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3D-Line

Q. 1 Find the vector equation of the line through A(3, 4, –7) and B(6, –1, 1),

also find the cartesian form.

Solution :

A

∴∴∴∴

≡ ≡ ≡ ≡ 3, 4, –7, B ≡ ≡ ≡ ≡ 6, –1, 1

AB = b – a =

λλλλ(3i −−−− 5j + 8k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

Direction ratios of the line are 3, –5, 8.

Equation of the line is

r = a +

3i −−−− 5j + 8k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

(3i + 4j – 7k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r = + λλλλ(3i −−−− 5j + 8k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

For cartesian form, put (xi −−−− yj + zk )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r =

3D-LINE

Group–A– Classwork Problems


3D-Line

x – 3

3

∴∴∴∴

= (3 + 3λλλλ) i∧∧∧∧

xi −−−− yj + zk∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ (4 – 5λλλλ) j∧∧∧∧

+ (–7 + 8λλλλ) k∧∧∧∧

∴∴∴∴ x = 3 + 3λλλλ, y = 4 – 5λλλλ, z = –7 + 8λλλλ

∴∴∴∴ = λλλλ, y – 4

3 = λλλλ,

z + 7

8 = λλλλ

x – 3

3 ∴∴∴∴ =

y – 4

3 =

z + 7

8

Which is the equation of a line in cartesian form.


3D-Line

Q. 2

Solution :

∴∴∴∴

Cartesian equation of the line is

Point A(6, –4, 5) lies on the line and direction

ratios of the line are 2, 7, 3.

The cartesian equation of a line is , find the vectorx – 6

2 =

y + 4

7 =

z – 5

3

equation of a line

x – 6

2 =

y + 4

7 =

z – 5

3

∴∴∴∴ Its equation in vector form is

λλλλ(2i + 7j + 3k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r = a +

∴∴∴∴ (6i – 4j + 5k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r = + λλλλ(2i + 7j + 3k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧


3D-Line

Q. 3 Find the vector equation of a line passing through the point (–1, –1, 2) and

parallel to the line 2x –2 = 3y + 1 = 6z – 2.

Solution :

∴∴∴∴

Equation of given line is 2x – 2 = 3y + 1 = 6z – 2

Dividing throughout by 6, we get

2(x – 1)

6=

3

6=

1

3y + 6

6

1

3z –

x – 1

3=

2=

1

3y +

1

1

3z –


3D-Line

Point (–1, –1, 2) lies on the line and direction ratios of the line are 3, 2, 1.

Its vector equation is

∴∴∴∴

r = a +

(–i + –j + 2k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r = + λλλλ(–i −−−− j + 2k )∧∧∧∧ ∧∧∧∧ ∧∧∧∧

λλλλ b


3D-Line

Q. 4

Solution :

Show that the points whose position vectors are 5i + 5k, –4i + 3j – k and ∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧

2i + j + 3k are collinear.∧∧∧∧ ∧∧∧∧ ∧∧∧∧

AB =

5i + 5k, ∧∧∧∧ ∧∧∧∧

Let a =

–9j + 3i –6k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

–4i + 3j – k,∧∧∧∧ ∧∧∧∧∧∧∧∧

b = = 2i + j + 3k∧∧∧∧ ∧∧∧∧∧∧∧∧

c =

∴∴∴∴ Direction ratios of line AB are –9, 3, –6 i.e. 3, –1, 2.

BC 6i – 2j + 4k∧∧∧∧ ∧∧∧∧∧∧∧∧

= c – b =

∴∴∴∴ Direction ratios of line BC are 6, –2, 4 i.e. 3, –1, 2.

Line AB and BC have same direction ratios.

∴∴∴∴ AB |||||||| BC

But, they have a common point i.e. point B.

∴∴∴∴ Points A, B, C are collinear.


3D-Line

Q. 5 If the points A(5, 5, λλλλ), B(–1, 3, 2) and C(–4, 2, –2) are collinear, find the value

of λλλλ.

Solution :

Direction ratios of line AB are –6, –2, 2 – λλλλ.

Direction ratios of line BC are –3, –1, –4.

