Class 12 Exam > Class 12 Videos > L9 : Example: Line through 2 points - Three Dimensional Geometry , Maths, Class 12
L9 : Example: Line through 2 points - Three Dimensional Geometry , Maths, Class 12 Video Lecture
FAQs on L9 : Example: Line through 2 points - Three Dimensional Geometry , Maths, Class 12 Video Lecture
| 1. How do you find the equation of a line passing through two points in three-dimensional geometry? |  |
Ans. To find the equation of a line passing through two points in three-dimensional geometry, we can use the formula:
(x - x₁) / (x₂ - x₁) = (y - y₁) / (y₂ - y₁) = (z - z₁) / (z₂ - z₁)
where (x₁, y₁, z₁) and (x₂, y₂, z₂) are the coordinates of the given points. This equation represents the line in three-dimensional space.
| 2. Is it necessary to have three points to determine a line in three-dimensional geometry? | |
Ans. No, it is not necessary to have three points to determine a line in three-dimensional geometry. Two distinct points are sufficient to determine a unique line in three-dimensional space. The line passing through these two points can be represented using the equation mentioned in the previous question.
| 3. Can the equation of a line be written in vector form in three-dimensional geometry? | |
Ans. Yes, the equation of a line can be written in vector form in three-dimensional geometry. The vector form of the equation is given by:
r = a + t * d
where r is the position vector of any point on the line, a is the position vector of a known point on the line, d is the direction vector of the line, and t is a parameter. This form is widely used to represent lines in three-dimensional space.
| 4. How can the direction vector of a line passing through two points be obtained? | |
Ans. The direction vector of a line passing through two points can be obtained by subtracting the coordinates of one point from the coordinates of the other point. Let's say the two points are (x₁, y₁, z₁) and (x₂, y₂, z₂). Then, the direction vector of the line is given by:
d = (x₂ - x₁, y₂ - y₁, z₂ - z₁)
The direction vector represents the direction in which the line is moving in three-dimensional space.
| 5. Can the equation of a line passing through two points be used to find the distance between them? | |
Ans. No, the equation of a line passing through two points cannot be directly used to find the distance between them. The equation of a line only represents the line itself, not the distance between the two points. To find the distance between two points, we can use the distance formula in three-dimensional space, which is derived from the Pythagorean theorem:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
By substituting the coordinates of the two points into this formula, we can calculate the distance between them.
Text Transcript from Video
hello friends this video on
three-dimensional geometry part 9 is
brought you by exam viacom no more
flavor of exam before watching this
video please make sure that you have
watched part 1 to part 8 next take some
examples of a line passing through the
three points so you define the vector
and the Cartesian equation for the line
that passes through reason that is 0 0 0
and this point so I have my let's
suppose a vector has 0 ie plus 0 Z plus
0 K and V vector as 5i minus 2j minus 3y
i minus TJ plus so I have to find in
vector in Cartesian let's mine vector
vector is e plus lambda V minus E is 0 0
I plus 0 j plus 0 k plus lambda B is Phi
I minus 2j plus 3k minus 0 vector 0 I is
0 J 0 K this is nothing but lambda into
Phi I minus 2 J plus and this is my
equation in vector form correct
now let's find the vector in Cartesian
so that is the formula I know that in my
equation passes through X 1 Y minus y1
by y2 minus y1 is equal to 0-7 by itself
in my physics this equation for the line
that passes through x1 y1 z1 and x2 y2
z2 so here X 1 minus n 0 0 0 x2 y2 z2 is
5 minus 2 n 3 so let's put the value so
with this I get X minus X bar is 0 right
and X 2 is 5 X 1 is again 0 is equal to
Y minus y1 is 0 why do it this guy minus
2 minus 0 Z minus z1 is again 0 by Z 2
is 3 so what you get is X by 5 is equal
to Y by minus 2 is equal to Z by 3 and
this is my equation of the line in
Cartesian so I have two equation this
guy in vector form and this guy in
condition so now we know how to form the
equation of the line that passes through
viewpoint
we'll take some more examples to clear
conserve a refinery in the vector in
Cartesian form of the line the pass is
to these two points here it doesn't pass
at origin it has two points let this be
a vector and this may be B so equation
I'm applying something but K vector plus
lambda into V vector minus a vector
correct
a vector is what 3 minus 2 minus 5 that
is 3i minus TJ minus 5
he his director last lambda into B
