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# 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. 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

$$\begin{array}{l}P\left( x,y,z \right)\end{array}$$
from origin
$$\begin{array}{l}\left( 0,0,0 \right)\end{array}$$
is
$$\begin{array}{l}\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\end{array}$$

### Distance between 2 points

Distance between 2 points

$$\begin{array}{l}P\left( x_1,y_1,z_1 \right)\end{array}$$
and
$$\begin{array}{l}Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\end{array}$$
is
$$\begin{array}{l}\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\end{array}$$

### Division of a line joining 2 points

Let

$$\begin{array}{l}P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\end{array}$$
and
$$\begin{array}{l}Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\end{array}$$
be 2 points. R derives the line segment PQ in ratio internally. Then R has coordinate

$$\begin{array}{l}\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\end{array}$$

### Projection in 3D Space Projection in 3D Space – 3D Geometry

Let AB be a line segment. It’s projection on a line PQ,

$$\begin{array}{l}AB\,\,\cos \theta .\end{array}$$
Where
$$\begin{array}{l}\theta\end{array}$$
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.

So if those angles are

$$\begin{array}{l}\alpha ,\beta ,\gamma\end{array}$$
then
$$\begin{array}{l}\cos \alpha ,cos\beta ,cos\gamma\end{array}$$
are the direction cosines of the line. They are denoted by
$$\begin{array}{l}l,m,n.\end{array}$$

$$\begin{array}{l}{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1.\end{array}$$
(Proof will be given)

Any 3 numbers,

$$\begin{array}{l}a,b,c\end{array}$$
which are proportional to direction cosines are called direction ratios.

Hence

$$\begin{array}{l}\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\end{array}$$

$$\begin{array}{l}l=\frac{a}{\sqrt{\sum{{{a}^{2}}}}},m=\frac{b}{\sqrt{\sum{{{a}^{2}}}}},n=\frac{c}{\sqrt{\sum{{{a}^{2}}}}}\end{array}$$

### Direction cosine of line joining two given points

Let

$$\begin{array}{l}P\left( x_1,y_1,z_1 \right)\end{array}$$
and
$$\begin{array}{l}Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\end{array}$$
be 2 points. Then direction cosines will be
$$\begin{array}{l}l=\frac{{{x}_{2}}-{{x}_{1}}}{\left| PQ \right|},m=\frac{{{y}_{2}}-{{y}_{1}}}{\left| PQ \right|},n=\frac{{{z}_{2}}-{{z}_{1}}}{\left| PQ \right|}\end{array}$$

### Projection of line segment joining 2 points, on another line

$$\begin{array}{l}P\left( {{x}_{1}}{{y}_{1}}{{z}_{1}} \right) and Q\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)\end{array}$$

Projection of PQ on a line whose direction cosines are

$$\begin{array}{l}l,m,n\end{array}$$
is
$$\begin{array}{l}l\left( {{x}_{2}}-{{x}_{1}} \right)+m\left( {{y}_{2}}-{{y}_{1}} \right)+n\left( {{z}_{2}}-{{z}_{1}} \right)\end{array}$$

### Angle between 2 lines in 3 Dimensional Space

2 lines having direction cosines

$$\begin{array}{l}\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)\end{array}$$
and
$$\begin{array}{l}\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right).\end{array}$$
Then angle between them is
$$\begin{array}{l}\theta ={{\cos }^{-1}}\left( {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} \right)\end{array}$$

### Projection of a plane area on 3 coordinate planes

Let

$$\begin{array}{l}\bar{A}\end{array}$$
be the vector area. If its direction cosines are
$$\begin{array}{l}\cos \alpha ,\cos \beta ,\cos \gamma .\end{array}$$
Then projections are
$$\begin{array}{l}{{A}_{1}}=A\cos \alpha ,{{A}_{2}}=A\cos \beta ,{{A}_{3}}=A\cos \gamma .\end{array}$$

$$\begin{array}{l}{{A}^{2}}=A_{1}^{2}+A_{2}^{2}+A_{3}^{2}.\end{array}$$

### Area of a triangle:

Using the projection formula, area of a triangle =

$$\begin{array}{l}\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}+\,\,\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|}^{2}}\end{array}$$

Check out more details about the area of a triangle in coordinate geometry , its derivation and problem solving strategies, etc.

