Three Dimensional Geometry
Introduction
We will learn about the 3-D geometryand also about the 3-D coordinate system.
How to make use of vector algebra in 3-D geometry.
We will discuss about the direction cosines and direction ratios of a line joining two points and also about the equations of plane and lines in space under different conditions.
In case of 3-D the three coordinateplanes divide the space into eight parts known as octants.
Some examples of where 3-D geometry is being used:
- The path followed an aeroplane is along the x, y and z coordinates.
- The path followed by a ball in a room is another example of 3-D.When a ball is thrown it will hit the walls and bounce back. The path followed by the ball was in all the 3 directions (x, y,and z).
Represent a point in Cartesian and Vector form
Consider a point P (4, 5, 6) and in Cartesian form it will be represented as:
- (x,y,z)=(4,5, 6).
In vector form point P will be represented as (4i+5j+6k).
Representing a line in 2Dwhose end points are given as (3,5) and (2,8):-
In Cartesian form:-
(y-8)/(x-2)=(8-5)/(2-3)=-3
(y-8)=-3(x-2) (equation(1))
In Vector form:-
Representing a line in 3D whose end points are given as:
In Cartesian form:-
Direction Cosines of a Line
Consider a line passing through origin and making angle α with x-axis, β angle with y-axis and γ angle with z-axis,then cosα,cosβ and cosγareknown as direction cosines of the line.
Direction cosines are the cosine of the angles which a line makes with XY and Z-axis.
They are denoted by(l,mand n).Therefore, cos α =l,cosβ=m and cos γ= n.
To prove: - l2+m2+n2=1 i.e. cos2α + cos2 β+ cos2 γ =1.
Proof:- Consider a line PQ and its equation is given as:-
PQ->= (a î+bĵ+ck̂)
=r(cos α î+ cosβ ĵ+ cos γ k̂)
where r=√(a2+b2+c2)
IPQ-->I2=(a2+b2+c2)
=r2 (cos2 α +cos2β +cos2γ)= (a2+b2+c2) =r2 (cancelling r2 from both the sides).
=(cos2 α + cos2β + cos2γ) =1
l2+m2+n2 =1.
The above proof holds good for any line.
Hence Proved.
Direction Ratios of a Line
Any 3 numbers which are proportional to the direction cosines of a line are known as direction ratios of that line.
Direction ratios are denoted by (a,b and c).
Consider a vector OP-->=a î +bĵ +ck̂ (Where a, b and c are direction ratios. )
=r (cos α î + cosβ ĵ + cos γ k̂) where r=√ (a2+b2+c2)
cosα,cosβ and cosγ are direction cosines.
r cosα=a,r cosβ=b, r cosγ=c are direction ratios.
Relation between Direction Ratios and Direction Cosines
Consider a line whose one point is at originand another is at P. Considering a vector OP-->=a î +bĵ +ck̂ (equation(1))
=r cos αî+r cosβĵ+r cos γk̂
where r=√ (a2+b2+c2) (equation(2))
Comparing equations (1) and (2) :-
a=(rcos α) ,b =(r cosβ) and c=(rcosγ)(equation(3))
cos α =(a/r), cosβ=(b/r) and cosγ=(c/r) (equation(4))
Squaring and adding equation(4) :-
(a2/r2) + (b2/r2)+(c2/r2) =1
using (cos2 α + cos2 β + cos2 γ) =1
=>(a2+ b2+c2) = r2 (equation(5))
From equation(4) cos α =(a/r)= (±) (a)/(√( a2+ b2+c2)
using equation(5)
Similarly cos β=(b/r) =(±)(b)/ (√( a2+ b2+c2) and
cos γ=(c/r) =(±)(c)/ (√( a2+ b2+c2)
usingequation (5)
Therefore cos α, cos β and cosγare the direction cosines which are written in the form of a, b and c which are direction ratios.
Relationship between direction ratios and cosines when a line passes through 2 points
One and only one line passes through two given points.
Consider a line PQ and the coordinates of points P and Q are given as P(x1, y1, z1) and Q(x2,y2,z2).
Let l,m and n are the direction cosines of the line PQ and which makes angleα,βand γ with the x,y and z-axis respectively.
If we draw perpendiculars from the point P and Q as shown in the figure (b), to XY-plane to meet at R and S. Also drawing a perpendicular from P to QS to meet at N.
Therefore cosα= (x2-x1)/(PQ)
Similarly cosβ =(y2-y1)/(PQ) and cosγ =(NQ)/(PQ) =(z2-z1)/(PQ)
Therefore the direction cosines of the line segment joining the points P(x1, y1, z1) and Q(x2, y2, z2) will be:-
PQ=(x2-x1)/(PQ) ,(y2-y1)/(PQ) +(z2-z1)/(PQ) equation(1)
Where PQ=√( x2-x1)2+( y2-y1)2+( z2-z1)2
Note:- The direction ratios of the line segment joining point P(x1, y1, z1) and
Q(x2, y2, z2) can be considered as – (x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2)
Problem:- If a line makes an angle of 90°, 135o, 45o with the positive direction ofx, y, z-axes, respectively, then find its direction cosines.
