Important Three Dimensional Geometry Formulas for JEE Maths
Three-dimensional geometry plays a major role as a lot of questions are included in the JEE exam. Here is a list of all the three-dimensional geometry formulas which will help students to go through and revise them quickly before the exam.
Three-dimensional Geometry Formulas
1. Vector representation of a point: Position vector of a point P(x, y, z) is xi^+yj^+zk^
Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) is PQ =
AB = ∣∣∣∣OB−OA∣∣∣∣
3. Distance of P from coordinate axes: PA = (y2+z2)
PB = (z2+x2)
PC = (x2+y2)
x = m+nmx2+nx1
y = m+nmy2+ny1
z = m+nmz2+nz1
x = 2x1+x2
y = 2y1+y2
z = 2z1+z2
5. Centroid of a triangle:
G = (3x1+x2+x3,3y1+y2+y3,3z1+z2+z3)
6. Incentre of triangle ABC: (a+b+cax1+bx2+cx3,a+b+cay1+by2+cy3,a+b+caz1+bz2+cz3)
7. Centroid of a tetrahedron:(4x1+x2+x3+x4,4y1+y2+y3+y4,4z1+z2+z3+z4)
8. Direction cosines and direction ratios:
(i) Direction cosines: let α, β, γ be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos α, cos β and cos γ are called the direction cosines of the line. The direction cosines are usually denoted by (l, m, n).
Therefore l = cos α, m = cos β, n = cos γ.
(ii) l2+m2+n2 = 1
(iii) If a, b, c are the direction ratios of any line L then ai^+bj^+ck^ will be a vector parallel to the line L.
(iv) If l, m, n are the direction cosines of any line L then li^+mj^+nk^ is a unit vector parallel to the line L.
(v) If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then
l = a2+b2+c2a
m = a2+b2+c2b
n = a2+b2+c2c
or l =a2+b2+c2−a,
m = a2+b2+c2−b
n = a2+b2+c2−c
(vi) If OP = r, the direction cosines of OP are l, m, n then the coordinates of P are ( lr, mr, nr).
If the direction cosines of the line AB are l, m, n, |AB| = r and the coordinates of A is (x1, y1, z1) then the coordinates of B are given as (x1+rl, y1+rm, z1+rn).
(vii) If the coordinates P and Q are (x1, y1, z1) and (x2, y2, z2) then the direction ratios of line PQ are a = x2-x1 , b = y2-y1 and c = z2-z1 and the direction cosines of line PQ are l = ∣PQ∣x2−x1, m = ∣PQ∣y2−y1 and n = ∣PQ∣z2−z1
(vii) Direction cosines of the x-axis is (1, 0, 0).
Direction cosines of the y-axis is (0, 1, 0).
Direction cosines of the z-axis is (0, 0, 1).
9. Angle between two line segments:
If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and θ is the acute angle between them, then
cos θ = ∣∣∣∣∣a12+b12+c12a22+b22+c22a1a2+b1b2+c1c2∣∣∣∣∣
The line will be perpendicular if a1a2+b1b2+c1c2 = 0, and parallel if a1/a2 = b1/b2 = c1/c2.
10. Projection of a line segment on a line:
If P(x1, y1, z1) and Q(x2, y2, z2) then the projection of PQ on a line having direction cosines l, m, n is |l(x2-x1)+m(y2-y1)+n(z2-z1)|
11. Equation of a plane: General form: ax+by+cz+d = 0, where a, b, c are not all zero, a, b, c, d ∈ R.
(i) Normal form: lx+my+nz = p
(ii) Plane through the point (x1, y1, z1): a(x-x1)+b(y-y1)+c(z-z1) = 0
(iii) Intercept form: ax+by+cz=1
(iv) vector form: (r−a).n=0 or r.n=a.n
(v) Planes parallel to the axes :
(a) plane parallel to X-axis is by+cz+d = 0
(b) plane parallel to Y-axis is ax+cz+d = 0
(c) plane parallel to Z-axis is ax+by+d = 0
(vi) Plane through origin: Equation of the plane passing through the origin is ax+by+cz = 0.
(vii) Transformation of the equation of a plane to the normal form: ax+by+cz-d = 0 in normal form is ±a2+b2+c2ax+±a2+b2+c2by+±a2+b2+c2cz=±a2+b2+c2d
(viii) Any plane parallel to the given plane ax+by+cz+d = 0 is ax+by+cz+λ = 0.
