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.
Distance from the Origin
Distance from the origin. By using pythagoras theorem. The distance of \(P\left( x,y,z \right)\) from origin \(\left( 0,0,0 \right)\) is \(\sqrt{{{x}^{2}}{{y}^{2}}{{z}^{2}}}\)
Distance between 2 points
Distance between 2 points \(P\left( x,y,z \right)\) and \(Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\) is \(\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\)
Division of a line joining 2 points
Let \(P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\) and \(Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\) be 2 points. R derives the line segment PQ in ratio internally. Then R has coordinate
\(\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\)
Projection in 3D Space
Let be a line segment. It’s projection on a line PQ is (CD is figure) \(AB\,\,\cos \theta .\) Where \(\theta\) 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 \(\alpha ,\beta ,\gamma\) then \(\cos \alpha ,cos\beta ,cos\gamma\) are the direction cosines of the line. They are denoted by \(l,m,n.\)
\({{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1.\) (Proof will be given)
Any 3 number \(a,b,c\) which are proportional to direction cosines are called direction ratios.
Hence \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{1}{\sqrt{{{a}^{2}}{{b}^{2}}+{{c}^{2}}}}\)
∴ \(l=\frac{a}{\sqrt{\sum{{{a}^{2}}}}},m=\frac{b}{\sqrt{\sum{{{a}^{2}}}}},n=\frac{c}{\sqrt{\sum{{{a}^{2}}}}}\)
Direction cosine of line joining two given points
Let \(P\left( x,y,z \right)\) and \(Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\) be 2 points. Then direction cosines will be \(l=\frac{{{x}_{2}}-{{x}_{1}}}{\left| PQ \right|},m=\frac{{{y}_{2}}-{{y}_{1}}}{\left| PQ \right|},n=\frac{{{z}_{2}}-{{z}_{1}}}{\left| PQ \right|}\)
Projection of line segment joining 2 points, on another line
\(P\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)Q\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)\)
Projection of PQ on a line whose direction cosines are \(l,m,n\) is \(l\left( {{x}_{2}}-{{x}_{1}} \right)+m\left( {{y}_{2}}-{{y}_{1}} \right)+n\left( {{z}_{2}}-{{z}_{1}} \right)\)
Angle between 2 lines in 3 Dimensional Space
2 lines having direction cosines \(\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)\) and \(\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right).\) Then angle between them is \(\theta ={{\cos }^{-1}}\left( {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} \right)\)
Projection of a plane area on 3 coordinate planes
Let \(\bar{A}\) be the vector area. If its direction cosines are \(\cos \alpha ,\cos \beta ,\cos \gamma .\) Then projections are \({{A}_{1}}=A\cos \alpha ,{{A}_{2}}=A\cos \beta ,{{A}_{3}}=A\cos \gamma .\)
∴ \({{A}^{2}}=A_{1}^{2}+A_{2}^{2}+A_{3}^{2}.\)
Area of a triangle:
Using the projection formula, area of a triangle =
\(\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}\)
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
\(ax+by+cz=0,{{z}^{2}}{{b}^{2}}+{{c}^{2}}\ne 0\) represents a plane.
Normal Form of a Plane
Let P be the length of the normal from the origin to the plane and \(l,m,n\) 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 \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)
Planes passing through 3 given points
Plane passing through \(\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right),\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right),\left( {{x}_{3}}{{y}_{3}}{{z}_{3}} \right)\) is
\(\left| \begin{matrix} x & y & z & 1 \\ {{x}_{1}} & {{y}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|=0\)
Angle between 2 planes
\({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\) is given by
\(\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)
Two sides of a plane
2 points \(A\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)\) and \(B\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)\) lie on the same side or opposite sides of a plane \(ax+by+cz+d=0\) accordingly as \(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d\) and \(a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d\) 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 \(\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)\) from \(ax+by+cz+d\) is \(\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\)
Equation of the planes bisecting the angle between 2 planes
Let \({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+d=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+d=0\) be 2 planes. The planes bisecting the angles between them are \(\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)
Position of origin
