Table of contents  
Basic Concepts of 3D Geometry  
Direction Cosines and Ratios of a Line  
PLANE  
STRAIGHT LINE 
The position vectors of the points P and Q are given by
Now we have
(for external division take –ve sign)
To determine the coordinates of a point R which divides the joining of two points P and internally in the ratio m1 : m2. Let OX, OY, OZ be a set of rectangular axes.
The position vectors of the two given points are given by
Also if the coordinates of the point R are (x, y, z), then
Now the point R divides the join of P and Q in the ratio m_{1} : m_{2}, so that
or [Using (1), (2) and (3)]
Remark : The middle point of the segment PQ is obtained by putting m_{1} = m_{2}. Hence the coordinates of the middle point of PQ are
Centroid of a Triangle :
Let ABC be a triangle. Let the coordinates of the vertices A, B and C be and respectively. Let AD be a median of the ΔABC. Thus D is the mid point of BC.
Now if G is the centroid of ΔABC, then G divides AD in the ratio 2 : 1. Let the coordinates of G be (x, y, z).
Centroid Of A Tetrahedron :
Let ABCD be a tetrahedron, the coordinates of whose vertices are = 1, 2, 3, 4.
Let be the centroid of the face ABC of the tetrahedron. Then the coordinates of are
The fourth vertex D of the tetrahedron does not lie in the plane of ΔABC. We know from statics that the centroid of the tetrahedron divides the line DG_{1} in the ratio 3 : 1. Let G be the centroid of the tetrahedron and if (x, y, z) are its coordinates, then
Ex.1 P is a variable point and the coordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively. Find the locus of P if 3PA = 2PB.
Sol. Let the coordinates of P be (x, y, z).
Now it is given that 3PA = 2PB i.e., 9PA^{2} = 4PB^{2}. ....(3)
Putting the values of PA and PB from (1) and (2) in (3), we get
9{(x + 2)^{2} + (y – 2)^{2} + (z – 3)^{2}} = 4 {(x – 13)^{2} + (y + 3)^{2} + (z – 13)^{2}}
or 9 {x^{2} + y^{2} + z^{2} + 4x – 4y – 6z + 17} = 4{x^{2} + y^{2} + z^{2} – 26x + 6y – 26z + 347}
or 5x^{2} + 5y^{2} + 5z^{2} + 140x – 60 y + 50 z – 1235 = 0
or x^{2} + y^{2} + z^{2} + 28x – 12y + 10z – 247 = 0
This is the required locus of P.
Ex.2 Find the ratio in which the xyplane divides the join of (–3, 4, –8) and (5, –6, 4). Also find the point of intersection of the line with the plane.
Sol. Let the xyplane (i.e., z = 0 plane) divide the line joining the points (–3,4, –8) and (5, –6, 4) in the ratio , in the point R. Therefore, the coordinates of the point R are
....(1)
But on xyplane, the z coordinate of R is zero
(4μ  8) / (μ + 1) = 0, or μ = 2. Hence μ : 1 = 2 : 1. Thus the required ratio is 2 : 1.
Again putting the coordinates of the point R become (7/3, –8/3, 0).
Ex.3 ABCD is a square of side length ‘a’. Its side AB slides between x and yaxes in first quadrant. Find the locus of the foot of perpendicular dropped from the point E on the diagonal AC, where E is the midpoint of the side AD.
Sol. Let vertex A slides on yaxis and vertex B slides on xaxis coordinates of the point A are (0, a sin θ) and that of C are (a cos θ + a sin θ, a cos θ)
...(2)
Form (1) and (2),
Let AB be a given line.Draw a line OP parallel to the line AB and passing through the origin O. Measure angles a, b, g, then cos a, cos β, cos g are the d.c.'s of the line AB. It can be easily seen that l, m, n, are the direction cosines of a line if and only if is a unit vector in the direction of that line.