Since points A, B, C are collinear

–6

–3 ∴∴∴∴ = =

2 – λλλλ–4

–2

–1

∴∴∴∴ =2 – λλλλ

–42

∴∴∴∴ –8 = 2 – λλλλ

∴∴∴∴ –10 = – λλλλ

∴∴∴∴ λλλλ = 10


3D-Line

Q. 6

Solution :

Find the vector equation of the passing through the point A(3, –4, 1) and

parallel to the vector 2i + j – 3k. ∧∧∧∧ ∧∧∧∧∧∧∧∧

Let a be the position vector of the point A w.r.t. the origin.

∴∴∴∴ a = 3i −−−− 4 j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

Let b = 2i −−−− j −−−− 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

The vector equation of the line passing through A( a ) and parallel to b is

r = a ++++ t b ,∧∧∧∧ ∧∧∧∧

where t is a scalar.

∴∴∴∴ vector equation of the required line is

r = 3i −−−− 4 j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 2i ++++ j −−−− 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧


3D-Line

Q. 7

Solution :

The line

Show line r = 3i −−−− j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 2i ++++ j∧∧∧∧ ∧∧∧∧

and r = i ++++ 2j – k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ s i – 2j + 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

are perpendicular to each other where t and s and scalars.

r = 3i −−−− j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 2i ++++ j∧∧∧∧ ∧∧∧∧

is parallel to the vector 2i ++++ j .∧∧∧∧ ∧∧∧∧

∴∴∴∴ The direction ratios of this line are

a1 = 2, b1 = 1, c1 = 0

The line r = i ++++ 2j – k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ s i – 2j + 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

is parallel to the vector i – 2j + 3k .∧∧∧∧ ∧∧∧∧ ∧∧∧∧

∴∴∴∴ the direction ratios of the line are


3D-Line

a2 = 1, b2 = –2, c2 = 3

Now,

a1a2 + b1b2 + c1c2 = 2(1) + 1(–2) + 0(3)

= 2 – 2 + 0 = 0

∴∴∴∴ the given lines are perpendicular to each other


3D-Line

Q. 8

Solution :

If the vector equation of a line is r = 2i −−−− j + 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 3i −−−− 4j + 5k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

then write the symmetric form of the equation of line.

The vector equation of the line is

r = 2i −−−− j + 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 3i −−−− 4j + 5k .∧∧∧∧ ∧∧∧∧ ∧∧∧∧

∴∴∴∴ the line is passing through the point A(2, –1, 3) and parallel to the vector

3i −−−− 4j + 5k .∧∧∧∧ ∧∧∧∧ ∧∧∧∧

∴∴∴∴ the direction ratios of the line are 3, –4 , 5.

The symmetric form of the equation of line passing through (x1 , y1, z1)

and having direction ratio a, b, c is


3D-Line

x – x1

a=

y – y1

b=

z – z1

c

∴∴∴∴ the symmetric form of the required line is

x – 2

3=

y + 1

– 4=

z – 3

5


3D-Line

Q. 9

Solution :

The direction ratios of the line

x – 1

–3= =

z

3

Show that the linex + 1

2=

y – 3

– 4=

z

3is perpendicular to the line

y – 3

– 4are

r = 5i −−−− j + 4k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 5i ++++ j – 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

a1 = 2, b1 = –4, c1 = 3

The line r = 5i −−−− j + 4k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 5i ++++ j – 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

is parallel to the vector 5i ++++ j – 2k .∧∧∧∧ ∧∧∧∧ ∧∧∧∧

∴∴∴∴ the direction ratios of this line are

a2 = 5, b2 = 1, c2 = –2


3D-Line

Now,

a1a2 + b1b2 + c1c2 = 2(5) + (–4)(1) + 3(–2)

= 10 – 4 – 6 = 0

∴∴∴∴ the given lines are perpendicular to each other .


3D-Line

Q. 10

Solution :

A line passes through the points A(–2, 0–1, 5) and B (1, 3 , –1), find the

equation of the line in

i) parametric form

ii) non- parametric form

iii) symmetric form

Let a and b be the position vectors of the points A and B respectively

w.r.t. the origin.