Answer:- Let l, m and be direction cosines of the line. Then l= cos90° = 0, m=cos 135o =-(1/√2) and n=cos 45o= (1/√2)
Therefore Direction cosines: [0,-(1/√2),(1/√2)]
Problem:- Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer:- Let the direction cosines of the line make an angle α with each of the coordinate axes.
l = cos α, m = cos α, n = cos α
l2+m2+n2 =1
=>cos2 α + cos2 β + cos2 γ) =1
=>3cos2 α =1
=>cos2 α = (1/3)
=>cosα = (±) (1/√3)
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,are(±)(1/√3),(±) (1/√3),and (±) (1/√3).
Problem:- If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Answer:- If a line has direction ratios of −18, 12, and −4, then its direction cosines are
= (-18)/ (√ ((18)2+ (12)2+ (-4)2),
= (12)/ ((√ ((18)2+ (12)2+ (-4)2),
= (-4)/(√ ((18)2+ (12)2+ (-4)2)
= i.e. (-18)/ (22), (12)/ (22) and (-4)/ (22)
= (-9/11), (6/11) and (-2)/ (11)
Thus, the direction cosines are (-9/11), (6/11) and (-2)/ (11).
Problem:- Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Answer:- The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).
It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2),are given by, (x2 − x1), (y2 − y1), and (z2 − z1).
The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., (−3, −5, and −3).
The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., (6, 10, and 6).
It can be seen that the direction ratios of BC are −2 times that of AB i.e., they areproportional.
Therefore, AB is parallel to BC. Since point B is common to AB and BC, points A, B, and C are collinear.
Problem:- Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−1, 1, 2) and (− 5, − 5, − 2)?
Answer:- The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).
The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., (−4), (−4), and(6).
Then,√ ((-4)2+ (-4)2+ (6)2) =√ (16+16+32)
=√ (68)
=2√ (17)
Therefore, the direction cosines of AB are
= (-4)/ (√ (-4)2 + (-4)2 + (6)2),
= (-4)/ (√ (-4)2 + (-4)2 + (6)2),
= (6)/ ((√ (-4)2 + (-4)2 + (6)2)
= (-4)/ (2√ (17), - (4)/2√ (17), (-6)/2√ (17)
= (-2)/√ (17), (-2)/√ (17), (3)/√ (17)
The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., (−4), (−6), and (−4). Therefore, the direction cosines of BC are
= (-4)/ (√ (-4)2 + (-4)2 + (6)2), (-4)/ (√ (-4)2 + (-4)2 + (6)2), (6)/ ((√ (-4)2 + (-4)2 + (6)2)
= (-4)/ (2√ (17), (-4)/(2√ (17)), (-6)/(2√ (17))
The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., (−8), (−10), and (2).
Therefore, the direction cosines of AC are
= (-8)/√ ((-8)2+ (10)2+ (2)2),
= (-5)/ ((-8)2+ (10)2+ (2)2),
= (2)/ ((-8)2+ (10)2+ (2)2)
=(-8)/ (2√ (42)), (-10)/ (2√ (42)), (2)/2√42))
Equation of a line in Space
A line is uniquely determined if:
- It passes through a given point and has direction, or
- It passes through two given points.
Line through a given point and parallel to a given vector:-
Derivation of Cartesian form from vector form
Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of the line be a, b, c.
Let the coordinates of the any point P be (x, y, z).
Consider r-> = xî +yĵ +zk̂, a=x1î+y1ĵ+z1k̂and b->= (a î +bĵ +ck̂) line vector L->=a î +bĵ +ck̂.
Equation of line: - r->= P-> + λL->.(equation(1))
Using equation (1) x1î+y1ĵ+z1k̂= x1î+y1ĵ+z1k̂+λ (aî +bĵ +ck̂).
Therefore, x=x1+λa => (x-x1)/(a)=λ
y=y1+λb => (y-y1)/(b) =λ
z=z1+λc => (z-z1)/(c)=λ
Therefore Equation of line in Cartesian form is given by :-
(x-x1)/ (a) = (y-y1)/ (b) = (z-z1)/(c)(By eliminatingλ).
Note: - If l, m and n are direction cosines of the line then the equation of line is given as:
=(x–x1)/l =(y-y1)/m= (z-z1)/n
Problem:-
Find the equation of the line which passes through the point (1, 2, and 3) and is parallel tothe vector 3î+2ĵ -2k̂.
Answer:-
It is given that the line passes through the point A (1, 2, and 3). Therefore, the position vector through A is a->=î+2ĵ +3k̂ and b-> =3î +2ĵ -2k̂.
It is known that the line which passes through point A and parallel to b-> is given by r->=a->+ λb->
=>r-> =î +2ĵ +3k̂+λ (3î +2ĵ -2k̂).
This is the required equation of the line.
Problem:-
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î-ĵ +4k̂ and is in the direction î +2ĵ -k̂?
Answer:-
It is given that the line passes through the point with position vector a->=2î -ĵ +4k̂(equation (1))
b-> = î +2ĵ -k̂(equation (2))
It is known that a line through a point with position vector a⃗and parallel tob ⃗ is givenby the equation, r-> = a-> + λb ->.
