3D Geometry

Introduction to Three Dimensional Geometry

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.

3D Geometry Coordinate System

Rectangular coordinate system – 3D Geometry

Distance from the Origin

Distance from the Origin in 3D Space

Distance from the Origin in 3D Space – 3D Geometry

Distance from the origin. By using pythagoras theorem. The distance of P(x,y,z)P\left( x,y,z \right) from origin (0,0,0)\left( 0,0,0 \right) is x2+y2+z2\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}

Distance between 2 points

Distance between 2 points P(x1,y1,z1)P\left( x_1,y_1,z_1 \right) and Q(x2,y2,z2)Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is (x2x1)2+(y2y1)2+(z2z1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}

Division of a line joining 2 points

Let P(x1,y1,z1)P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and Q(x2,y2,z2)Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) be 2 points. R derives the line segment PQ in ratio internally. Then R has coordinate

(mx2+nx1m+n,my2+ny1m+nmz2+nz1m+n)\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)

Projection in 3D Space

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.AB\,\,\cos \theta . Where θ\theta 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.

Learn More: Direction Cosines & Direction Ratios Of A Line

So if those angles are α,β,γ\alpha ,\beta ,\gamma then cosα,cosβ,cosγ\cos \alpha ,cos\beta ,cos\gamma are the direction cosines of the line. They are denoted by l,m,n.l,m,n.

l2+m2+n2=1.{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1. (Proof will be given)

Any 3 number a,b,ca,b,c which are proportional to direction cosines are called direction ratios.

Hence la=mb=nc=l2+m2+n2a2+b2+c2=1a2b2+c2\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{1}{\sqrt{{{a}^{2}}{{b}^{2}}+{{c}^{2}}}}

l=aa2,m=ba2,n=ca2l=\frac{a}{\sqrt{\sum{{{a}^{2}}}}},m=\frac{b}{\sqrt{\sum{{{a}^{2}}}}},n=\frac{c}{\sqrt{\sum{{{a}^{2}}}}}

Direction cosine of line joining two given points

Let P(x,y,z)P\left( x,y,z \right) and Q(x2,y2,z2)Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) be 2 points. Then direction cosines will be l=x2x1PQ,m=y2y1PQ,n=z2z1PQl=\frac{{{x}_{2}}-{{x}_{1}}}{\left| PQ \right|},m=\frac{{{y}_{2}}-{{y}_{1}}}{\left| PQ \right|},n=\frac{{{z}_{2}}-{{z}_{1}}}{\left| PQ \right|}

Projection of line segment joining 2 points, on another line

P(x1y1z1)Q(x2y2z2)P\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)Q\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)

Projection of PQ on a line whose direction cosines are l,m,nl,m,n is l(x2x1)+m(y2y1)+n(z2z1)l\left( {{x}_{2}}-{{x}_{1}} \right)+m\left( {{y}_{2}}-{{y}_{1}} \right)+n\left( {{z}_{2}}-{{z}_{1}} \right)

Angle between 2 lines in 3 Dimensional Space

2 lines having direction cosines (l1,m1,n1)\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right) and (l2,m2,n2).\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right). Then angle between them is θ=cos1(l1l2+m1m2+n1n2)\theta ={{\cos }^{-1}}\left( {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} \right)

Projection of a plane area on 3 coordinate planes

Let Aˉ\bar{A} be the vector area. If its direction cosines are cosα,cosβ,cosγ.\cos \alpha ,\cos \beta ,\cos \gamma . Then projections are A1=Acosα,A2=Acosβ,A3=Acosγ.{{A}_{1}}=A\cos \alpha ,{{A}_{2}}=A\cos \beta ,{{A}_{3}}=A\cos \gamma .

A2=A12+A22+A32.{{A}^{2}}=A_{1}^{2}+A_{2}^{2}+A_{3}^{2}.

Area of a triangle:

Using the projection formula, area of a triangle =

14x1y11x2y21x3y312+14y1z11y2z21y3z312+14x1z11x2z21x3z312\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}

Check out more details about the area of a triangle in coordinate geometry in this article like its deriavtion, problem solving strategies, etc.

Concept of Plane in 3 Dimensional Geometry

A first degree equation in x, y, z represents a plane in 3D

ax+by+cz=0,z2b2+c20ax+by+cz=0,{{z}^{2}}{{b}^{2}}+{{c}^{2}}\ne 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,nl,m,n be the direction cosines of that normal. Then the equation of the plane is given by lx+my+nz=P.lx+my+nz=P.

Intercept form

Let a plane cuts length a,b,ca,b,c from the coordinate axis.

Then equation of the plane is xa+yb+zc=1.\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.