The origin lies in the acute or obtuse angle between \({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\) according as \({{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}<or>0\) provided \({{d}_{1}}\) and \({{d}_{2}}\) 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 \(U+\lambda V=0\lambda\) 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 \(\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)\) and having direction cosines \(\left\{ l,m,n \right\}\) is given by
\(\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\)
- Two point form
Equation of a straight line passing through \(\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right)\) and \(\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)\) is \(\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\)
- Two plane form to symmetrical form
Let 2 planes be \({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\) 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 \(ax+by+cz+d=0\) is intersected by \(\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}.\)
To find the intersection point
Let \(\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}=t\)
∴ \(x={{x}_{1}}+lt,y={{y}_{1}}+mt,z={{z}_{1}}+nt\) put these in the equation of plane and solve for
Plane through a given straight line
Let the line be \(\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\) of the plane through this line be \(ax+by+cz+d=0\) then \(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d=0\) 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 \(\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\,\,\And \,\,\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\)
They are coplanar iff \(\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|=0\)
Distance of a point from a straight line
Let the line be \(\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}\)
AQ = projection of AP on the straight line \(=l\left( {{x}_{1}}\alpha \right)+m\left( {{y}_{1}}-\beta \right)+n\left( {{z}_{1}}-\gamma \right)\)
∴ \(\,\,PQ=\sqrt{A{{P}^{2}}-A{{Q}^{2}}}\)
Shortest distance between two skew lines
Let the 2 skew lines be \(\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\) and \(\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\)
The shortest distance is
\(\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|}{\sqrt{\sum{{{\left( {{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}} \right)}^{2}}}}}\)
The equation of the shortest distance is
\(\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ l & m & n \\ \end{matrix} \right|=0\)
and
\(\left| \begin{matrix} x-{{x}_{2}} & y-{{y}_{2}} & z-{{z}_{2}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ l & m & n \\ \end{matrix} \right|=0\)
Problems on 3D Geometery
Problem 1. If a variable plane forms a tetrahedron of constant volume \(64{{K}^{3}}\) with the coordinate planes then the lows of the centroid of the tetrahedron is \(xyz=u{{k}^{3}}.\) Find \(u.\)
Answer: Let the equation of the plane be \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
Centroid of the tetrahedron is \(\left( \frac{a}{4},\frac{b}{4},\frac{c}{4} \right)\)
Volume of the tetrahedron \(=\frac{abc}{6}=64{{K}^{3}}.\)
So letting \(\frac{a}{4}=x,\frac{b}{4}=y,\frac{c}{4}=z\)
We have \(\frac{abc}{6}=\frac{{{4}^{3}}xyz}{6}=64{{K}^{3}}.\)
∴ \(xyz=6{{K}^{3}}\)
∴ K=6.
Problem 2. The ration in which \(yz\) plane divides the line joining \(\left( 2,4,5 \right)\) and \(\left( 3,5,7 \right)\) is
Answer: Let the ratio be \(\lambda :1.\)
\(x\) coordinate \(=0=\frac{3\lambda +2}{\lambda +1}\)
\(\lambda =-\frac{2}{3}\)
∴ the ratio is 2:3
Problem 3. A line makes angles \(\alpha ,\beta ,\gamma ,\delta\) with the 4 diagonals of a cube then \(\sum{{{\cos }^{2}}\alpha =?}\)
Answer:
Let the direction cosine of that line be \(\left( l,m,n \right).\)
Direction cosine of 1st diagonal \(\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)\)
Direction cosine of 2nd diagonal \(\left( \frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)\)
Direction cosine of 3rd diagonal \(\left( \frac{1}{\sqrt{3}},\frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)\)and
Direction cosine of 4th diagonal \(\left( \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)\)
∴ \(\cos \alpha =\frac{l}{\sqrt{3}}+\frac{m}{\sqrt{3}}+\frac{n}{\sqrt{3}}\)
\(\cos \beta =\frac{l-m+n}{\sqrt{3}},\cos \gamma =\frac{+l+m-n}{\sqrt{3}},\cos \delta =\frac{l-m-n}{\sqrt{3}}\)
∴ \(\sum{{{\cos }^{2}}\alpha =\frac{4\left( {{l}^{2}}+{{m}^{2}}+{{n}^{2}} \right)}{3}}=\frac{4}{3}.\)
Problem 4. The angle between the lines whose direction cosines are given by 3l – 5m + n = 0 and 30mn – 2ln + 10lm = 0
Answer:Â 30mn + 2l( 5m – n) = 0
or, 30mn + 2l, 3l = 0
or, 5mn + l2 = 0
also 3l = 5m – n.