=>r-> =2î -ĵ +4k̂+ λ(î +2ĵ -k̂)
This is the required equation of the line in vector form.
r-> = x î +yĵ +zk̂=>(λ+2)î+ (2λ-1)ĵ+ (-λ+4)k̂
Eliminating λ, we obtain the Cartesian form equation as:
(x-2)/ (1) =(y+2)/ (2) = (z-4)/ (-1)
This is the required equation of the given line in Cartesian form.
Problem:-
Find the Cartesian equation of the line which passes through the point(−2, 4, −5) and parallel to the line given by (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)?
Answer:-
It is given that the line passes through the point (−2, 4, −5) and is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)
The direction ratios of the line,(x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6), are 3, 5, and 6.
The required line is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6).
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0
It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios, a, b, c, is given by (x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)
Therefore the equation of the required line is (x+2)/ (3k) =(y-4)/ (5k) = (z+5)/ (6k)
=>(x+3)/ (3) = k
=> (y-4)/ (5) = k
=> (z+8)/ (6) =k.
Equation of a line passing through two given points
Derivation of Cartesian form from vector form
Consider r-> = x î +yĵ +zk̂, a=x1 î +y1ĵ +z1k̂ and b ->= (x2 î +y2ĵ +z2k̂)
Substituting the values in equation r=a+λ(b->-a->) ,Therefore
x î +yĵ +zk̂=x1 î +y1ĵ +z1k̂+ λ((x2-x1)î+(y2-y1)ĵ +(z2-z1)k̂)
Equating the coefficients of î,ĵ,k̂we get,
x=x1+ λ(x2-x1); y=y1+λ(y2-y1);z=z1+λ(z2-z1)
On eliminating λ, we get,
Therefore equation of the line in Cartesian form will be :-
(x-x1)/(x2-x1) = (y-y1)/(y2-y1) =(z-z1)/(z2-z1)
Problem:-
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Answer:-
Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.
Since PQ passes through P (3, −2, −5), its position vector is given by,a->=3î-2ĵ-5k̂
The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is: -b->=0.î-0ĵ+11k̂=11k̂
The equation of PQ in vector form is given by,r-> = a-> + λb->,whereλ∈ R.
=>r->= (3î -2ĵ -5k̂) +11λk̂
The equation of PQ in Cartesian form is=(x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)
=(x-3)/ (0) =(y+2)/ (0) = (z+5)/ (11)
Problem:-
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Answer:-
The required line passes through the origin. Therefore, its position vector is given by,
a->=0 (equation (1))
The direction ratios of the line through origin and (5, −2, 3) are
(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3
The line is parallel to the vector given by the equation, b->= 5î-2ĵ +3k̂
The equation of the line in vector form through a point with position vectors a-> and parallel to
b->= r->=0 +λ (5î -2ĵ +3k̂)
r->=λ(5î -2ĵ +3k̂)
The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given by,
(x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)
Therefore, the equation of the required line in the Cartesian form is:-
(x-0)/ (5) =(y-0)/ (-2) = (z-0)/ (3)
=>(x/5) =(y/-2)/ (z/3)
Problem:-
The Cartesian equation of a line is (x-5)/ (3) =(y+4)/ (7) = (z-6)/ (2). Write its vector form.
Answer:-
The Cartesian equation of the line is (x-5)/ (3) =(y+4)/ (7) = (z-6)/ (2) (equation (1)).
The given line passes through the point (5, −4, 6). The position vector of this point is a->=5î-4ĵ +6k̂.
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector, b-> =3î+7ĵ +2k̂.
It is known that the line through position vector a⃗ and in the direction of the vector b-> is given by the equation, r->= a-> + λb->, whereλ∈ R.
=>r-> = (5î -4ĵ +6k̂) + λ (3î +7ĵ +2k̂)
This is the required equation of the given line in vector form.
Angle between two lines (in terms of Direction ratios)
Let L1 and L2 be two lines passing through the origin and with direction ratios (a1, b1, c1) and (a2, b2, c2) respectively.
Let P be a point on L1 and Q be a point on line L2.
Let ϴ = angle between OP and OQ.
Note:- In case the lines L1 and L2do not pass through the origin, we may takelines L’1 and L’2 which are parallel to L1 andL2 respectively and pass through the origin.
Angle between two lines (in terms of Direction cosines)
Considering 2 line vectorsP and Q and the direction cosines of P (l1, m1, n1) and of
Q(l2, m2, n2). Magnitude of P =r1 and magnitude of Q = r2.
Therefore P->=r1(l1î+m1ĵ +n1k̂) and Q-> =r2(l2î+m2ĵ +n2k̂)
P->.Q->= IPIIQIcosϴ
cosϴ= (P->.Q->)/(IPIIQI)
=r1 (l1î +m1ĵ +n1k̂). r2(l2î +m2ĵ +n2k̂)/(r1√l1)2+(m1)2+(n1)2) (r2√((l2)2+(m2)2+(n2)2)
cosϴ =Il1l2+m1m2+n1n2I(using for any line (l) 2+ (m) 2+ (n) 2 =1)
Perpendicular and parallel Test between 2 line vectors
Two lines with direction ratios a1, b1, c1and a2,b2,c2.
When Perpendicularϴ=900:-
When direction ratios are given then a1a2+b1b2+c1c2=0,and when direction cosines are given then l1l2+m1m2+n1n2=0.