Planes passing through 3 given points

Plane passing through (x1y1z1),(x2y2z2),(x3y3z3)\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right),\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right),\left( {{x}_{3}}{{y}_{3}}{{z}_{3}} \right) is

xyz1x1y1z11x2y2z21x3y3z31=0\left| \begin{matrix} x & y & z & 1 \\ {{x}_{1}} & {{y}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|=0

Angle between 2 planes

a1x+b1y+c1z+d1=0{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0 and a2x+b2y+c2z+d2=0{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0 is given by

cosθ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

Two sides of a plane

2 points A(x1y1z1)A\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right) and B(x2y2z2)B\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right) lie on the same side or opposite sides of a plane ax+by+cz+d=0ax+by+cz+d=0 accordingly as ax1+by1+cz1+da{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d and ax2+by2+cz2+da{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+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)\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right) from ax+by+cz+dax+by+cz+d is ax1+by1+cz1+da2+b2+c2\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|

Equation of the planes bisecting the angle between 2 planes

Let a1x+b1y+c1z+d=0{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+d=0 and a2x+b2y+c2z+d=0{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+d=0 be 2 planes. The planes bisecting the angles between them are a1x+b1y+c1z+d1a12+b12+c12=±a2x+b2y+c2z+d2a22+b22+c22\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

Position of origin

The origin lies in the acute or obtuse angle between a1x+b1y+c1z+d1=0{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0 and a2x+b2y+c2z+d2=0{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0 according as a1a2+b1b2+c1c2<0 or >0{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}<0\ or\ >0 provided d1{{d}_{1}} and d2{{d}_{2}} are both positive.

Two Intersecting plane

If U=0U=0 and V=0V=0 be 2 planes then the plane passing through the line of their intersection is U+λV=0λU+\lambda V=0\lambda 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}\left\{ l,m,n \right\} is given by

xx1l=yy1m=zz1n\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}

  • Two point form

Equation of a straight line passing through (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and (x2y2z2)\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right) is xx1x2x1=yy1y2y1=zz1z2z1\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}

  • Two plane form to symmetrical form

Let 2 planes be a1x+b1y+c1z+d1=0{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0 and a2x+b2y+c2z+d2=0{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=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=0ax+by+cz+d=0 is intersected by xx1l=yy1m=zz1n.\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}.

To find the intersection point

Let xx1l=yy1m=zz1n=t\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}=t

x=x1+lt,y=y1+mt,z=z1+ntx={{x}_{1}}+lt,y={{y}_{1}}+mt,z={{z}_{1}}+nt put these in the equation of plane and solve for

Plane through a given straight line

Let the line be xx1l=yy1m=zz1n\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n} of the plane through this line be ax+by+cz+d=0ax+by+cz+d=0 then ax1+by1+cz1+d=0a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d=0 and al+bm+cn=0al+bm+cn=0 and from other given conditions a,b,ca,b,c are determined.

Distance of a point from a straight line

Let 2 lines are xx1l1=yy1m1=zz1n1&xx2l2=yy2m2=zz2n2\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\,\,\And \,\,\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}

They are coplanar iff x2x1y2y1z2z1l1m1n1l2m2n2=0\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|=0

Distance of a point from a straight line

Distance of a point from a straight line

Distance of a point from a straight line – 3D Geometry

Let the line be xαl=yβm=zγn\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}

AQ = projection of AP on the straight line =l(x1α)+m(y1β)+n(z1γ)=l\left( {{x}_{1}}\alpha \right)+m\left( {{y}_{1}}-\beta \right)+n\left( {{z}_{1}}-\gamma \right)

PQ=AP2AQ2\,\,PQ=\sqrt{A{{P}^{2}}-A{{Q}^{2}}}

Shortest distance between two skew lines

Let the 2 skew lines be xx1l1=yy1m1=zz1n1\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}} and xx2l2=yy2m2=zz2n2\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}

The shortest distance is

x2x1y2y1z2z1l1m1n1l2m2n2(m1n2m2n1)2\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|}{\sqrt{\sum{{{\left( {{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}} \right)}^{2}}}}}

The equation of the shortest distance is

xx1yy1zz1l1m1n1lmn=0\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ l & m & n \\ \end{matrix} \right|=0

and

xx2yy2zz2l2m2n2lmn=0\left| \begin{matrix} x-{{x}_{2}} & y-{{y}_{2}} & z-{{z}_{2}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ l & m & n \\ \end{matrix} \right|=0

Problems on 3D Geometry

Problem 1. If a variable plane forms a tetrahedron of constant volume 64K364{{K}^{3}} with the coordinate planes then the lows of the centroid of the tetrahedron is xyz=uk3.xyz=u{{k}^{3}}. Find u.u.

Answer: Let the equation of the plane be xa+yb+zc=1\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

Centroid of the tetrahedron is (a4,b4,c4)\left( \frac{a}{4},\frac{b}{4},\frac{c}{4} \right)

Volume of the tetrahedron =abc6=64K3.=\frac{abc}{6}=64{{K}^{3}}.

