3D geometry involves the mathematics of shapes in 3D space and involving 3 coordinates which are x-coordinate, y-coordinate and z-coordinate. In a 3d space, three parameters are required to find the exact location of a point. For JEE, three-dimensional geometry plays a major role as a lot of questions are included in the exam. Here, the basic concepts of geometry involving 3-dimensional coordinates are covered which will help to understand different operations on a point in 3d plane.
Coordinate System in 3D Geometry
In 3 dimensional geometry, a coordinate system refers to the process of identifying the position or location of a point in the coordinate plane. To understand more about coordinate planes and system, refer to the coordinate geometry lesson which covers all the basic concepts, theorems, and formulas related to coordinate or analytic geometry.
Rectangular coordinate system
Three lines perpendicular to each other pass through a common point. That common point is called the origin, the 3 lines the axes. They are x-axis, y-axis, z-axis respectively. O is the observer with respect to his position of any other point is measured. The position or coordinates of any point in 3D space is measured by how much he has moved along x, y and z-axis respectively. So if a point has a position (3, -4, 5) means he has moved 3 unit along positive x-axis, 4 unit along negative y-axis, 5 unit along positive z-axis.
Rectangular coordinate system – 3D Geometry
Distance from the Origin
Distance from the Origin in 3D Space – 3D Geometry
Distance from the origin. By using pythagoras theorem. The distance of P(x,y,z) from origin (0,0,0) is x2+y2+z2
Distance between 2 points
Distance between 2 points P(x1,y1,z1) and Q(x2,y2,z2) is (x2−x1)2+(y2−y1)2+(z2−z1)2
Division of a line joining 2 points
Let P(x1,y1,z1) and Q(x2,y2,z2) be 2 points. R derives the line segment PQ in ratio internally. Then R has coordinate
(m+nmx2+nx1,m+nmy2+ny1m+nmz2+nz1)
Projection in 3D Space
Projection in 3D Space – 3D Geometry
Let be a line segment. It’s projection on a line PQ is (CD is figure) ABcosθ. Where θ is angle between AB and PQ or CD).
Direction cosines and direction Ratios of a Line in Cartesian Plane
Cosines of the angles a line makes with the positive x, y and z axis respectively, are called direction cosines of that line.
A first degree equation in x, y, z represents a plane in 3D
ax+by+cz=0,z2b2+c2=0 represents a plane.
Normal Form of a Plane
Let P be the length of the normal from the origin to the plane and l,m,n be the direction cosines of that normal. Then the equation of the plane is given by lx+my+nz=P.
Intercept form
Let a plane cuts length a,b,c from the coordinate axis.
Then equation of the plane is ax+by+cz=1.
Planes passing through 3 given points
Plane passing through (x1y1z1),(x2y2z2),(x3y3z3) is
2 points A(x1y1z1) and B(x2y2z2) lie on the same side or opposite sides of a plane ax+by+cz+d=0 accordingly as ax1+by1+cz1+d and ax2+by2+cz2+d are of same sign or opposite sign.
Distance from a point to a plane
Distance of a point (x1,y1,z1) from a plane.
Distance of (x1y1z1) from ax+by+cz+d is ∣∣∣∣a2+b2+c2ax1+by1+cz1+d∣∣∣∣
Equation of the planes bisecting the angle between 2 planes
Let a1x+b1y+c1z+d=0 and a2x+b2y+c2z+d=0 be 2 planes. The planes bisecting the angles between them are a12+b12+c12a1x+b1y+c1z+d1=±a22+b22+c22a2x+b2y+c2z+d2
Position of origin
The origin lies in the acute or obtuse angle between a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 according as a1a2+b1b2+c1c2<0or>0 provided d1 and d2 are both positive.
Two Intersecting plane
If U=0 and V=0 be 2 planes then the plane passing through the line of their intersection is U+λV=0λ to be determined from given condition.
Straight lines in 3D
2 intersecting planes together represent a straight line.
Equations of a straight line:
Equation of a straight line in symmetrical form
A straight line passing through (x1, y1, z1) and having direction cosines {l,m,n} is given by
lx−x1=my−y1=nz−z1
Two point form
Equation of a straight line passing through (x1,y1,z1) and (x2y2z2) is x2−x1x−x1=y2−y1y−y1=z2−z1z−z1
Two plane form to symmetrical form
Let 2 planes be a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 eliminate x to get a relation between y and z. Eliminate y to get relation between y and z. Then find in terms of x and find y in terms of z. Then equate them.
Intersection of a straight line and a plane
Let ax+by+cz+d=0 is intersected by lx−x1=my−y1=nz−z1.
To find the intersection point
Let lx−x1=my−y1=nz−z1=t
∴ x=x1+lt,y=y1+mt,z=z1+nt put these in the equation of plane and solve for
Plane through a given straight line
Let the line be lx−x1=my−y1=nz−z1 of the plane through this line be ax+by+cz+d=0 then ax1+by1+cz1+d=0 and al+bm+cn=0 and from other given conditions a,b,c are determined.