When Parallelϴ=0:-
(a1/a2)=(b1/b2)=(c1/c2) or (l1/l2)/(m1/m2)=(n1/n2)=0
Angle between lines when line equations are given (Vector)
Consider the equation of a line passing through a1⃗ (point vector)and parallel to b1⃗ such that r1⃗ = a1⃗ + λ b1⃗ andanother line passing through a2⃗ (point vector)and parallel to b2⃗ such that r2⃗= a2⃗ +μ b2⃗.
From the figure we can say that b1⃗ and b2⃗ are line vectors.
Angles between lines r1⃗ and r2⃗ are same as angle between line vectors b1⃗ and b2⃗
Problem:-
Find the angle between the following pairs of lines:
r ⃗=2î-5ĵ+k̂+λ (3î-2ĵ +6k̂) and r ⃗=7î-6k̂+μ (î+2ĵ +2k̂)?
Answer:-
=3+4+12
=19
=> cosϴ = (19)/ (7x3)
=>ϴ = cos-1(19/21)
Problem:-
Find the angle between the following pairs of lines:
(x-2)/ (2) =(y-1)/ (5)/ (z+3)/ (-3) and (x+2)/ (-1) =(y-4)/ (8) = (z-5)/ (4).
Answer:-
(x-2)/ (2) =(y-1)/ (5)/ (z+3)/ (-3) and (x+2)/ (-1) =(y-4)/ (8) = (z-5)/ (4)
Therefore b1⃗=2î+5ĵ -3k̂andb2⃗=-î +8ĵ +4k̂
Ib1⃗I=√ ((2)2+ (5)2+ (-3)2) =√38
Ib2⃗I= √ ((-1)2+ (8)2+ (4)2) =√81 =9
b1⃗.b2⃗ = (2î+5ĵ -3k̂). (-î+8ĵ +4k̂)
=2(-1) +5x8+ (-3).4
=-2+40-12
=26
=>cosϴ= (26)/ (9√38)
=>ϴ =cos-1(26)/ (9√38))
Problem:-
Find the values of p so the line (1-x)/(3)=(7y-14)/(2p)=(z-3)/(2)and (7-7x)/(3p)=(y-5)/(1)=(6-z)/(-5)are atright angles.
Answer:-
The given equations can be written in the standard form as:
(1-x)/(3)=(7y-14)/(2p)=(z-3)/(2)and (7-7x)/(3p)=(y-5)/(1)=(6-z)/(-5)
The direction ratios of the lines are −3, (2p)/ (7), 2 and (-3p)/ (7), 1,-5 respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other,
If a1a2 + b1 b2 + c1c2 = 0.
Therefore ((-3)x(-3p)/(7))+((2p)/(1))+((2).(-5))=0
=> (9p)/ (7) + (2p)/ (7) =10
=>11p=70
=>p=(70/11).
Thus, the value of p is (70/11).
Shortest Distance between two lines
If two lines intersect at a point, then the shortest distance between is 0.
If two lines are parallel, then the shortest distance between will be given by the length of the perpendicular drawn from a point on one line form another line.
Skew Lines
Skew lines are the lines which are neither intersecting nor parallel.
In 2-D lines are either parallel or intersecting. There are no skew lines in 2-D.
But in case of 3-D there are lines which are neither intersecting nor parallel to each other.
For skew lines, the line of the shortest distance will be perpendicular to both the lines.
From the figure we can see when we consider one line in xy plane and one in xz plane.We can see that these lines will never meet.
Shortest Distance between Two Skew Lines (Vector form)
Shortest Distance between Two Skew lines (Cartesian form)
Distance between parallel lines
Problem:-
Answer:-
Plane
A plane is determined uniquely if any one of the following is known:
(i) The normal to the plane and its distance from the origin is given, i.e., equation ofa plane in normal form.
(ii) It passes through a point and is perpendicular to a given direction.
(iii) It passes through three given non collinear points.
Equation of a plane in normal form: Vector
Consider a plane whose distance from the origin is‘d’.
Equation of a plane in normal form: Cartesian form
Let P(x, y, z) be any point on the plane.
Note:- Equation (3) shows that if r⃗ = xî+ yĵ+ zk̂ =d is the vector equation of the plane, then ax+by+cz=d is the Cartesian equation of the plane, where a,b and c are the direction ratios of the normal to the plane.
Problem:
Determine the direction cosines of the normal to the plane and the distance from origin:
x+y+z=1 , 2.2x+3y-z=5
Answer:-
x + y + z = 1 ... (1)
The direction ratios of normal are 1, 1, and 1.
Therefore √ ((1)2+ (1)2+ (1)2)
Dividing both sides of equation (1) by √3, we obtain
(1/√3) x+ (1/√3) y+ (1/√3) z= (1/√3) … (2)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal are (1/√3),(1/√3) and (1/√3) the distance ofnormal from the origin is (1/√3) units.
2x + 3y − z = 5 ... (1)
The direction ratios of normal are 2, 3, and −1.
Therefore, √ (2)2+ (3)2+ (-1)2 =√ (14)
Dividing both sides of equation (1) by √ (14), we obtain
(2/√ (14)) x+ (3/√ (14)) y– (1/√ (14)) z= (5/√ (14))
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are (2/√ (14)),(3/√ (14)) and (-1/√ (14)) and the distance of normal from the origin is (5/√ (14)) units.