∴∴∴∴ a = –2i −−−− j + 5k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

and b = i + 3+ 3+ 3+ 3j –k .∧∧∧∧ ∧∧∧∧ ∧∧∧∧

i) The parametric form of the equation of line passing through A ( a )

and B(b) is

r = a + t b – a

where b – a = i ++++ 3j – k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

– – 2i – j + 5k∧∧∧∧ ∧∧∧∧ ∧∧∧∧


3D-Line

= 3i ++++ 4j – 6k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

∴∴∴∴ the parametric form of the equation of required line is

r = – 2i – j + 5k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 3i ++++ 4j – 6k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

ii) The non-parametric form of the equation of the line passing through

A( a ) and B( b ) is

r – a × × × × b – a = 0

∴∴∴∴ r × × × × b – a = a × × × × b – a


3D-Line

where a × × × × b – a

i∧∧∧∧

j∧∧∧∧

k∧∧∧∧

– 2 – 1 5

3 4 – 6

=

= i∧∧∧∧

(6 – 20) – j∧∧∧∧

(12 – 15) – k∧∧∧∧

(– 8 + 3)

= i∧∧∧∧

–14 + j∧∧∧∧

3 – k∧∧∧∧

5

∴∴∴∴ the non-parametric form of the equation of required line is

r × × × × 3i ++++ 4j – 6k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

= i∧∧∧∧

–14 + j∧∧∧∧

3 – k∧∧∧∧

5


3D-Line

iii) The symmetric form of the equation of line passing through the points

A(x1 , y1, z1) and B (x2 , y2, z2) is

x – x1

x2 – x1

=y – y1

y2 – y1

=z – z1

z2 – z1

∴∴∴∴ the symmetric form of the equation of required line passing through

A(–2, –1, 5) and B(1, 3, –1) is

x – (– 2)

1 – (– 2)=

y – (– 1)

3 – (– 1)=

z – 5

–1 – 5

i.e.x + 2

3= =

z – 5

– 6

y + 1

4


3D-Line

Q. 11

Solution :

Show line passing through the point P(2, – 14) and Q(1, 3, 2) is parallel to

line passing through the points R(–1, 5, 1) and S(1, – 3, 5)

Direction ratios of the line PQ are

a1 = 1 – 2 = – 1, b1 = 3 – (–1) = 4,

c1 = 2 – 4 = – 2

Direction ratios of the line RS are

a2 = 1 – (–1) = 2, b2 = – 3 – 5 = – 8,

c2 = 5 – 1 = 4

∴∴∴∴a1

a2

=– 1

2,

b1

b2

=4

– 8=

– 1

2and

c1

c2

=– 2

4=

– 1

2, i.e.,

a1

a2=

b1

b2=

c1

c2

Hence, the line PQ is parallel to the line RS.


3D-Line

Q. 12

Solution :

Find the coordinates of the point where the line

The symmetric form of the equation of the line

r = 4i −−−− 3 j + 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 5i ++++ 2j −−−− k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

crosses XY-plane.

r = 4i −−−− 3 j + 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

+ t 5i ++++ 2j −−−− k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

is

x – 4

5= =

z – 2

– 1

y + 3

2= t

The coordinates of any point on this line are (4 + 5t, – 3 + 2t, 2 – t)

The point where this line crosses CY–plane, has its z-coordinate = 0.

∴∴∴∴ 2 – t = 0 ∴∴∴∴ t = 2

∴∴∴∴ The coordinates of the point where the line crosses XY–plane are

(4 + 5(2), –3 + 2(2), 2 – 2)

i.e., (14 , 1 ,0)


3D-Line

Q. 13 Prove that the line joining the points A and B having position vectors

Solution :

∴∴∴∴

6 a – 4b + 4c and –4c and the line joining the points C and D having

Find the position vector of their point of intersection.

– a – 2b – 3c and a + 2b – 5c intersect.position vector

Direction ratios of line AB are –6, 4, –8 i.e. 3, –2, 4

x – 6

3 =

y + 4

–2 =

z – 4

4 Equation line AB is

Direction ratios of line CD is 2, 4, –2 i.e. 1, 2, –1.