So letting a4=x,b4=y,c4=z\frac{a}{4}=x,\frac{b}{4}=y,\frac{c}{4}=z

We have abc6=43xyz6=64K3.\frac{abc}{6}=\frac{{{4}^{3}}xyz}{6}=64{{K}^{3}}.

xyz=6K3xyz=6{{K}^{3}}

∴ K=6.

Problem 2. The ration in which yzyz plane divides the line joining (2,4,5)\left( 2,4,5 \right) and (3,5,7)\left( 3,5,7 \right) is

Answer: Let the ratio be λ:1.\lambda :1.

xx coordinate =0=3λ+2λ+1=0=\frac{3\lambda +2}{\lambda +1}

λ=23\lambda =-\frac{2}{3}

∴ the ratio is 2:3

Problem 3. A line makes angles α,β,γ,δ\alpha ,\beta ,\gamma ,\delta with the 4 diagonals of a cube then cos2α=?\sum{{{\cos }^{2}}\alpha =?}

Answer:

3D Geometry Problem

3D Geometry Problem

Let the direction cosine of that line be (l,m,n).\left( l,m,n \right).

Direction cosine of 1st diagonal (13,13,13)\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)

Direction cosine of 2nd diagonal (+13,13,13)\left( \frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)

Direction cosine of 3rd diagonal (13,+13,13)\left( \frac{1}{\sqrt{3}},\frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)and

Direction cosine of 4th diagonal (13,13,13)\left( \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)

cosα=l3+m3+n3\cos \alpha =\frac{l}{\sqrt{3}}+\frac{m}{\sqrt{3}}+\frac{n}{\sqrt{3}}

cosβ=lm+n3,cosγ=+l+mn3,cosδ=lmn3\cos \beta =\frac{l-m+n}{\sqrt{3}},\cos \gamma =\frac{+l+m-n}{\sqrt{3}},\cos \delta =\frac{l-m-n}{\sqrt{3}}

cos2α=4(l2+m2+n2)3=43.\sum{{{\cos }^{2}}\alpha =\frac{4\left( {{l}^{2}}+{{m}^{2}}+{{n}^{2}} \right)}{3}}=\frac{4}{3}.

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+2(5mn)23=030mn+2\frac{{{\left( 5m-n \right)}^{2}}}{3}=0

or, 90mn+50m2+2n220mn=090mn+50{{m}^{2}}+2{{n}^{2}}-20mn=0

50m2+2n2+70mn=0.\Rightarrow \,\,\,\,\,\,\,\,\,\,50{{m}^{2}}+2{{n}^{2}}+70mn=0.

25m2+n2+35mn=0.25{{m}^{2}}+{{n}^{2}}+35mn=0.

Problem 5. The angle between the lines x23=y+12=c20\frac{x-2}{3}=\frac{y+1}{-2}=\frac{c-2}{0} and a11=2b+33=c+52\frac{a-1}{1}=\frac{2b+3}{3}=\frac{c+5}{2} is

Answer: cosθ=3.1+(2)32+0\cos \theta =3.1+\left( -2 \right)\frac{3}{2}+0

θ=π2\theta =\frac{\pi }{2}

Problem 6. Find the equation of the plane containing the lines x+13=y22=z1\frac{x+1}{3}=\frac{y-2}{2}=\frac{z}{1} and x33=y+42=z11\frac{x-3}{3}=\frac{y+4}{2}=\frac{z-1}{1}

Answer: The plane be

ax+by+cz+d=0ax+by+cz+d=0

∴ 3a + 2b + c = 0

a+2b+d=04a6b+c=0.-a+2b+d=0\,\,\,\,\,\Rightarrow 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 x12=y2x1=z3x2\frac{x-1}{2}=\frac{y-2}{{{x}_{1}}}=\frac{z-3}{{{x}_{2}}} and x23=y34=z45\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} lies in the same plane then for the equation x1t2+(x2+2)t+a=0{{x}_{1}}{{t}^{2}}+\left( {{x}_{2}}+2 \right)t+a=0 prove sum of roots =2=-2

Answer: Let the plane be ax+by+cz+d=0ax+by+cz+d=0

a+2b+3c+d=0(i)a+2b+3c+d=0\rightarrow{{}}\left( i \right)

2a+3b+4c+d=0(ii)2a+3b+4c+d=0\rightarrow{{}}\left( ii \right)

2a+x1b+x2c=0(iii)2a+{{x}_{1}}b+{{x}_{2}}c=0\rightarrow{{}}\left( iii \right)

3a+4b+5c=0(iv)3a+4b+5c=0\rightarrow{{}}\left( iv \right)

Sum of the roots =(x2+2)x1=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}

from (i)\left( i \right) and(ii)\left( ii \right) a+b+c=0a+b+c=0

from (iii)\left( iii \right) and (iv)\left( iv \right) a+b+cd=0a+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\,\,{{x}_{1}}=3{{x}_{2}}=4

sum of roots =(x2+2)x1=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}

=63=2=-\frac{6}{3}=-2

Problem 8. The line xK=y2=712\frac{x}{K}=\frac{y}{2}=\frac{7}{-12} makes an isosceles triangle with the planes 2x+y+3z1=02x+y+3z-1=0 and x+2y3z1=0x+2y-3z-1=0 then K=?K=?