vector 3 minus 2 6 3 I minus 2 ej + 6 k
in all caps
-
a 3-2 - v 3i - TJ - correct so this is
my equation and if you solve this this
becomes 3i minus 3i becomes zero this is
3i minus TJ minus Phi k plus lambda into
3a minus 3i becomes 0 2 J minus 2
becomes 0 6 K minus minus Phi is 6 5 6
plus 5 is 11 right so with this I get 3i
minus 2j plus 11 lambda minus 5 okay
this is my equation of line in the
vector form correct I have to find the
equation of the line in the Cartesian
form so we know this is nothing but X
minus x1 by x2 minus x1 is equal to Y
minus y1 by y2 - y1 is equal to Z minus
Z 1 by Z 2 minus Z so let's put the
value X - X minus 3 so let me put the
value ray exposures and it will try to
expand by one side 1 3 minus 2 minus 5
and x 2 y 2 Z 3 minus 2 X 6 so X 1 is 3
by X 2 is 3 minus X 1 is 3 this is equal
to Y minus y1 is minus 2 so y plus 2 it
becomes Y 2 is minus 2 minus y 1 is
again minus 2 equal to Z 2 minus Z 1 is
fine so this becomes plus 5 and this is
y 2 Z 2 - n to the 6 minus minus 5 that
is 6
this becomes X minus three by zero is
equal to y plus two pi 0 is 8 plus 5 and
that is masked
this is an equation of line in Cartesian
form for this condition thank you
visit exam viacom to watch free
educational videos try free on my deaths
get the best quality study materials
study from the best tutors and mentors
and much more
thanks again
three-dimensional geometry part 9 is
brought you by exam viacom no more
flavor of exam before watching this
video please make sure that you have
watched part 1 to part 8 next take some
examples of a line passing through the
three points so you define the vector
and the Cartesian equation for the line
that passes through reason that is 0 0 0
and this point so I have my let's
suppose a vector has 0 ie plus 0 Z plus
0 K and V vector as 5i minus 2j minus 3y
i minus TJ plus so I have to find in
vector in Cartesian let's mine vector
vector is e plus lambda V minus E is 0 0
I plus 0 j plus 0 k plus lambda B is Phi
I minus 2j plus 3k minus 0 vector 0 I is
0 J 0 K this is nothing but lambda into
Phi I minus 2 J plus and this is my
equation in vector form correct
now let's find the vector in Cartesian
so that is the formula I know that in my
equation passes through X 1 Y minus y1
by y2 minus y1 is equal to 0-7 by itself
in my physics this equation for the line
that passes through x1 y1 z1 and x2 y2
z2 so here X 1 minus n 0 0 0 x2 y2 z2 is
5 minus 2 n 3 so let's put the value so
with this I get X minus X bar is 0 right
and X 2 is 5 X 1 is again 0 is equal to
Y minus y1 is 0 why do it this guy minus
2 minus 0 Z minus z1 is again 0 by Z 2
is 3 so what you get is X by 5 is equal
to Y by minus 2 is equal to Z by 3 and
this is my equation of the line in
Cartesian so I have two equation this
guy in vector form and this guy in
condition so now we know how to form the
equation of the line that passes through
viewpoint
we'll take some more examples to clear
conserve a refinery in the vector in
Cartesian form of the line the pass is
to these two points here it doesn't pass
at origin it has two points let this be
a vector and this may be B so equation
I'm applying something but K vector plus
lambda into V vector minus a vector
correct
a vector is what 3 minus 2 minus 5 that
is 3i minus TJ minus 5
he his director last lambda into B
vector 3 minus 2 6 3 I minus 2 ej + 6 k
in all caps
-
a 3-2 - v 3i - TJ - correct so this is
my equation and if you solve this this
becomes 3i minus 3i becomes zero this is
3i minus TJ minus Phi k plus lambda into
3a minus 3i becomes 0 2 J minus 2
becomes 0 6 K minus minus Phi is 6 5 6
plus 5 is 11 right so with this I get 3i
minus 2j plus 11 lambda minus 5 okay
this is my equation of line in the
vector form correct I have to find the
equation of the line in the Cartesian
form so we know this is nothing but X
minus x1 by x2 minus x1 is equal to Y
minus y1 by y2 - y1 is equal to Z minus
Z 1 by Z 2 minus Z so let's put the
value X - X minus 3 so let me put the
value ray exposures and it will try to
expand by one side 1 3 minus 2 minus 5
and x 2 y 2 Z 3 minus 2 X 6 so X 1 is 3
by X 2 is 3 minus X 1 is 3 this is equal
to Y minus y1 is minus 2 so y plus 2 it
becomes Y 2 is minus 2 minus y 1 is
again minus 2 equal to Z 2 minus Z 1 is
fine so this becomes plus 5 and this is
y 2 Z 2 - n to the 6 minus minus 5 that
is 6
this becomes X minus three by zero is
equal to y plus two pi 0 is 8 plus 5 and
that is masked
this is an equation of line in Cartesian
form for this condition thank you
visit exam viacom to watch free
educational videos try free on my deaths
get the best quality study materials
study from the best tutors and mentors
and much more
thanks again
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Jan 04, 2025
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