∴∴∴∴ Equation of line CD is

x + 1

1 =

y +2

2 =

z + 3

–1


3D-Line

x – 6

3 =

y + 4

–2 =

z – 4

4 Let = k

∴∴∴∴ x = 3k + 6, y = –2k – 4, z = 4k + 4

∴∴∴∴ (3k + 6, –2k – 4, 4k + 4)

Also, Let x + 1

1 =

y +2

2 =

z + 3

–1= λλλλ

∴∴∴∴ (λλλλ – 1, 2λλλλ – 2, –λλλλ –3)

∴∴∴∴ 3k + 6 = λλλλ – 1 and –2k –4 = 2λλλλ – 2

∴∴∴∴ 3k – λλλλ = – 7 …(i)

2λλλλ + 2k = – 2

…(ii)∴∴∴∴ λλλλ + k = –1


3D-Line

Adding equations (i) and (ii), we get

∴∴∴∴

4k = –8

k = –2

∴∴∴∴ λλλλ – 2 = –1

∴∴∴∴ λλλλ = 1

∴∴∴∴ 4k + 4 = 4(–2) + 4 = –4

–λλλλ –3 = –1 –3 = –4

∴∴∴∴ These lines intersect and their point of intersection is (0, 0, –4) i.e. –4c


3D-Line

Q. 14

Solution :

∴∴∴∴

The equation of the lines are

Direction ratios of two lines are

Find the value of λλλλ, so that the lines and 1– x

3=

7y – 14

2λλλλ=

z – 3

2

x – 1

–3= =

z – 3

2

are at right angle.7 – 7x

3λλλλ=

y – 5

1=

6 – 3

5

y – 2

2

7λλλλ

andx – 2

–3

7λλλλ

=y – 5

1=

z – 6

–5

2

7λλλλ,–3, 2 and

–3

7λλλλ, 1, –5

Since, the lines are perpendicular


3D-Line

∴∴∴∴ (–3) +–3

7λλλλ (1) + (2)(–5) = 0

2

7λλλλ

∴∴∴∴ 9

7λλλλ +

2

7λλλλ λλλλ – 10 = 0

∴∴∴∴ 11λλλλ – 70 = 0

∴∴∴∴ 70

11λλλλ =


3D-Line

Q. 15

Solution :

∴∴∴∴

Direction ratios of the given lines are (1, 2, 2) and (3, 2, 6).

1(3) + 2(2) + 2(6)

Find the angle between the pair of = 3i + 2j – 4k + λλλλ ( i + 2j + 2k) and r∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧

= 5i – 2k + µµµµ (3i + 2j + 6k). r∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧ ∧∧∧∧

cosθθθθ =1 + 4 + 4√√√√ . 9 + 4 + 36√√√√

=19

7(3)

∴∴∴∴ θθθθ = cos–119

21


3D-Line

1) Find the length of the perpendicular from (2, –3, 1) to the line

x + 1

2=

y – 3

3=

z + 2

–1

Solution : A (2, – 3, 1)

N (– 1, 3, – 2)M

The line x + 1

2=

y – 3

3=

z + 2

–1. . . (1) passes through

The point N (– 1, 3, – 2) and has d.r.s. 2, 3, – 1.

Group–B– Classwork Problems


3D-Line

∴∴∴∴ It has d.c.s2

√√√√14,

3

√√√√14,

– 1

√√√√14

∴∴∴∴ (x1 + r cos αααα, y1 + r cos ββββ, z1 + r cos γγγγ)

i.e. (x1 + rl, y1 + rm, z1 + rn)

Denote the coordinates of a point on the line at a distance

r from N. Let M be the foot of perpendicular from A.

Let l(MN) = r

∴∴∴∴ M ≡≡≡≡ – 1 +2

√√√√14r , 3 +

3

√√√√14r , – 1 +

– 1

√√√√14r


3D-Line

∴∴∴∴ d.r.s. at AM are

– 1 +2

√√√√14r , – – 1 +

3

√√√√14r , 1 – 2 –

r

√√√√14r 2

i.e. –2r

√√√√14, – –

3r

√√√√14, 3 +

r

√√√√143

Since, AM ⊥⊥⊥⊥ line (1)

3 –2r

√√√√14+ 3 – 6 –

3r

√√√√14– 1 3 +

r

√√√√14=2 0

– 3 – –

6


3D-Line

∴∴∴∴ –4r

√√√√14– 18 –

9r

√√√√14– 3 –

r

√√√√14=6 0

– 15

√√√√14=r

∴∴∴∴ M ≡≡≡≡ – 1 +2

√√√√14××××

– 15

√√√√14, 3 +

3

√√√√14××××

– 15

√√√√14, –

– 1

√√√√14××××

– 15

√√√√142

= – 1 –30

14, 3 –

45

14, – 2 +

15

14


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∴∴∴∴ AM = 2 +44

14+ 3 +

3

14+ 1 +

13

14

2

–

2 2

√=

(72)2

142

+ (39)2 + (27)2

√=

1

4√√√√7434 units


3D-Line

2) Find the shortest distance between the lines

r = + λλλλ i + 2j – 3k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

4i – j∧∧∧∧ ∧∧∧∧

and r = i – j + 2k∧∧∧∧ ∧∧∧∧

+ µµµµ i + 4j – 5k∧∧∧∧ ∧∧∧∧ ∧∧∧∧∧∧∧∧

Solution :