Clearly, OP'(i.e. the line through O and parallel to BA) makes angle 180^{o}  α, 180^{o } β, 180^{o}  γ, with OX, OY and OZ respectively.
Hence d.c.'s of the line BA are cos (180^{o}  α), cos (180^{o}  β), cos (180^{o}  γ) i.e., are cos α, cos β ,  cos γ.
If the length of a line OP through the origin O be r, then the coordinates of P are (ℓr, mr, nr) where ℓ, m, n are the d c.'s of OP.
If ℓ, m, n are direction cosines of any line AB, then they will satisfy ℓ ^{2} + m^{2} + n^{2} = 1.
If the direction cosines ℓ, m, n of a given line be proportional to any three numbers a, b, c respectively, then the numbers a, b, c are called direction ratios (briefly written as d.r.'s of the given line.
Let a, b, c be the direction ratios of a line whose d.c.’s are ℓ, m, n. From the definition of d.r.'s. we have ℓ/a = m/b = n/c = k (say). Then ℓ = ka, m = kb, n = kc. But ℓ^{2} + m^{2} + n^{2} = 1.
Taking the positive value of k, we get
Again taking the negative value of k, we get
Remark. Direction cosines of a line are unique. But the direction ratios of a line are by no means unique. If a, b, c are direction ratios of a line, then ka, kb, kc are also direction ratios of that line where k is any nonzero real number. Moreover if a, b, c are direction ratios of a line, then a ˆi + b ˆj + c kˆ is a vector parallel to that line.
Ex.4 Find the direction cosines ℓ + m + n of the two lines which are connected by the relation ℓ + m + n = 0 and mn  2nℓ 2ℓm = 0.
Sol.
The given relations are ℓ+ m + n = 0 or ℓ = m  n ....(1)
and mn  2nℓ  2ℓm = 0 ...(2)
Putting the value of ℓ from (1) in the relation (2), we get mn  2n (m n)  2(m  n) m = 0 or 2m_{2} + 5mn + 2n_{2} = 0 or (2m + n) (m + 2n) = 0.
From (1), we have ...(3)
Now when ,
(3) given
∴
∴ The d.c.’s of one line are
i.e.
∴ The d.c.’s of the one line are
Again when
i.e.
∴ The d.c.’s of the other line are
To find the projection of the line joining two points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) on the another line whose d.c.’s are ℓ, m, n
Let O be the origin. Then
Now the unit vector along the line whose d.c.’s are ℓ,m,n
∴ projection of PQ on the line whose d.c.’s are ℓ, m, n
The angle q between these two lines is given by
If l_{1}, m_{1}, n_{1} and l_{2} , m_{2}, n_{2} are two sets of real numbers, then
(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) (l_{2}^{2 }+ m_{2}^{2} + n_{2}^{2})  (l_{1}l_{2} + m_{1}m_{2 }+ n_{1}n_{2})^{2} = (m_{1}n_{2} – m_{2}n_{1})^{2 }+ (n_{1}l_{2 } n_{2}l_{1})^{2} + (l_{1}m_{2}  l_{2}m_{1})^{2}
Now, we have
sin^{2} θ = 1  cos^{2} θ = 1  (l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2})^{2} = (l_{1}^{2} + m_{1}^{2 }+ n_{1}^{2}) (l_{2}^{2} + m_{2}^{2} + n_{2}^{2})  (l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2})^{2}
= (m_{1}n_{2} – m_{2}n_{1})^{2} + (n_{1}l_{2}  n_{2}l_{1})^{2} + (l_{1}m_{2}  l_{2}m_{1})^{2 } =
Condition for perpendicularity
Condition for parallelism
Ex.5 Show that the lines whose d.c.’s are given by l + m + n = 0 and 2mn + 3ln  5lm = 0 are at right angles.