5y + 8 = 0
=> 0 x − 5y + 0z = 8................. (1)
The direction ratios of normal are (0, −5, and 0).
Therefore √ (0+ (-5)2+0) =5.
Dividing both sides of equation (1) by 5, we obtain
-y= (8/5)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are (0, −1, and 0) and thedistance of normal from the origin is (8/5) units.
Problem:-
Find the distance of the plane 2x-3y+4z-6=0 from the origin.
Answer:-
Since the direction ratios of the normal to the plane are 2,-3, 4; the direction cosines of it are
(2)/ (√ (2)2+ (-3)2+ (4)2),
(-3)/(√ (2)2+ (-3)2+ (4)2),
(4)/(√ (2)2+ (-3)2+ (4)2),
i.e. (2)/√ (29),
(-3)/√ (29),
(4)/√ (29)
Hence, dividing the equation (2x-3y+4z-6)=0 i.e. 2x-3y+4z=6 throughout by √ (29), we get
((2)/√ (29))x + ((-3)/√ (29))y + ((4)/√ (29)) z = (6)/√ (29)
This is the form lx+my+nz=d, where d is the distance of the plane from the origin.So, the distance of the plane from the origin is (6)/√ (29)
Problem:-
Find the Cartesian equation of the following planes:
Answer:-
It is given that equation of the plane is
r-> . (î+ ĵ-2k̂)=2 (1)
For any arbitrary point P(x, y, z) on the plane, position vector r⃗is given by,
r-> = xî+ yĵ+ zk̂
Substituting the value of r⃗ in equation (1), we obtain
(xî+ yĵ+ zk̂)(î+ ĵ-2k̂) =2
=>x+y-z=2.
This is the Cartesian equation of the plane.
r->. (2î+3 ĵ-4k̂)=1 (1)
For any arbitrary point P (x, y, z) on the plane, position vector r-> is given by,
r-> = xî+ yĵ+ zk̂
Substituting the value of r-> in equation (1), we obtain
(xî+ yĵ+ zk̂)(2î+3 ĵ-4k̂) = 1
=>2x+3y-4z=2.
This is the Cartesian equation of the plane.
It is given that equation of the plane is
r->. [(s-2t)î +(3-t)ĵ +(2s+t)k̂] (1)
For any arbitrary point P (x, y, z) on the plane, position vector r-> is given by,
r-> = xî+ yĵ+ zk̂
Substituting the value of r⃗ in equation (1), we obtain
(xî+ yĵ+ zk̂) [(s-2t)î +(3-t)ĵ +(2s+t)k̂]
⇒ (s − 2t) x + (3 − t) y + (2s + t) z = 15
This is the Cartesian equation of the plane.
Problem:-
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x-3y+4z-6=0?
Answer:-
Let the coordinates of the foot of the perpendicular P from the origin to the plane is (x1, y1, z1).
Then, the direction ratios of the line OP are (x1, y1, z1).
Writing the equation of the plane in the normal form, we have
(2/√29) x-(3/√29) y+ (4/√29) z= (6/√29) where (2/√29), (3/√29) y, (4/√29) are the direction cosines of the OP.
Since d.c.’s and direction ratios of a line are proportional, we have
x1/ (2/√29)) = (y1/ -3/√29)) = (z1/ (4/√29)) = k
i.e. x1=(2k)/(√29), y1 =((-3k)/(√29), z1 =(4k)/(√29)
Substituting these in the equation of the plane, we get k = (6/√29)
Hence, the foot of the perpendicular is (12/29), (- 18/29), (24/29).
Equation of a plane perpendicular to a given vector and passing through a given point
There can be many planes that are perpendicular to the given vector, but through a given point P(x1,y1,z1) only one such plane exists.
Let a plane pass through a point A with position vector a-> and perpendicular to vector N->.
Let r-> be the position vector of any point P(x, y,z) in the plane.
Then the point P lies in the plane if and only if
AP-> is perpendicular to N->.i.e. AP->. N-> = 0.
But AP-> = r->. a->. Therefore (r->- a->). N->=0. ..(1)
This is the vector equation of the plane.
Cartesian form
Let the given point A (x1, y1, z1), P(x, y, z) and direction ratios of N⃗ are A,B and C.
Then a-> = x1î+y1ĵ+z1 k̂, r->=xî+yĵ+zk̂ and N->=Aî+Bĵ+C k̂
Now (r-> - a->). N-> =0
So [(x-x1) î +(y-y1) ĵ+ (z-z1) k̂].(Aî+Bĵ+C k̂)=0
e.A(x-x1)+B(y-y1)+C(z-z1) = 0
Problem:-
Find the vector and Cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios (2, 3, – 1).
Answer:-
We have the position vector of point (5, 2,-4) as a-> = 5î+2ĵ-4k̂ and the normal vector N-> perpendicular to the plane as N-> =2î+3ĵ-k̂.
Therefore, the vector equation of the plane is given by ( r->- a->). N-> =0
Or [r-> - (5î+2ĵ-4k̂)]. (2î+3ĵ-k̂)=0… (1)
Transforming (1) into Cartesian form, we have
[(x-5) î +(y-2) ĵ + (z+4) k̂]. (2î+3ĵ-k̂)= 0
Or 2(x-5) +3(y-2)-1(z+4) =0
i.e. 2x+3y-z=20, which is the Cartesian equation of the plane.