The distance between r = a + tb and r = c + td is given by

[ c – a b d ]

| b ×××× d |

In the given problem,

=a 4i – j,∧∧∧∧ ∧∧∧∧

=b i + 2j – 3k,∧∧∧∧ ∧∧∧∧ ∧∧∧∧

=c =d i + 4j – 5k,∧∧∧∧ ∧∧∧∧ ∧∧∧∧

i – j + 2k,∧∧∧∧ ∧∧∧∧ ∧∧∧∧


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[ c – a b d ]∴∴∴∴ =

– 3 0 2

1 2 – 3

1 4 – 5

= – 3 (– 10 + 12) + 2(4 – 2)

= – 6 + 4

= – 2

b ×××× d∴∴∴∴ = 1 2 – 3

1 4 – 5

i∧∧∧∧

j∧∧∧∧

k∧∧∧∧

= i∧∧∧∧

(– 10 + 12) – (– 5 + 3) + (– 5 + 3)∧∧∧∧kj

∧∧∧∧


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= 2 i∧∧∧∧

+ 2 j∧∧∧∧

+ 2 k∧∧∧∧

| b ×××× d |∴∴∴∴ =

2√√√√ 3∴∴∴∴

[ c – a b d ]

| b ×××× d |=

– 2

=√√√√ 3

1units.

2√√√√ 3


3D-Line

3) Find the shortest distance between the lines

x + 1

7=

y + 1

– 6=

z + 1

1and

x – 3

1=

y – 5

– 2=

z – 7

1

Solution :

The linex + 1

7=

y + 1

– 6=

z + 1

1passes through the

point (– 1, – 1, – 1) and has d.r.s. 7, – 6, 1.

The linex – 3

1=

y – 5

– 2=

z – 7

1passes through the

point (3, 5, 7) and has d.r.s. 1, – 2, 1.


3D-Line

| D2 |=

D1

S.D.

=Where D1

x2 – x1 y2 – y1 z2 – z1

a1 b1 c1

a2 b2 c2

=

3 + 1 5 + 1 7 + 1

7 – 6 1

1 – 2 1

= 4 (– 6 + 2)

= – 16 – 36 – 64

– 6 (7 – 1) + 8(– 14 + 6)

= – 116.


3D-Line

D2 = a1 b1 c1

a2 b2 c2

i∧∧∧∧

j∧∧∧∧

k∧∧∧∧

= 7 – 6 1

1 – 2 1

i∧∧∧∧

j∧∧∧∧

k∧∧∧∧

= i∧∧∧∧

(– 6 + 2) – (7 – 1) + (– 14 + 6)∧∧∧∧kj

∧∧∧∧

= – 4 i∧∧∧∧

– 6 j∧∧∧∧

– 8 k∧∧∧∧

| D2 | = √√√√ (– 4)2 + (– 6)2 + (– 8)2

= √√√√116

∴∴∴∴


3D-Line

S.D. =∴∴∴∴| D2 |

D1

=– 116

√√√√116

= √√√√116 units

= 2√√√√ 29 units


3D-Line

4) A (1, 0, 4), B (0, –11, 13), C (2, –3, 1) are three points and D is the foot of the

perpendicular from A on BC. Find coordinates of D.

