that can be stretched straight inside a rectangular room is going to be the diagonal of the room.

Length of diagonal of a rectangular room of length ‘l’, breadth ‘w’ and height ‘h’ is given by –

∴ longest piece of string that can be stretched =

= √(100+169+64)

= √333

a = -3 or a = 5

Explanation:

Let P be the point whose coordinate is (a, 2, 1) and Q represents the point (1, -1, 1).

Given, PQ = 5

From distance formula-

PQ =

⇒

Squaring both sides-

a² -2a + 10 = 25

⇒ a² -2a - 15 = 0

⇒ a² -5a + 3a - 15 = 0

⇒ a(a-5) + 3(a-5) = 0

⇒ (a + 3)(a – 5) = 0

∴ a = -3 or a = 5

(1,1,-2)

Explanation:

∵ Mid-points of sides of ΔABC are D( 1, 2, -3), E(3, 0, 1) and F(-1, 1, -4).

From basic geometry we know that, centroid of Δ ABC and Δ DEF will be the same.

∴ centroid of Δ ABC is given by

Hence, centroid of ΔABC is = (1, 1, -2)

(a) – (iii), (b) – (i), (c) – (ii), (d) – (vi), (e) – (iv),

(f) – (v), (g) – (viii), (h) – (vii), (i) – (x), (j) – (ix)

Explanation:

a) In xy - plane all points lie on it. So distance from xy plane remains 0.

∴ we can say that z-coordinate is 0

b) Point (2,3,4) has all coordinates positive. So it implies that the point lies in the 1^st octant.

c) If x – coordinate of a point is zero. This means that its distance from yz plane is zero or we can say that point lies on yz plane.

∴ Locus of point having x coordinate 0 is yz plane

d) For a line to be parallel to x-axis , it must remain parallel to xz and xy plane. As they remain parallel to these plane, all points on the line has a fixed or equal y and z coordinate.

e) x = 0 and y = 0 is the locus of z-axis.

∴ They represent z – axis.

f) As all points on the plane z = c has fixed z coordinate or we can say that they are at a constant distance from xy plane.

∴ z = c is a plane parallel to xy plane.

g) The plane x = a is parallel to yz-plane.

Plane y = b is parallel to xz-plane.

So, planes x = a and y = b gives the set of coordinates that satisfies both the equation of plane.

It means these points are the point of intersection of these plane and these points constitute a line called line of intersection of planes

Now, line of intersection of yz-plane and xz-plane is z-axis.

∵ x = a and y = b are planes parallel to yz and xz. So their line of intersection will be parallel to z-axis.

h) By basic definition of coordinates we know that coordinates of a point are the distances from the origin to the feet of perpendicular from the point on the respective axis.

(i) A ball is the solid region in the space enclosed by a sphere.

(j) The region in the plane enclosed by a circle is known as a disc.