Problem:-
Find the vector and Cartesian equation of the planes
(a) That passes through the point (1, 0, −2) and the normal to the plane is î+ ĵ-k̂.
(b) That passes through the point (1, 4, and 6) and the normal vector to the plane is î-2ĵ+ k̂.
Answer:-
The position vector of point (1, 0,-2) is a->= î-2k̂.
The normal vector N-> perpendicular to the plane is N->= î+ ĵ-k̂
The vector equation of the plane is given by, (r-> - a->). N-> =0
=>[r-> (î-2k̂)]. (î+ ĵ-k̂) =0 … (1)
r-> is the position vector of any point P(x,y,z) in the plane.
Therefore r-> =xî+yĵ+zk̂
Therefore equation (1) becomes
[(xî+yĵ+zk̂)-(î-2k̂)]. (î+ ĵ-k̂) =0
=> [(x-1) î + yĵ+ (z+2) k̂]. (î+ ĵ-k̂)=0
=>(x-1) +y-(z+2) =0
=>x+y-z-3=0
=>x+y-z=3.
This is the Cartesian equation of the required plane.
(a) The position vector of the point (1, 4, 6) is a->= î+4ĵ+6k̂.
The normal vector N-> perpendicular to the plane is N-> = î-2ĵ+k̂
The vector equation of the plane is given by, (r-> - a->). N-> =0
=>[r->- (î+4ĵ+6k̂)]. (î- 2ĵ +k̂) =0 … (1)
r-> is the position vector of any point P(x,y,z) in the plane.
Therefore r-> =xî+yĵ+zk̂
Therefore equation (1) becomes
[(xî+yĵ+zk̂)-(î+4ĵ+6k̂)]. (î- 2ĵ +k̂) =0
=> [(x-1)î + (y-4) ĵ+ (z-6)k̂].(î- 2 ĵ+k̂)=0
=>(x-1) -2(y-4) + (z-6) =0
=>x-2y+z+1=0.
This is the Cartesian equation of the plane.
Equation of a plane passing through 3 non collinear points
Let R, S and T be three non collinear points on the plane with position vectors a->, b-> and c-> respectively.
This is the equation of plane in vector form passing through three noncollinear points.
These planes resemble the pages of the book where the line containing the points R, S and T are present in the binding of the book.
Note:-Three points has to be non-collinear else there will be many planes that will contain them.
Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T respectively.
Let (x,y,z) be the coordinates of the point P on the plane with position vector r->.
Then
Substituting these values in equation(1) of the vector form and writing in the determinant form:
The above equation is the equation of the plane in Cartesian form passing through three non collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3).
Problem:-
Find the vector equations of the plane passing through 3 points (1,1,-1),(6,4,5) ,(-4,-2,3)?
Answer:-
The given points are (1,1,-1),(6,4,5) ,(-4,-2,3).
=(12-10)-(18-20)-(-12+16)
=2+2-4=0.
Since 3 points are collinear points,there will be infinite number of planes passing through the given points.
Problem:-
Find the vector equations of the plane passing through 3 points (1,1,0),(1,2,1) ,(2,2,-1)?
Answer:-
The given points are (1,1,0),(1,2,1) ,(2,2,-1).
=(-2-2)-(2+2)=-8≠0.
Therefore, a plane will pass through the points A,B and C.
It is known that the equation of the plane through the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
=>(-2) (x-1)-3(y-1) +3z=0
=>-2x-3y+3z+2+3=0
=>2x-3y+3z=-5
=>2x+3y-3z=5.
This is the Cartesian equation of the required plane.
Intercept form of the equation of a plane
Let the equation of the plane be Ax+By+Cz+ D=0 (where D≠0). (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively.
The plane meets x, y and z-axes at(a,0,0), (0,b,0) ,(0,0,c) respectively.
Therefore putting (0,0,c) we get, A(0)+B(0)+C(c)+D =0
=>C= - (D/c).
Similarly by putting (0,b,0) we get, A(0)+B(b)+C(0)+D=0
=> B=(-D/b)
And by putting (a,0,0) we get, A(a)+B(0)+C(0)+D=0
=>A=(-D/a)
Substituting the values of A,B and C in (1) we get,
(-D/a)x +(-D/b)y+(-D/c)z +D =0
=> (-D) ((x/a) + (y/b) + (z/c) – 1) =0.
(x/a)+(y/b) + (z/c) =1. Hence Proved.
Problem:-
Find the intercepts cut off by the plane 2x+y-z=5.
Answer:-
Considering the equation 2x+y-z=5. Dividing both sides of the equation by 5, we obtain:
(2/5) x+(y/5)-(z/5) =1
=>(x)/ (5/2) + (y/5) + (z/-5) =1 (2)
It is known that the equation of a plane in intercept form is, (x/a) +(y/b) +(c/z) =1, wherea, b, c are the intercepts cut off by the plane at x, y, z axes respectively.
Therefore a=(5/2), b =5, and c=-5.
Thus the intercepts cut off by the plane are (5/2),(5) and (-5).
Problem:-
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Answer:-
The equation of the plane ZOX is y = 0.
Any plane parallel to it is of the form, y = a.