Equation of line BC

x – x1

x2 – x1

y – y1

y2 – y1

z – y1

z2 – z1

= =

A(1,0,4)

B(0,-11,13)B(2,-3,1)

D

Solution :


3D-Line

x – 0

2 – 0

y – 11

-3 + 11

z – 13

1 – 13= =

x

2

y + 11

8

z – 13

−−−− 12= = t – say∴∴∴∴

∴∴∴∴

x = 2t , y = 8t – 11 , z = - 12t +13∴∴∴∴

(2t , 8t -11 , -12t + 13), ∴∴∴∴ Denote co – ordinates of a point on line BC

We take D = (2t , 8t – 11, -12 + 13 )

d.r.s of AD are∴∴∴∴

2t – 1 , ∴∴∴∴ 8t -11 – 0 , – 12t + 13 – 4


3D-Line

2t – 1 , i.e 8t -11 , -12t + 9

d.r.s of BC are 2, 8 , -12

since AD. ⊥⊥⊥⊥ BC

(2t – 1)2 , + (-12t + 9) (8t -11)8 + (- 12) = 0

4t −−−− 2∴∴∴∴ + 64t – 88 + 144t – 108 = 0

∴∴∴∴ 212t – 198 = 0

∴∴∴∴ t = 198

212

99

106=

∴∴∴∴ D = 99

106=

99

1062 × 8 ×××× - 11,-12 ××××

99

106+ 13


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198

106

-374

106

190

106, ,=

99

53

187

53

95

53, ,=


3D-Line

5) If the linesx – 1

2=

y + 1

3=

Z – 1

4and

x – 3

1=

y – k

2=

z

1

intersect, then find the value of k.

Let x – 1

2=

y + 1

3=

z – 1

4(say)

∴∴∴∴ x = 2r + 1 , y = 3r – 1 , z = 4r + 1

∴∴∴∴ (2r + 1 , 3r – 1 , 4r + 1) are the coordinates of point on first line

Let x – 3

1=

y + k

2=

z – 1

1s (say)

Solution :


3D-Line

∴∴∴∴ (s + 3 , 2s + k, s) are the coordinates of point on second line

Sinces , the two lines intersect at the point of intersection

x = s + 3 ,

(2r + 1 = 3r – 1 , 4x + 1 ) = (s + 3 , 2s + k , s )

2r + 1 = s + 3 , 3r – 1 = 2s + k , 4x + 1 = s

2r + 1 = 4 ∴∴∴∴

∴∴∴∴

….(i)

3r – 2r = k + 1

2r - s = - 1

Solving Eqs. (i) and (iii) simultaneous

y = 2s + k , z = s


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r = -3

2s = 5,

Putting these values in Eq . (ii)

3 -3

2- 2(5) = k + 1

-9

2+ 10 = k + 1-

9

2k =


3D-Line

6) Find the shortest distance between the following pair of line

r = + µµµµ 2i + j + 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

2i – j – k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

r = + λλλλ i – j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

andi + 2j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

Let

a = i + 2j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

b = i −−−− 2j + k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

c = 2i −−−− j −−−− k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

d =

Then equation of the lines are

r = a + λλλλ b

2i + j + 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

Solution :


3D-Line

The shortest distances between these line is given by

[c – a b d ]

[b ×××× d]SD =

a c−−−− i + 3j + 2k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

=

[c – a b d ]∴∴∴∴

1 -3 -2

1 -1 1

2 -1 2

=

=

1(- 2 -1 ) + 3 (2 - 2) - 2(1 + 2)

= - 3 + 0 - 6 = -9


3D-Line

[b ×××× d] =

i j k∧∧∧∧ ∧∧∧∧ ∧∧∧∧

1 -1 1

2 -1 2

= + -i (- 2 -1 )∧∧∧∧

j (2 - 2)∧∧∧∧

k (1 + 2)∧∧∧∧

= = -9- 3i + 3k∧∧∧∧ ∧∧∧∧

[b ×××× d] = √9 + 9 = 3√ 2∴

=∴ S.D- 9

3√ 2

3

√ 2units=

GROUP (A) – HOME WORK PROBLEMS

Q-1) Find the vector equation of the line passing

through the points whose position vectors

with respect to origin are �� 2 + +i j k�� and

�� – 3 +i j k��

Ans. Given :

� ��= 2 + +a i j k

� ��= – 3 +b i j k

� �( ) � �( )� �– = – 3 + – 2 + +b a i j k i j k

�( )�– = – – 4b a i j

( )= + –a t b bππππ

� �( ) �( )� �= 2 + + + – – 4i j k t i jππππ

3D-Line

Q-2) Find the angle between lines

–1 +1 –1= =

2 –3 1

x y z and

+1 – 2= =

3 4 3

x y z

Ans. The direction ratios of the given lines are 2,

–3, 1 and 3, 4, 3 respectively.