∴ \(30mn+2\frac{{{\left( 5m-n \right)}^{2}}}{3}=0\)
or, \(90mn+50{{m}^{2}}+2{{n}^{2}}-20mn=0\)
\(\Rightarrow \,\,\,\,\,\,\,\,\,\,50{{m}^{2}}+2{{n}^{2}}+70mn=0.\)
\(25{{m}^{2}}+{{n}^{2}}+35mn=0.\)
Problem 5. The angle between the lines \(\frac{x-2}{3}=\frac{y+1}{-2}=\frac{c-2}{0}\) and \(\frac{a-1}{1}=\frac{2b+3}{3}=\frac{c+5}{2}\) is
Answer: \(\cos \theta =3.1+\left( -2 \right)\frac{3}{2}+0\)
∴ \(\theta =\frac{\pi }{2}\)
Problem 6. Find the equation of the plane containing the lines \(\frac{x+1}{3}=\frac{y-2}{2}=\frac{z}{1}\) and \(\frac{x-3}{3}=\frac{y+4}{2}=\frac{z-1}{1}\)
Answer: The plane be
\(ax+by+cz+d=0\)
∴ 3a + 2b + 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 \(\frac{x-1}{2}=\frac{y-2}{{{x}_{1}}}=\frac{z-3}{{{x}_{2}}}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) lies in the same plane then for the equation \({{x}_{1}}{{t}^{2}}+\left( {{x}_{2}}+2 \right)t+a=0\) prove sum of roots \(=-2\)
Answer: Let the plane be \(ax+by+cz+d=0\)
\(a+2b+3c+d=0\rightarrow{{}}\left( i \right)\)
\(2a+3b+4c+d=0\rightarrow{{}}\left( ii \right)\)
\(2a+{{x}_{1}}b+{{x}_{2}}c=0\rightarrow{{}}\left( iii \right)\)
\(3a+4b+5c=0\rightarrow{{}}\left( iv \right)\)
Sum of the roots \(=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}\)
from \(\left( i \right)\) and\(\left( ii \right)\) \(a+b+c=0\)
from \(\left( iii \right)\) and \(\left( iv \right)\) \(a+b+c-d=0\)
∴ 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
∴ \(\,\,{{x}_{1}}=3{{x}_{2}}=4\)
sum of roots \(=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}\)
∴ \(=-\frac{6}{3}=-2\)
Problem 8. The line \(\frac{x}{K}=\frac{y}{2}=\frac{7}{-12}\) 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 \(\frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}\)
i.e. \(2x+y+3z-1=x+2y-3z+1\)
or \(x-y+6z=0\rightarrow{{}}\left( i \right)\)
and \(2x+y+3z-1=-x-2y+3z+1\)
i.e. \(3x+3y-2=0\rightarrow{{}}\left( ii \right)\)
so the given line must be parallel to \(\left( i \right)\) or \(\left( ii \right)\)
∴ \(K-2-74=0or3K+6=0\)
∴ \(\,\,K-2(K=76\) is discarded
Problem 9. Direction cosine of normal to the plane containing lines \(x=y=z\) and \(x-1=y-1=\frac{z-1}{d}\) are
Answer:
Let \(d.{{c}^{s}}\) be \(l,m,n\)
∴ l + m + n = 0 and l + m + dn = 0
i.e. n=0
∴ l = -m from \({{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\)
We have \(2{{l}^{2}}=1\)
∴ \(l=\pm \frac{1}{\sqrt{2}}\)
∴ \(\left( l,m,n \right)\) is \(\left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0 \right)\) or \(\left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \right).\)
Problem 10. If the line \(x=y=z\) intersect the line \(\sin Ax+\sin By+\sin Cz=2{{d}^{2}},\)
\(\sin 2A\,\,x+\sin 2B\,\,y+\sin 2C\,\,y={{d}^{2}}\) then \(\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{c}{2}=?\) \(\left( A+B+C=\pi \right).\)
Answer: Let the point of intersection be \(\left( t,t,t \right)\)
∴ \(\left( \sin A+\sin B+\sin C \right)t=2{{d}^{2}}\)
\(\left( \sin 2A+\sin 2B+\sin 2C \right)t={{d}^{2}}\)
\(\sin 2A+\sin 2B-\sin \left( 2A+2B \right)=\frac{{{d}^{2}}}{t}.\)
\(\Rightarrow 2.\sin \left( A+B \right).\cos \left( A-B \right)-2\sin \left( A+B \right).\cos \left( A+B \right)\)
\(\Rightarrow 4\sin C\,\,\sin A.\,\,\sin B=\frac{{{d}^{2}}}{t.}\rightarrow{{}}\left( i \right)\)
\(\sin A+\sin B+\sin C\)
\(=2.\sin \frac{A+B}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}\)
\(=2\cos \frac{C}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}\)
\(=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2}+\cos \frac{A+B}{2} \right)\)
\(=2\cos \frac{C}{2}.2cos\frac{A}{2}.\cos \frac{B}{2}=\frac{2{{d}^{2}}}{t}\rightarrow{{}}\left( ii \right)\)
dividing \(\left( i \right)\div \left( ii \right)\)
\(\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}=\frac{1}{16}.\)
Problem 11. Equation of the sphere having center at \(\left( 3,6,-4 \right)\) and touching the plane \(\bar{r}.\left( 2\hat{i}-2\hat{j}-\hat{k} \right)=10\) is \({{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}+{{\left( z+4 \right)}^{2}}={{K}^{2}},\) then
Answer: Equation of the plane is \(2x-2y-z-10=0\)
distance from \(\left( 3,6,-4 \right)\) to it is \(\left| \frac{6-12+4-10}{\sqrt{9}} \right|=\left| \frac{12}{3} \right|=4\)
∴ K = 4.