Sol. From the first relation, we have l = m  n. ..(i)
Putting this value of l in the second relation, we have
2mn + 3 (–m –n) n – 5 (–m –n) m = 0 or 5m^{2} + 4mn – 3n^{2} = 0 or 5(m/n)^{2} + 4(m/n) – 3 = 0 ....(2)
Let l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} be the d,c's of the two lines. Then the roots of (2) are m_{1}/n_{1} and m_{2}/n_{2}.
product of the roots ...(3)
Again from (1), n = – l  m and putting this value of n in the second given relation, we have
2m (–l  m) + 3l(l  m)  5lm = 0
or 3(l/m)^{2} + 10 (l/m) + 2 = 0.
From (3) and (4) we have
l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = (2 + 3  5) k = 0 . k = 0. ⇒ The lines are at right angles.
Remarks :
(a) Any three numbers a, b, c proportional to the directi on cosi nes are called the direction ratios
same sign either +ve or –ve should be taken throughout.
(b) If θ is the angle between the two lines whose d.c's are l_{ 1} , m_{1}, n_{1} and l_{ 2 }, m_{2}, n_{2}
cos θ = l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}
Hence if lines are perpendicular then l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0.
if lines are parallel then
Note that if three lines are coplanar then
(c) Projection of the join of two points on a line with d.c’s l, m, n are l (x_{2} – x_{1}) + m(y_{2} – y_{1}) + n(z_{2} – z_{1})
(d) If l_{1}, m_{1}, n_{1 }and l_{2}, m_{2}, n_{2} are the d.c.s of two concurrent lines, show that the d.c.s of two lines bisecting the angles between them are proportional to l_{1} ± l_{2}, m_{1} ± m_{2}, n_{1 }± n_{2}.
Show that the area of a triangle whose vertices are the origin and the points A(x_{1}, y_{1}, z_{1}) and B( x_{2}, y_{2}, z_{2}) is
The direction ratios of OA are x_{1}, y_{1}, z_{1} and those of OB are x_{2}, y_{2}, z_{2}.
Also OA =
and OB =
∴ the d.c.' s of OA are
and the d.c.’s of OB are
Hence if q is the angle between the line OA and OB, then
Hence the area of ΔOAB
Ex.6 Find the area of the triangle whose vertices are A(1, 2, 3), B(2, –1, 1)and C(1, 2, –4).
Sol. Let Δx, Δy, Δz be the areas of the projecti ons of the area Δ of triangle ABC on the yz, zx and xyplanes respectively. We have
∴ the required area
Ex.7 A plane is passing through a point P(a, –2a, 2a), a ≠ 0, at right angle to OP, where O is the origin to meet the axes in A, B and C. Find the area of the triangle ABC.
Sol.
Equation of plane passing through P(a, –2a, 2a) is A(x – a) + B(y + 2a) + C(z – 2a) = 0.
∵ the direction cosines of the normal OP to the plane ABC are proportional to a – 0, –2a – 0, 2a – 0 i.e. a, –2a, 2a.
⇒ equation of plane ABC is a(x – a) – 2a(y + 2a) + 2a(z – 2a) = 0 or ax – 2ay + 2az = 9a2 ....(1)
Now projection of area of triangle ABC on ZX, XY and YZ
planes are the triangles AOC, AOB and BOC respectively.
∴ (Area ΔABC)^{2} = (Area ΔAOC)^{2} + (Area ΔAOB)^{2} + (Area ΔBOC)^{2}
(ii) Equation of a plane passing through (x_{1}, y_{1}, z_{1}) is a(x – x_{1}) + b (y – y_{1}) + c(z – z_{1}) = 0
where a, b, c are the direction ratios of the normal to the plane.
(iii) Equation of a plane if its intercepts on the coordinate axes are
(iv) Equation of a plane if the length of the perpendicular from the origin on the plane is ‘p’ and d.c’s of the perpendiculars as ℓ , m, n is lx + my + nz = p
(v) Parallel and perpendicular planes : Two 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 are
Perpendicular if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0, parallel if
and Coincident if
(vi) Angle between a plane and a line is the complement of the angle between the normal to the plane and the line. If
where θ is the angle between the line and normal to the plane.
