Basic Concepts of Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced PDF Download

Basic Concepts of 3D Geometry

Distance Between Two Points

The position vectors of the points P and Q are given by

Now we have

Section Formula

(for external division take –ve sign)
To determine the co-ordinates 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 co-ordinates of the point R are (x, y, z), then

Now the point R divides the join of P and Q in the ratio m₁ : m₂, so that

or    [Using (1), (2) and (3)]

Remark : The middle point of the segment PQ is obtained by putting m₁ = m₂. Hence the co-ordinates of the middle point of PQ are

Centroid:

Centroid of a Triangle :

Let ABC be a triangle. Let the co-ordinates 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 co-ordinates of G be (x, y, z). 

Centroid Of A Tetrahedron :

Let ABCD be a tetrahedron, the co-ordinates of whose vertices are  = 1, 2, 3, 4.

Let  be the centroid of the face ABC of the tetrahedron. Then the co-ordinates 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 DG₁ in the ratio 3 : 1. Let G be the centroid of the tetrahedron and if (x, y, z) are its co-ordinates, then

Ex.1 P is a variable point and the co-ordinates 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 co-ordinates of P be (x, y, z).

Now it is given that 3PA = 2PB i.e., 9PA² = 4PB². ....(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 {x² + y² + z² + 4x – 4y – 6z + 17} = 4{x² + y² + z² – 26x + 6y – 26z + 347}

or 5x² + 5y² + 5z² + 140x – 60 y + 50 z – 1235 = 0

or x² + y² + z² + 28x – 12y + 10z – 247 = 0

This is the required locus of P.

Ex.2 Find the ratio in which the xy-plane 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 xy-plane (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 co-ordinates of the point R by section formula which is as follow:

 ....(1)

But on xy-plane, the z - coordinate of R is 0.

(4μ - 8) / (μ + 1) = 0, or μ = 2. Hence  μ : 1 = 2 : 1. Thus the required ratio is 2 : 1.
Again putting  the co-ordinates of the point R become

The point of intersection is (7/3, –8/3, 0).

Ex.3 ABCD is a square of side length ‘a’. Its side AB slides between x and y-axis 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 y-axis and vertex B slides on x-axis 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),

Direction Cosines and Ratios of a Line

Direction Cosines

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 co-ordinates 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 + m² + n² = 1.

Direction Ratios :

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.

Relation Between Direction Cosines And Direction Ratios :

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 ℓ² + m² + n² = 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 non-zero 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  2m₂ + 5mn + 2n₂ = 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

Ex.5 To find the projection of the line joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) on the another line whose d.c.’s are ℓ, m, n

Sol. 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  

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 5m² + 4mn – 3n² = 0 

or 5(m/n)² + 4(m/n) – 3 = 0 ....(2)

Let l₁, m₁, n₁ and l₂, m₂, n₂ be the d,c's of the two lines. Then the roots of (2) are m₁/n₁ and m₂/n₂.

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  

l₁l₂ + m₁m₂ + n₁n₂ = (2 + 3 - 5) k = 0 . k = 0.  ⇒  The lines are at right angles.

Remarks :

(a) Any three numbers a, b, c proportional to the direction cosines 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 , m₁, n₁ and l ₂ , m₂, n₂

cos θ = l₁l₂ + m₁m₂ + n₁n₂

Hence if lines are perpendicular then l₁l₂ + m₁m₂ + n₁n₂ = 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 (x₂ – x₁) + m(y₂ – y₁) + n(z₂ – z₁)

(d) If l₁, m₁, n₁ and l₂, m₂, n₂ 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 l₁ ± l₂, m₁ ± m₂, n₁ ± n₂.

Area Of A Triangle

Ex. 6 Show that the area of a triangle whose vertices are the origin and the points A(x₁, y₁, z₁) and B( x₂, y₂, z₂) is 

Sol. The direction ratios of OA are x₁, y₁, z₁ and those of OB are x₂, y₂, z₂.

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.7 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 projections of the area Δ of triangle ABC on the yz, zx and xy-planes respectively. We have 

  ∴ the required area,  

Ex.8 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

PLANE

(ii) Equation of a plane passing through (x₁, y₁, z₁) is a(x – x₁) + b (y – y₁) + c(z – z₁) = 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 co-ordinate 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 a₁ x + b₁ y + c₁z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are

  Perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 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 (x₁, y₁, z₁) to a plane ax + by + cz + d = 0 is p =

(viii) Distance between two parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is

(ix) Planes bisecting the angle between two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂ x + b₂y + c₂ z + d₂ = 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 P₁ and P₂ is given by P₁ + λP₂ = 0

Ex.9 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.10 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.

STRAIGHT LINE

(i) Equation of a line through A(x₁, y₁, z₁) and having direction cosines ℓ , m , n are and the lines through (x₁, y₁, z₁) and (x₂, y₂, z₂)

(ii) Intersection of two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 together represent  the un symmetrical form of the straight line.

(iii) General equation of the plane containing the line  is A(x – x₁) + B(y – y₁) + c(z – z₁) = 0 where A ℓ + bm + cn = 0.

(iv) Line of Greatest Slope AB is the line of intersection of G-plane 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.11 Show that the distance of the point of intersection 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 co-ordinates 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 co-ordinates 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.12 Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0. Find also the co-ordinates 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 co-ordinates 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 co-ordinates 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 (x₁, y₁, z₁) be the co-ordinates of the point Q. Clearly P is the middle point of OQ.
Hence we have 

or x₁ = 6/61, y₁ = –8/61, z₁ = 12/61.

∴ The co-ordinates of Q are (–6/61, –8/61, 12/61).

Ex.13 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 co-ordinates 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.14 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.15 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.16 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 l₁, m₁, n₁ be the d.c.'s of the first line. Then 3l₁ + 2m₁ + n₁ = 0, l₁ + m₁ - 2n₁ = 0.

 Solving, we get

Again let l₂, m₂,n₂ be the d.c.'s of the second line, then 2l₂ - m₂ - n₂ = 0, 7l₂ + 10m₂ - 8n₂ = 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.