Distance between ax+by+cz+d1 = 0 and ax+by+cz+d2 = 0 is a2+b2+c2∣d1−d2∣
(ix) A plane ax+by+cz+d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio (−ax2+by2+cz2+dax1+by1+cz1+d)
(x) Coplanarity of four points: The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are coplanar if ∣∣∣∣∣∣∣x2−x1x3−x1x4−x1y2−y1y3−y1y4−y1z2−z1z3−z1z4−z1∣∣∣∣∣∣∣=0
12. A point and a plane:
(i) distance of the point (x’, y’, z’) from the plane ax+by+cz+d = 0 is given by a2+b2+c2ax‘+by‘+cz‘+d
(ii) Length of the perpendicular from a point a to the plane r.n=d is given by p = ∣n∣∣a.n−d∣.
(iii) Foot (x’, y’, z’) of perpendicular drawn from the point (x1, y1, z1) to the plane ax+by+cz+d = 0 is given by ax‘−x1=by‘−y1=cz‘−z1=−a2+b2+c2ax1+by1+cz1+d
(iv) To find image of a point with respect to a plane:
Let P (x1, y1, z1) be a given point and ax+by+cz+d = 0 is given plane. Let (x’, y’, z’) is the image point. Then ax‘−x1=by‘−y1=cz‘−z1=−2a2+b2+c2(ax1+by1+cz1+d).
(v) The distance between two parallel planes ax+by+cz+d = 0 and ax+by+cz+d’ = 0 is given by a2+b2+c2∣d−d’∣
(ii) Planes are perpendicular if aa’+bb’+cc’ = 0 and parallel if a/a’ = b/b’ = c/c’.
(iii) The angle θ between the planes r.n1=d1 and r.n2=d2 is given by cos θ = ∣n1∣.∣n2∣n1.n2
(iv) Planes are perpendicular if n1.n2 = 0 and planes are parallel if n1=λn2, λ is a scalar.
14. Angle bisectors:
(i) The equations of a planes bisecting the angle between two given planes a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 are a12+b12+c12a1x+b1y+c1z+d1 = ±a22+b22+c22a2x+b2y+c2z+d2
(ii) Bisector of acute or obtuse angle: First make both the constant terms positive. Then a1a2+b1b2+c1c2 > 0 ⇒ origin lies on obtuse angle.
a1a2+b1b2+c1c2 < 0 ⇒ origin lies on acute angle.
15. Area of a triangle:
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle, then Δ = Δx2+Δy2+Δz2 where Δx=21∣∣∣∣∣∣∣y1y2y3z1z2z3111∣∣∣∣∣∣∣, Δy=21∣∣∣∣∣∣∣z1z2z3x1x2x3111∣∣∣∣∣∣∣ and Δz=21∣∣∣∣∣∣∣x1x2x3y1y2y3111∣∣∣∣∣∣∣
Vector method: from two vector AB and AC the area is given by 21∣∣∣∣AB×AC∣∣∣∣
16. Volume of a tetrahedron:
Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) is given by V = 61∣∣∣∣∣∣∣∣∣x1x2x3x4y1y2y3y4z1z2z3z41111∣∣∣∣∣∣∣∣∣
17. Equation of a line:
(i) A straight line is the intersection of two planes. It is represented by two planes a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 .
(ii) Symmetric form: a(x−x1)=b(y−y1)=c(z−z1)=r
(iii) vector equation:r=a+λb
(iv) Reduction of cartesian form of equation of a line to vector form and vice versa ax−x1=by−y1=cz−z1 ⇔r=(x1i^+y1j^+z1k^)+λ(ai^+bj^+ck^)
18. To find image of a point with respect to a line:
Let L = ax−x2=by−y2=cz−z2 is a given line.
Let (x’, y’, z’) is the image of the point (x1, y1, z1) with respect to the line L. Then
(i) a(x1-x’)+b(y1-y’)+C(z1-z’) = 0
From (ii) get the value of x’, y’, z’ in terms of λ as x’= 2aλ+2x2-x1, y’= 2bλ+2y2-y1,
z’= 2cλ+2z2-z1 . Then put the values of x’, y’, z’ in (i) get λ and substitute the value of λ to get (x’, y’,z’).
19. Angle between a line and a plane:
(i) If θ is the angle between a line lx−x1=my−y1=nz−z1 and the plane ax+by+cz+d = 0, then sinθ=∣∣∣∣a2+b2+c2l2+m2+n2al+bm+cn∣∣∣∣
(ii) Vector form : If θ is the angle between a line r=a+λb and r.n=d then sin θ = ∣b∣∣n∣b.n