Since the y-intercept of the plane is 3,
a = 3
Thus, the equation of the required plane is y = 3.
Plane passing through intersection of 2 planes:Vector
Let π1and π2 are two planes with equations r->. n̂1 = d1 and r-> .n̂2=d2
The position vector of any point on the line of intersection satisfies both the equations.
If t-> is the position vector of a point on the line, then
t->. n̂1 = d1 and t->.n̂2=d2
Therefore, for all real values of λ, we have
t->.( n̂1 + λ n̂2)=d1+ λd2
Since t-> is arbitrary, it satisfies for any point on the line.
Hence the equation (r->. (n1⃗+ λn2⃗))=d1+ λd2 represents a plane π3 which is such that if any vector r-> satisfies both the equations π1 and π2.
Then it will also satisfy the equation π3e., any plane passing through the intersection of the planes.
Therefore r->. n̂1 = d1 and r->.n̂2=d2 has the equation,
r⃗.( n̂1 + λ n̂2)=d1+ λd2 (1)
Cartesian form:-
Let n1⃗= (A1î+ B1ĵ+ C1k̂) and n2⃗= (A2î+ B2ĵ+ C2k̂)
r->= xî+ y ĵ+ zk̂
Then equation(1) becomes,
x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2
or (A1 x + B1 y + C1z – d1) + λ(A2 x + B2 y + C2 z – d2) = 0 ... (2)
This is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.
Problem:-
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and
x + y + z – 2 = 0 and the point (2, 2, 1).
Answer:-
The equation of any plane through the intersection of the planes,
3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is
(3x − y + 2z – 4)+ λ (x + y + z – 2) = 0 where λ ∈ R. (1)
The plane passes through the point (2, 2, 1).
Therefore, this point will satisfy equation (1).
Therefore (3x2-2+2x1-4) + λ(2+2+1-2)=0.
=>2+3λ=0.
=> λ= (-2/3)
Substituting λ= (-2/3) in equation (1), we obtain
(3x − y + 2z – 4)(-2/3)(x + y + z – 2)=0
=>3(3x − y + 2z – 4)-2(x + y + z – 2) =0
=> (9x-3y+6z-12)-2(x+y+z-2)=0
=>7x-5y+4z-8=0.
This is the required equation of the plane.
Problem:-
Find the vector equation of the plane passing through the intersection of the planes r⃗. (î+ĵ+k̂)=6 and r⃗.(2 î+3ĵ+4k̂)=-5, and the point(1, 1, 1).
Answer:-
n1⃗=(î+ ĵ+ k̂) and n2⃗= (2 î+3ĵ+4k̂) andd1=6, d2=-5
Using the relation (r->. (n1⃗+ λn2⃗))=d1+ λd2, we get,
r->. [(î+ĵ+k̂) +λ (2 î+3ĵ+4k̂)]=6-5λ
Or r->. (1+2 λ) x+ (1+3 λ) y + (1+4 λ) z=6-5 λ (1)
Where, λ is some real number.
Taking r->= (xî+ yĵ+ zk̂), we get
(xî+ y ĵ+ zk̂). [(1+2 λ) î+ (1+3 λ) ĵ+ (1+4 λ) k̂]=6-5 λ
Or(1+2 λ)x+(1+3 λ)y + (1+4 λ) z=6-5 λ
Or(x+y+z-6) + λ (2x+3y+4z+5) =0 (2)
Given that the plane passes through the point (1, 1, 1), it must satisfy (2), i.e.
(1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
Or λ = (3/14)
Putting the values of λ in (1), we get
r-> [((1+ (3/7)) î+ (1+ (9/14)) ĵ+ (1+ (6/7)) k̂)] = (6)-(15/14)
r-> ((10/7) î + (23/14) ĵ+ (13/14) k̂) = (69/14)
Or r->. (20 î+23 ĵ+26 k̂)=69 which is the required vector equation of the plane.
Problem:-
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and
2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Answer:-
The equation of the plane through the intersection of the planes, x + y + z = 1 and
2x + 3y + 4z = 5, is
(x + y + z -1) +λ (2x + 3y + 4z- 5)=0.
=> (2λ+1)x+(3λ+1)y+(4λ+1)z-(5λ+1)=0 (1)
The direction ratios, a1, b1, c1, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).
The plane in equation (1) is perpendicular to x – y + z=0.
Its direction ratios a2, b2, c2, are 1, −1, and 1.
Since the planes are perpendicular,
a1a2+b1b2+c1c2=0
=> (2λ + 1)-(3λ + 1) + (4λ + 1) =0
=>3λ + 1=0
=>λ= (-1/3)
Substituting λ= (-1/3) in equation (1), we obtain
This is the required equation of the plane.
(1/3) x-(1/3) z+92/3) =0
=>x-z+2=0.
Coplanarity of Two lines:Vector
Consider 2 lines such that: -First line be represented as: - r->=a1 ⃗+λ b1⃗, it passes through a point A, having position vector a1⃗as and it is parallel to b1⃗.
Second line r-> = a2⃗+µ b2⃗, it passes through a point B,having position vector a2⃗and is parallel to b2⃗.
Thus AB ⃗= (a2⃗- a1⃗). The given lines are coplanar iff AB ⃗is perpendicular to (b1⃗x b2⃗).