Answer: Equation of the bisector planes are 2x+y+3z114=±x+2y3z114\frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}

i.e. 2x+y+3z1=x+2y3z+12x+y+3z-1=x+2y-3z+1

or xy+6z=0(i)x-y+6z=0\rightarrow{{}}\left( i \right)

and 2x+y+3z1=x2y+3z+12x+y+3z-1=-x-2y+3z+1

i.e. 3x+3y2=0(ii)3x+3y-2=0\rightarrow{{}}\left( ii \right)

so the given line must be parallel to (i)\left( i \right) or (ii)\left( ii \right)

K274=0or3K+6=0K-2-74=0or3K+6=0

K2(K=76\,\,K-2(K=76 is discarded

Problem 9. Direction cosine of normal to the plane containing lines x=y=zx=y=z and x1=y1=z1dx-1=y-1=\frac{z-1}{d} are

Answer:

Problems in 3D Geometry

Problems in 3D Geometry

Let d.csd.{{c}^{s}} be l,m,nl,m,n

∴ l + m + n = 0 and l + m + dn = 0

i.e. n=0

∴ l = -m from l2+m2+n2=1{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1

We have 2l2=12{{l}^{2}}=1

l=±12l=\pm \frac{1}{\sqrt{2}}

(l,m,n)\left( l,m,n \right) is (12,12,0)\left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0 \right) or (12,12,0).\left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \right).

Problem 10. If the line x=y=zx=y=z intersect the line sinAx+sinBy+sinCz=2d2,\sin Ax+\sin By+\sin Cz=2{{d}^{2}},

sin2Ax+sin2By+sin2Cy=d2\sin 2A\,\,x+\sin 2B\,\,y+\sin 2C\,\,y={{d}^{2}} then sinA2.sinB2.sinc2=?\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{c}{2}=? (A+B+C=π).\left( A+B+C=\pi \right).

Answer: Let the point of intersection be (t,t,t)\left( t,t,t \right)

(sinA+sinB+sinC)t=2d2\left( \sin A+\sin B+\sin C \right)t=2{{d}^{2}}

(sin2A+sin2B+sin2C)t=d2\left( \sin 2A+\sin 2B+\sin 2C \right)t={{d}^{2}}

sin2A+sin2Bsin(2A+2B)=d2t.\sin 2A+\sin 2B-\sin \left( 2A+2B \right)=\frac{{{d}^{2}}}{t}.

2.sin(A+B).cos(AB)2sin(A+B).cos(A+B)\Rightarrow 2.\sin \left( A+B \right).\cos \left( A-B \right)-2\sin \left( A+B \right).\cos \left( A+B \right)

4sinCsinA.sinB=d2t.(i)\Rightarrow 4\sin C\,\,\sin A.\,\,\sin B=\frac{{{d}^{2}}}{t.}\rightarrow{{}}\left( i \right)

sinA+sinB+sinC\sin A+\sin B+\sin C

=2.sinA+B2.cosAB2+2sinC2.cosC2=2.\sin \frac{A+B}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}

=2cosC2.cosAB2+2sinC2.cosC2=2\cos \frac{C}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}

=2cosC2(cosAB2+cosA+B2)=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2}+\cos \frac{A+B}{2} \right)

=2cosC2.2cosA2.cosB2=2d2t(ii)=2\cos \frac{C}{2}.2cos\frac{A}{2}.\cos \frac{B}{2}=\frac{2{{d}^{2}}}{t}\rightarrow{{}}\left( ii \right)

dividing (i)÷(ii)\left( i \right)\div \left( ii \right)

sinA2.sinB2.sinC2=116.\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}=\frac{1}{16}.

Problem 11. Equation of the sphere having center at (3,6,4)\left( 3,6,-4 \right) and touching the plane rˉ.(2i^2j^k^)=10\bar{r}.\left( 2\hat{i}-2\hat{j}-\hat{k} \right)=10 is (x3)2+(y6)2+(z+4)2=K2,{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}+{{\left( z+4 \right)}^{2}}={{K}^{2}}, then

Answer: Equation of the plane is 2x2yz10=02x-2y-z-10=0