## Concept of Plane in 3 Dimensional Geometry

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

$$\begin{array}{l}ax+by+cz=0,{{z}^{2}}{{b}^{2}}+{{c}^{2}}\ne 0\end{array}$$
represents a plane.

### Normal Form of a Plane

Let P be the length of the normal from the origin to the plane and

$$\begin{array}{l}l,m,n\end{array}$$
be the direction cosines of that normal. Then the equation of the plane is given by
$$\begin{array}{l}lx+my+nz=P.\end{array}$$

### Intercept form

Let a plane cuts length

$$\begin{array}{l}a,b,c\end{array}$$
from the coordinate axis.

Then equation of the plane is

$$\begin{array}{l}\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\end{array}$$

### Planes passing through 3 given points

Plane passing through

$$\begin{array}{l}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)\end{array}$$
is

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

### Angle between 2 planes

$$\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\end{array}$$
and
$$\begin{array}{l}{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\end{array}$$
is given by

$$\begin{array}{l}\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\end{array}$$

### Two sides of a plane

Consider 2 points

$$\begin{array}{l}A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\end{array}$$
and
$$\begin{array}{l}B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\end{array}$$
lie on the same side or opposite sides of a plane
$$\begin{array}{l}ax+by+cz+d=0\end{array}$$
, accordingly as
$$\begin{array}{l}a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d\end{array}$$
and
$$\begin{array}{l}a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d\end{array}$$
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

$$\begin{array}{l}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\end{array}$$
from
$$\begin{array}{l}ax+by+cz+d\end{array}$$
is
$$\begin{array}{l}\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\end{array}$$

### Equation of the planes bisecting the angle between 2 planes

Let

$$\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+d=0\end{array}$$
and
$$\begin{array}{l}{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+d=0\end{array}$$
be 2 planes. The equation of plane bisecting the angles between them is

$$\begin{array}{l}\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\end{array}$$

### Position of origin

The origin lies in the acute or obtuse angle between

$$\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\end{array}$$
and
$$\begin{array}{l}{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\end{array}$$
according as
$$\begin{array}{l}{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}<0\ or\ >0\end{array}$$
provided
$$\begin{array}{l}{{d}_{1}}\end{array}$$
and
$$\begin{array}{l}{{d}_{2}}\end{array}$$
are both positive.

### Two Intersecting plane

If

$$\begin{array}{l}U=0\end{array}$$
and
$$\begin{array}{l}V=0\end{array}$$
be 2 planes then the plane passing through the line of their intersection is
$$\begin{array}{l}U+\lambda V=0\lambda\end{array}$$
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

$$\begin{array}{l}\left\{ l,m,n \right\}\end{array}$$
is given by

$$\begin{array}{l}\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\end{array}$$

• Two point form

Equation of a straight line passing through

$$\begin{array}{l}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\end{array}$$
and
$$\begin{array}{l}\left( {{x}_{2}}{{y}_{2}}{{z}_{2}} \right)\end{array}$$
is
$$\begin{array}{l}\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\end{array}$$

• Two plane form to symmetrical form

Let 2 planes be

$$\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\end{array}$$
and
$$\begin{array}{l}{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\end{array}$$
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

$$\begin{array}{l}ax+by+cz+d=0\end{array}$$
is intersected by
$$\begin{array}{l}\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}.\end{array}$$

To find the intersection point

Let

$$\begin{array}{l}\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}=t\end{array}$$

$$\begin{array}{l}x={{x}_{1}}+lt,y={{y}_{1}}+mt,z={{z}_{1}}+nt\end{array}$$
put these in the equation of plane and solve for it.

### Plane through a given straight line

Let the line be

$$\begin{array}{l}\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\end{array}$$
of the plane through this line be
$$\begin{array}{l}ax+by+cz+d=0\end{array}$$
then
$$\begin{array}{l}a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d=0\end{array}$$
and
$$\begin{array}{l}al+bm+cn=0\end{array}$$
and from other given conditions
$$\begin{array}{l}a,b,c\end{array}$$
are determined.