Cartesian form:-
Let the coordinates of A =(x1, y1, z1) and of B= (x2, y2, z2).
Direction ratios of b1⃗= (a1, b1, c1) and of b2⃗ =(a2,b2,c2).
AB ⃗=(x2-x1) î +(y2-y1) ĵ +(z2-z1) k̂
b1⃗= (a1 î+b1 ĵ+c1 k̂) and b2⃗= (a2 î,b2 ĵ,c2 k̂).
The lines will be coplanar iff AB ⃗. (b1⃗x b2⃗) =0.
Therefore in the Cartesian form ,it can be given as:-
Angle between Two Planes: Vector
The angle between 2 planes is defined as the angle which they make with their normal.
Let θ = angle between the two planes (Fig (a)) then (1800- θ) = angle between the two planes (Fig (b)).
r->.n1⃗=d1 and r->.n2⃗=d Where n1⃗ and n2⃗ = normal to the planes, θ =Angle between the planes.
θ= angle between the normal to the planes drawn from some common point.
Therefore,
Note: -The planes are perpendicular to each other if n1⃗.n2⃗=0 and parallel ifn1⃗ is parallel to n2⃗.
Cartesian Form:-
Let θ =angle between the planes.
A1x+B1y+C1z+D1=0 and A2x+B2y+C2z+D2=0 where A1,B1,C1 and A2,B2,C2
Therefore ,
Note: -
If the planes are at right angles, then θ=900 and so cosθ =0. Therefore ,
cosθ= (A1A2 + B1B2+ C1C2).
If the planes are parallel, then (A1/A2) = (B1/B2) = (C1/C2).
Problem:-
Find the angle between the planes whose vector equations are: - r->. (2 î+2ĵ-3 k̂)=5 and
r->. (3 î - 3ĵ+5 k̂)=3.
Answer:-
The equations of the given planes are r->. (2 î+2ĵ-3 k̂)=5 and r->. (3 î - 3ĵ+5 k̂)=3.
It is known that if n1⃗ and n2⃗ are r->.n1⃗=d1 and r⃗.n2⃗=d2normal to the planes, then the angle between them, θ, is given by,
Here, n1⃗= (2î+2ĵ-3k̂) and n2⃗= (3î - 3ĵ+5k̂)
Therefore, n1⃗.n2⃗= (2î+2ĵ-3k̂). (3î - 3ĵ+5k̂) =2.3+2. (-3)+ (-3).5=-15
Substituting the value
cosθ = (15)/ (√ (731)
Therefore θ =cos-1(15)/ (√ (731)
Problem:Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.
Answer: - Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get A1 = 3, B1 = – 6, C1 = 2 A2 = 2, B2 = 2, C2 = – 2
= (5)/ (7√3)
= (5√3)/ (21)
Therefore, θ = cos-1 ((5√3)/ (21))
Distance of a Point from a Plane: vector
Consider a point P with position vector a ⃗ and a plane π1 whose equation is r->. nˆ=d.
Consider a plane π2 through P parallel to the plane π1.
The unit vector normal to π2 = nˆ.
Therefore it equation will be (r-> - a->). nˆ=0. i.e. r->. nˆ=a ⃗. nˆ
Therefore, the distance PQ from the plane π1 is (Fig. (a)), i.e.
ON –ON’= | d - (a ⃗. nˆ) |
This gives the length of the perpendicular from a point to the given plane.
Note:-
If the equation of the plane π2 is in the form r->. N =d, where N= normal to the plane, then the perpendicular distance is given as | ((a ⃗.N)-d)/N|.
The length of the perpendicular from the origin O to the plane
r->. N=d =| d |/| N| (since a ⃗=0).
Cartesian form:-
Let P(x1, y1, z1) is the given point with position vector a ⃗.
The Cartesian equation is given by Ax+By+Cz+D=0.
Then a ⃗ = x1î +y1ĵ +z1k̂ and N=A î +B1 ĵ +C k̂
Therefore by using the result | ((a ⃗.N)-d)/N|, the perpendicular from P to the plane is
Problem: -Find the distance of a point (2, 5, – 3) from the plane r->. (6î -3ĵ +2k̂)=4?
Answer: - Here, a ⃗=2î +5ĵ -3k̂, N =6î -3ĵ +2k̂ and d=4.
Therefore, the distance of the point (2, 5, – 3) from the given plane is
= 13/7
Angle between a Line and a Plane
The angle between a line and a plane is defined as the complement of the angle between the line and normal to the plane.
Vector form:-
Equation of the line r->=a ⃗+λb-> and the equation of the plane r->.n->=d.
Then the angle between the line and the normal to the plane θ is given as,
The angle between line and the plane is φ= (900 – θ).
Using sin (900 – θ) =cosθ.
Problem:-
Find the angle between the line (x+1)/ (2) =(y/3) = (z-3)/ (6) and the plane 10x+2y-11z=3?
Answer:-
Let θ =angle between the line and the normal to the plane.
In Vector form: -r->= (-î+3k̂) +λ (2î+3ĵ + 6k̂)
And r->. (10 î +2 ĵ -11 k̂)=3
Here b->=2 î+3 ĵ+6 k̂ and nˆ=10 î+2ĵ-11k̂
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