(vii) Length of the ⊥ar from a point (x_{1}, y_{1}, z_{1}) to a plane ax + by + cz + d = 0 is p =
(viii) Distance between two parallel planes ax + by + cz + d_{1} = 0 and ax + by + cz + d_{2} = 0 is
(ix) Planes bisecting the angle between two 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
of these two bisecting planes, one bisects the acute and the other obtuse angle between the given planes.
(x) Equation of a plane through the intersection of two planes P_{1} and P_{2} is given by P_{1} + λP_{2} = 0
Ex.8 Reduce the equation of the plane x + 2y – 2z – 9 = 0 to the normal form and hence find the length of the perpendicular drawn form the origin to the given plane.
Sol. The equation of the given plane is x + 2y – 2z – 9 = 0
Bringing the constant term to the R.H.S., the equation becomes x + 2y – 2z = 9 ...(1)
[Note that in the equation (1) the constant term 9 is positive. If it were negative, we would have changed the sign throughout to make it positive.]
Now the square root of the sum of the squares of the coefficients of x, y, z in (1)
Dividing both sides of (1) by 3, we have
....(2)
The equation (2) of the plane is in the normal form ℓx + my + nz = p.
Hence the d.c.’s ℓ, m, n of the normal to the plane are 1/2,2/3,2/3 and the length p of the perpendicular from the origin to the plane is 3.
Ex.9 Find the equation to the plane through the three points (0, –1, –1), (4, 5, 1) and (3, 9, 4).
Sol. The equation of any plane passing through the point (0, –1, –1) is given by a(x – 0) + b{y – (–1)} + c{z – (–1)} = 0 or ax + b(y + 1) + c (z + 1) = 0 ....(1)
If the plane (1) passes through the point (4, 5, 1), we have 4a + 6b + 2c = 0 ....(2)
If the plane (1) passes through the point (3, 9, 4), we have 3a + 10b + 5c = 0....(3)
Now solving the equations (2) and (3), we have
∴ a = 10λ, b = 14λ, c = 22λ.
Putting these value of a, b, c in (1), the equation of the required plane is given by
λ[10x  14(y + 1) + 22(z + 1)] = 0 or 10x  14(y + 1) + 22(z + 1) = 0 or 5x  7y + 11z + 4 = 0.
(i) Equation of a line through A(x_{1}, y_{1}, z_{1}) and having direction cosines ℓ , m , n are and the lines through (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2})
(ii) Intersection of two 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 together represent the un symmetrical form of the straight line.
(iii) General equation of the plane containing the line is A(x – x_{1}) + B(y – y_{1}) + c(z – z_{1}) = 0 where A ℓ + bm + cn = 0.
(iv) Line of Greatest Slope AB is the line of intersection of Gplane and H is the h orizontal plane. Line of greatest slope on a given plane, drawn through a given point on the plane, is the line through the point 'P'perpendicular to the line of intersection of the given plane with any horizontal plane.
Ex.15 Show that the di stance of the point of inter section of the line and the plane x – y + z = 5 from the point (–1, –5, –10) is 13.
Sol. The equation of the given line are
= r (say) ....(1)
The coordinates of any point on the line (1) are (3r + 2, 4r  1, 12 r + 2).
If this point lies on the plane x – y + z = 5, we have 3r + 2 – (4r – 1) + 12r + 2 = 5, or 11r = 0, or r = 0.
Putting this value of r, the coordinates of the point of intersection of the line (1) and the given plane are (2, –1, 2).
∴ The required distance = distance between the points (2, –1, 2) and (–1, –5, –10)
=
Ex.16 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0. Find also the coordinates of the point on the line which is at the same distance from the foot of the perpendicular as the origin is.
Sol. The equation of the plane is 3x + 4y – 6z + 1 = 0. ....(1)
The direction ratios of the normal to the plane (1) are 3, 4, –6.