### Coplanarity of Two Lines In 3D Geometry

Let 2 lines are

$$\begin{array}{l}\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\,\,\And \,\,\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\end{array}$$

Two lines are coplanar iff

$$\begin{array}{l}\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|=0\end{array}$$

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

Let the line be

$$\begin{array}{l}\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}\end{array}$$

AQ = projection of AP on the straight line

$$\begin{array}{l}=l\left( {{x}_{1}}\alpha \right)+m\left( {{y}_{1}}-\beta \right)+n\left( {{z}_{1}}-\gamma \right)\end{array}$$

$$\begin{array}{l}\,\,PQ=\sqrt{A{{P}^{2}}-A{{Q}^{2}}}\end{array}$$

### Shortest distance between two skew lines

Let the 2 skew lines be

$$\begin{array}{l}\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\end{array}$$
and
$$\begin{array}{l}\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\end{array}$$

The shortest distance is

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

The equation of the shortest distance is

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

and

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

## Problems on 3D Geometry

Problem 1. If a variable plane forms a tetrahedron of constant volume

$$\begin{array}{l}64{{K}^{3}}\end{array}$$
with the coordinate planes then the lows of the centroid of the tetrahedron is
$$\begin{array}{l}xyz=u{{K}^{3}}.\end{array}$$
Find
$$\begin{array}{l}u.\end{array}$$

Answer: Let the equation of the plane be

$$\begin{array}{l}\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\end{array}$$

Centroid of the tetrahedron is

$$\begin{array}{l}\left( \frac{a}{4},\frac{b}{4},\frac{c}{4} \right)\end{array}$$

Volume of the tetrahedron

$$\begin{array}{l}=\frac{abc}{6}=64{{K}^{3}}.\end{array}$$

So letting

$$\begin{array}{l}\frac{a}{4}=x,\frac{b}{4}=y,\frac{c}{4}=z\end{array}$$

We have

$$\begin{array}{l}\frac{abc}{6}=\frac{{{4}^{3}}xyz}{6}=64{{K}^{3}}.\end{array}$$

$$\begin{array}{l}xyz=6{{K}^{3}}\end{array}$$

On comparing, we have u=6.

Problem 2. The ration in which

$$\begin{array}{l}yz\end{array}$$
plane divides the line joining
$$\begin{array}{l}\left( 2,4,5 \right)\end{array}$$
and
$$\begin{array}{l}\left( 3,5,7 \right)\end{array}$$
is

$$\begin{array}{l}\lambda :1.\end{array}$$

$$\begin{array}{l}x\end{array}$$
coordinate
$$\begin{array}{l}=0=\frac{3\lambda +2}{\lambda +1}\end{array}$$

$$\begin{array}{l}\lambda =-\frac{2}{3}\end{array}$$

∴ the ratio is 2:3

Problem 3. A line makes angles

$$\begin{array}{l}\alpha ,\beta ,\gamma ,\delta\end{array}$$
with the 4 diagonals of a cube then
$$\begin{array}{l}\sum{{{\cos }^{2}}\alpha =?}\end{array}$$ Let the direction cosine of that line be

$$\begin{array}{l}\left( l,m,n \right).\end{array}$$

Direction cosine of 1st diagonal

$$\begin{array}{l}\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)\end{array}$$

Direction cosine of 2nd diagonal

$$\begin{array}{l}\left( \frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)\end{array}$$

Direction cosine of 3rd diagonal

$$\begin{array}{l}\left( \frac{1}{\sqrt{3}},\frac{+1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)\end{array}$$
and

Direction cosine of 4th diagonal

$$\begin{array}{l}\left( \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}} \right)\end{array}$$

$$\begin{array}{l}\cos \alpha =\frac{l}{\sqrt{3}}+\frac{m}{\sqrt{3}}+\frac{n}{\sqrt{3}}\end{array}$$

$$\begin{array}{l}\cos \beta =\frac{l-m+n}{\sqrt{3}},\cos \gamma =\frac{+l+m-n}{\sqrt{3}},\cos \delta =\frac{l-m-n}{\sqrt{3}}\end{array}$$

$$\begin{array}{l}\sum{{{\cos }^{2}}\alpha =\frac{4\left( {{l}^{2}}+{{m}^{2}}+{{n}^{2}} \right)}{3}}=\frac{4}{3}.\end{array}$$