Distance of a point from a straight line
Let 2 lines are l1x−x1=m1y−y1=n1z−z1&l2x−x2=m2y−y2=n2z−z2
They are coplanar iff ∣∣∣∣∣∣∣x2−x1l1l2y2−y1m1m2z2−z1n1n2∣∣∣∣∣∣∣=0
Distance of a point from a straight line
Distance of a point from a straight line – 3D Geometry
Let the line be lx−α=my−β=nz−γ
AQ = projection of AP on the straight line =l(x1α)+m(y1−β)+n(z1−γ)
∴ PQ=AP2−AQ2
Shortest distance between two skew lines
Let the 2 skew lines be l1x−x1=m1y−y1=n1z−z1 and l2x−x2=m2y−y2=n2z−z2
Problem 1. If a variable plane forms a tetrahedron of constant volume 64K3 with the coordinate planes then the lows of the centroid of the tetrahedron is xyz=uk3. Find u.
Answer: Let the equation of the plane be ax+by+cz=1
Centroid of the tetrahedron is (4a,4b,4c)
Volume of the tetrahedron =6abc=64K3.
So letting 4a=x,4b=y,4c=z
We have 6abc=643xyz=64K3.
∴ xyz=6K3
∴ K=6.
Problem 2. The ration in which yz plane divides the line joining (2,4,5) and (3,5,7) is
Answer: Let the ratio be λ:1.
x coordinate =0=λ+13λ+2
λ=−32
∴ the ratio is 2:3
Problem 3. A line makes angles α,β,γ,δ with the 4 diagonals of a cube then ∑cos2α=?
Answer:
3D Geometry Problem
Let the direction cosine of that line be (l,m,n).
Direction cosine of 1st diagonal (31,31,31)
Direction cosine of 2nd diagonal (3+1,3−1,31)
Direction cosine of 3rd diagonal (31,3+1,3−1)and
Direction cosine of 4th diagonal (31,3−1,3−1)
∴ cosα=3l+3m+3n
cosβ=3l−m+n,cosγ=3+l+m−n,cosδ=3l−m−n
∴ ∑cos2α=34(l2+m2+n2)=34.
Problem 4. The angle between the lines whose direction cosines are given by 3l – 5m + n = 0 and 30mn – 2ln + 10lm = 0
Answer: 30mn + 2l( 5m – n) = 0
or, 30mn + 2l, 3l = 0
or, 5mn + l2 = 0
also 3l = 5m – n.
∴ 30mn+23(5m−n)2=0
or, 90mn+50m2+2n2−20mn=0
⇒50m2+2n2+70mn=0.
25m2+n2+35mn=0.
Problem 5. The angle between the lines 3x−2=−2y+1=0c−2 and 1a−1=32b+3=2c+5 is
Answer:cosθ=3.1+(−2)23+0
∴ θ=2π
Problem 6. Find the equation of the plane containing the lines 3x+1=2y−2=1z and 3x−3=2y+4=1z−1
Answer: The plane be
ax+by+cz+d=0
∴ 3a + 2b + c = 0
−a+2b+d=0⇒4a−6b+c=0.
3a – 4b + c + d = 0
-26b = c
∴ 8b + 2b + d = 0
∴ d = 6b
∴ plane is ax + by + cz + d = 0
or 8bx + by – 26bz + 6b = 0
or 8x + y – 26z + 6 = 0
Problem 7. If lines 2x−1=x1y−2=x2z−3 and 3x−2=4y−3=5z−4 lies in the same plane then for the equation x1t2+(x2+2)t+a=0 prove sum of roots =−2
Answer: Let the plane be ax+by+cz+d=0
a+2b+3c+d=0→(i)
2a+3b+4c+d=0→(ii)
2a+x1b+x2c=0→(iii)
3a+4b+5c=0→(iv)
Sum of the roots =x1−(x2+2)
from (i) and(ii)a+b+c=0
from (iii) and (iv)a+b+c−d=0
∴ d = 0
∴ a + 2b + 3c = 0
∴ b + 2c = 0
2a + 3b + 4c = 0
∴ b = 2c
3a + 4b + 5c = 0
2a – bc + 4c = 2c
2a = 2c
∴ a = c
equation of the plane is ax + by + cz = 0 or cx + 2cy + cz = 0
or x – 2y = z = 0
∴ x1=3x2=4
sum of roots =x1−(x2+2)
∴ =−36=−2
Problem 8. The line Kx=2y=−127 makes an isosceles triangle with the planes 2x+y+3z−1=0 and x+2y−3z−1=0 then K=?
Answer: Equation of the bisector planes are 142x+y+3z−1=±14x+2y−3z−1
i.e. 2x+y+3z−1=x+2y−3z+1
or x−y+6z=0→(i)
and 2x+y+3z−1=−x−2y+3z+1
i.e. 3x+3y−2=0→(ii)
so the given line must be parallel to (i) or (ii)
∴ K−2−74=0or3K+6=0
∴ K−2(K=76 is discarded
Problem 9. Direction cosine of normal to the plane containing lines x=y=z and x−1=y−1=dz−1 are
Answer:
Problems in 3D Geometry
Let d.cs be l,m,n
∴ l + m + n = 0 and l + m + dn = 0
i.e. n=0
∴ l = -m from l2+m2+n2=1
We have 2l2=1
∴ l=±21
∴ (l,m,n) is (21,−21,0) or (−21,21,0).
Problem 10. If the line x=y=z intersect the line sinAx+sinBy+sinCz=2d2,
sin2Ax+sin2By+sin2Cy=d2 then sin2A.sin2B.sin2c=?(A+B+C=π).