Hence the line normal to the plane (1) has d.r.’s 3, 4, –6, so that the equations of the line through (0, 0, 0) and perpendicular to the plane (1) are x/3 = y/4 = z/–6 = r (say) ....(2)
The coordinates of any point P on (2) are (3r, 4r, – 6r) ....(3) If this point lies on the plane (1), then 3(3r) + r(4r) – 6(–6r) + 1 = 0, or r = –1/61.
Putting the value of r in (3), the coordinates of the foot of the perpendicular P are (–3/61, –4/61, 6/61).
Now let Q be the point on the line which is at the same distance from the foot of the perpendicular as the origin. Let (x_{1}, y_{1}, z_{1}) be the coordinates of the point Q. Clearly P is the middle point of OQ.
Hence we have
or x_{1} = 6/61, y_{1} = –8/61, z_{1} = 12/61.
∴ The coordinates of Q are (–6/61, –8/61, 12/61).
Ex.17 Find in symmetrical form the equations of the line 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0 and find its direction cosines.
Sol. The equations of the given line in general form are 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0 ..(1) Let l, m, n be the d.c.s of the line. Since the line is common to both the planes, it is perpendicular to the normals to both the planes. Hence we have 3l + 2m – n = 0, 4l + m – 2n = 0.
Solving these, we get
or
∴ the d.c.’s of the line are
Now to find the coordinates of a point on the line given by (1), let us find the point where it meets the plane z = 0.
Putting z = 0 i the equations given by (1), we have 3x + 2y – 4 = 0, 4x + y + 3 = 0.
Solving these, we get , or x = –2, y = 5.
Therefore the equation of the given line in symmetrical form is
Ex.18 Find the equation of the plane through the line 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.
Sol. The equation of the given line are 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 ...(1)
The equation of any plane through the line (1) is (3x – 4y + 5z – 10) + l (2x + 2y  3z  4) = 0
or (3 + 2λ)x + (4 +2λ) y + (5  3λ) z  10  4λ = 0. ...(2)
The plane (1) will be parallel to the line x = 2y = 3z
(3 + 2λ) . 6 + (4 + 2λ). 3 + (5  3λ).2 = 0 or λ(12 + 6  6) + 18  12 + 10 = 0 or λ =4/3
Putting this value of l in (2), the required equation of the plane is given by
or x – 20y + 27z = 14.
Ex.19 Find the equation of a plane passing through the line and making an angle of 30° with the plane x + y + z = 5.
Sol. The equation of the required plane is (x – y + 1) + λ (2y + z  6) = 0 Þ x + (2l  1) y + λz + 1  6l = 0
Since it makes an angle of 30° with x +y + z = 5
are two required planes.
Ex.20 Prove that the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 3 and 2x – y – z = 0 = 7x + 10y – 8z – 15 are perpendicular.
Sol. Let l_{1}, m_{1}, n_{1} be the d.c.s of the first line. Then 3l_{1} + 2m_{1} + n_{1} = 0, l_{1} + m_{1}  2n_{1} = 0. Solving, we get
Again let l_{2}, m_{2},n_{2} be the d.c.s of the second line, then 2l_{2}  m_{2}  n_{2} = 0, 7l_{2} + 10m_{2}  8n_{2} = 0.
Hence the d.c.’s of the two given lines are proportional to –5, 7, 1 and 2, 1, 3.
We have –5.2 + 7.1 + 1.3 = 0
Therefore, the given lines are perpendicular.
209 videos443 docs143 tests

1. What are the direction cosines of a line in threedimensional geometry? 
2. How are direction ratios related to direction cosines in threedimensional geometry? 
3. How can the equation of a plane be represented in threedimensional geometry? 
4. What is the significance of the concept of a straight line in threedimensional geometry? 
5. How do the concepts of planes and straight lines intersect in threedimensional geometry? 
209 videos443 docs143 tests


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