Problem 4. The angle between the lines

$$\begin{array}{l}\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}\end{array}$$
and
$$\begin{array}{l}\frac{x-1}{1}=\frac{2y+3}{3}=\frac{z+5}{2}\end{array}$$
is equal to

$$\begin{array}{l}\cos \theta =\frac{3\times1 – 2\times3/2+0\times2}{\sqrt{3^2+(-2)^2+0^2}\sqrt{1^2+(3/2)^2+2^2} }\end{array}$$

$$\begin{array}{l}\theta =\frac{\pi }{2}\end{array}$$

Problem 5. If lines

$$\begin{array}{l}\frac{x-1}{2}=\frac{y-2}{{{x}_{1}}}=\frac{z-3}{{{x}_{2}}}\end{array}$$
and
$$\begin{array}{l}\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\end{array}$$
lies in the same plane then for the equation
$$\begin{array}{l}{{x}_{1}}{{t}^{2}}+\left( {{x}_{2}}+2 \right)t+a=0\end{array}$$
prove sum of roots
$$\begin{array}{l}=-2\end{array}$$

$$\begin{array}{l}ax+by+cz+d=0\end{array}$$

$$\begin{array}{l}a+2b+3c+d=0\rightarrow{{}}\left( i \right)\end{array}$$

$$\begin{array}{l}2a+3b+4c+d=0\rightarrow{{}}\left( ii \right)\end{array}$$

$$\begin{array}{l}2a+{{x}_{1}}b+{{x}_{2}}c=0\rightarrow{{}}\left( iii \right)\end{array}$$

$$\begin{array}{l}3a+4b+5c=0\rightarrow{{}}\left( iv \right)\end{array}$$

Sum of the roots

$$\begin{array}{l}=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}\end{array}$$

from

$$\begin{array}{l}\left( i \right)\end{array}$$
and
$$\begin{array}{l}\left( ii \right)\end{array}$$
$$\begin{array}{l}a+b+c=0\end{array}$$

from

$$\begin{array}{l}\left( iii \right)\end{array}$$
and
$$\begin{array}{l}\left( iv \right)\end{array}$$
$$\begin{array}{l}a+b+c-d=0\end{array}$$

∴ 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

$$\begin{array}{l}\,\,{{x}_{1}}=3{{x}_{2}}=4\end{array}$$

sum of roots

$$\begin{array}{l}=\frac{-\left( {{x}_{2}}+2 \right)}{{{x}_{1}}}\end{array}$$

$$\begin{array}{l}=-\frac{6}{3}=-2\end{array}$$

Problem 6. The line

$$\begin{array}{l}\frac{x}{K}=\frac{y}{2}=\frac{z}{-12}\end{array}$$
makes an isosceles triangle with the planes
$$\begin{array}{l}2x+y+3z-1=0\end{array}$$
and
$$\begin{array}{l}x+2y-3z-1=0\end{array}$$
then
$$\begin{array}{l}K=?\end{array}$$

Answer: Equation of the bisector planes are

$$\begin{array}{l}\frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}\end{array}$$

i.e.

$$\begin{array}{l}2x+y+3z-1=x+2y-3z-1\end{array}$$

or

$$\begin{array}{l}x-y+6z=0\rightarrow{{}}\left( i \right)\end{array}$$

and

$$\begin{array}{l}2x+y+3z-1=-x-2y+3z+1\end{array}$$

i.e.

$$\begin{array}{l}3x+3y-2=0\rightarrow{{}}\left( ii \right)\end{array}$$

so the given line must be parallel to

$$\begin{array}{l}\left( i \right)\end{array}$$
or
$$\begin{array}{l}\left( ii \right)\end{array}$$

The co-efficient of x and y of required straight line equation is -1 and 1.
Therefore, the value of K is -2.

Problem 7. Direction cosine of normal to the plane containing lines

$$\begin{array}{l}x=y=z\end{array}$$
and
$$\begin{array}{l}x-1=y-1=\frac{z-1}{d}\end{array}$$
are Problems in 3D Geometry

Let directions cosines be l,m,n.

∴ l + m + n = 0 and l + m + dn = 0

i.e. n=0

∴ l = -m from

$$\begin{array}{l}{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\end{array}$$

We have

$$\begin{array}{l}2{{l}^{2}}=1\end{array}$$

$$\begin{array}{l}l=\pm \frac{1}{\sqrt{2}}\end{array}$$

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

Problem 8. If the line

$$\begin{array}{l}x=y=z\end{array}$$
intersect the line
$$\begin{array}{l}x\sin A+y\sin B+z\sin C=2{{d}^{2}},\end{array}$$

$$\begin{array}{l}x\sin 2A\,+y\sin 2B\,+z\sin 2C\,={{d}^{2}}\end{array}$$
then

$$\begin{array}{l}\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}=?\end{array}$$

where,

$$\begin{array}{l}\left( A+B+C=\pi \right).\end{array}$$

Answer: Let the point of intersection be

$$\begin{array}{l}\left( t,t,t \right)\end{array}$$

$$\begin{array}{l}\left( \sin A+\sin B+\sin C \right)t=2{{d}^{2}}\end{array}$$

$$\begin{array}{l}\left( \sin 2A+\sin 2B+\sin 2C \right)t={{d}^{2}}\end{array}$$

$$\begin{array}{l}\sin 2A+\sin 2B-\sin \left( 2A+2B \right)=\frac{{{d}^{2}}}{t}.\end{array}$$

$$\begin{array}{l}\Rightarrow 2.\sin \left( A+B \right).\cos \left( A-B \right)-2\sin \left( A+B \right).\cos \left( A+B \right)=\frac{{{d}^{2}}}{t}\end{array}$$

$$\begin{array}{l}\Rightarrow 4\sin C\,\,\sin A.\,\,\sin B=\frac{{{d}^{2}}}{t.}\rightarrow{{}}\left( i \right)\end{array}$$

Again,

$$\begin{array}{l}\sin A+\sin B+\sin C\end{array}$$

$$\begin{array}{l}=2.\sin \frac{A+B}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}\end{array}$$

$$\begin{array}{l}=2\cos \frac{C}{2}.\cos \frac{A-B}{2}+2\sin \frac{C}{2}.\cos \frac{C}{2}\end{array}$$

$$\begin{array}{l}=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2}+\cos \frac{A+B}{2} \right)\end{array}$$

$$\begin{array}{l}=2\cos \frac{C}{2}.2cos\frac{A}{2}.\cos \frac{B}{2}=\frac{2{{d}^{2}}}{t}\rightarrow{{}}\left( ii \right)\end{array}$$

dividing

$$\begin{array}{l}\left( i \right)\div \left( ii \right)\end{array}$$

$$\begin{array}{l}\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}=\frac{1}{16}.\end{array}$$

Problem 9. Equation of the sphere having center at

$$\begin{array}{l}\left( 3,6,-4 \right)\end{array}$$
and touching the plane
$$\begin{array}{l}\bar{r}.\left( 2\hat{i}-2\hat{j}-\hat{k} \right)=10\end{array}$$
is
$$\begin{array}{l}{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}+{{\left( z+4 \right)}^{2}}={{K}^{2}},\end{array}$$
then k =

Answer: Equation of the plane is

$$\begin{array}{l}2x-2y-z-10=0\end{array}$$

distance from

$$\begin{array}{l}\left( 3,6,-4 \right)\end{array}$$
to it is
$$\begin{array}{l}\left| \frac{6-12+4-10}{\sqrt{9}} \right|=\left| \frac{12}{3} \right|=4\end{array}$$

∴ k = 4.

### Vector Algebra and 3D Geometry – Important Topics ### Vector Algebra and 3D Geometry – Important Questions ### Vector Algebra and 3D Geometry – Important Topics Part 2 ### Vector Algebra and 3D Geometry – Important Questions Part 2 ### 3D Geometry – Top 10 Most Important and Expected JEE Questions ### 3-D Geometry – JEE Advanced Problems ### What do you mean by Direction Cosines of a line?

The direction cosine of a line is the cosine of the angle subtended by this line with the x-axis, y-axis, and z-axis respectively.

### How do you represent a point in 3D Geometry?

We can represent a point in three-dimensional geometry either in cartesian form or a vector form. The cartesian form of representation of any point in 3D geometry is (x, y, z) and is with reference to the x-axis, y-axis, and z-axis respectively.

### Give the formula to find the distance of a point P(x, y, z) from the origin.

The distance of a point P(x, y, z) from the origin (0, 0, 0) is given by √(x2+y2+z2).

### What is the relation between direction cosines of a line?

If l, m, n denotes the direction cosines of a line, then l2+m2+n2=1.

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