Let a and b be the vectors along these lines

respectively.

∴∴∴∴ � ��= 2 – 3 +a i j k and � ��= 3 + 4 +3b i j k

Let θθθθ be the required angle between the lines.

Then cos θθθθ = .a b

ab

where .a b =� �( ) � �( )� �2 – 3 +3 . 3 + 4 +3i j k i j k

= 2(3) + (–3)(4) + 1(3)

= 6 – 12 + 3 = –3

a = ( )2 + –3 +122 2

= 4 + 9 +1 = 14

and b = 3 + 4 +32 2 2

= 9 +16 +9

= 34

∴∴∴∴ cos θθθθ = –3

14 13 =

–3

2 119

θθθθ = –3

cos2 119

–1

Q-3) Find the angle between the line

��(((( )))) ��(((( ))))= 2 – + 3 –r i k t i j� �� and the line joining

the points A(2, 1, –3) and B(1, –1, 2).

Ans. The line �( ) �( )� �= 2 – + 3 +r i k t i j is parallel to the

vector ��3 +i j .

∴∴∴∴ the direction ratios of this line are a1 = 3,

b1 = 1, c

1 = 0

The direction ratios of the line joining the

points

A(2, 1, –3) and B(1, –1, 2) are a2 = 2 – 1 = 1,

b2 = 1–(–1) = 2, c

2 = –3 – 2 = –5

Let θ θ θ θ be the angle between the lines.

∴∴∴∴ cos θθθθ =+ +

+ + . + +

a a b b c c

a b c a b c

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

±±±±

=( ) ( ) ( )

( )

3 1 +1 2 +0 –5

3 +1 +0 . 1 + 2 + –522 2 2 2 2

±±±±

=3+2 – 0

9+1+0. 1+ 4+25±±±±

∴∴∴∴ cos θθθθ =5 1

=10 3 2 3

± ±± ±± ±± ±

∴∴∴∴ θ θ θ θ =

1cos

2 3

–1 or θθθθ =

–1cos

2 3

–1

Q-4) If Cartesian equations of line is

–1 – 2 – 3= =

2 3 4

x y z then find vector

equation of the line in parametric and non-

parametric form .

3D-Line

50 Mahesh Tutorials Science

Ans. The cartesian form of the equation of the line

is –1 – 2 – 3

= =2 3 4

x y z

∴∴∴∴ the line passing through the point

A(1, 2, 3) and having direction ratios 2, 3, 4.

Let a be the position vector of the point A

w.r.t. the origin and b be the vector parallel

to the line.

∴∴∴∴ � ��= + 2 +3a i j k and � ��= 2 +3 + 4b i j k

The vector equation of the line in parametric

form passing through the point ( )A a and

parallel to b is

= +r a tb

∴∴∴∴ the parametric form of the equation of

required line is

� �( ) � �( )� �= +2 +3 + 2 + 3 + 4r i j k t i j k

Q-5) Find the Cartesian equation of the line

which passes through the point (–2, 4, – 5)

and parallel to the line given by

+ 2 – 3 +5= = .

3 5 6

x y z

Ans. Line passes through the point (–2, 4, –5).

Ans direction ratios of given line are 3, 5, 6

Required line is parallel to it.

∴∴∴∴ Direction ratios of required line are 3, 5, 6.

∴∴∴∴ Its equation in cartesian form is

– – –= =

x x y y z z

a b c

1 1 1

i.e. + 2 – 4 +5

= =3 5 6

x y z

Q-6) Find angle between the line

�� (((( )))) �� (((( ))))= 2 – + + + +r i j k t i j k� �� and

�� (((( )))) �� (((( ))))= – +2 + 2 + – 3r i j k s i j k� ��

Ans. The line � �( ) � �( )� �= 2 – +2 + + +r i j k t i j k is

parallel to the vector � �� + +i j k .


1 1,a = 1 1,b = 1 1c =

The line � �( ) � �( )� �= + – 2 + 2 + +3r i j k s i j k is

parallel to the vector � ��2 + + 3i j k .


2 2,a = 2 1,b = 2 1c =

Let θθθθ be the required angle between the lines.

∴∴∴∴+ +

cos =+ + . + +

a a b b c c

a b c a b c

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

θ ±θ ±θ ±θ ±

= ( ) ( ) ( )1 2 +1 1 +1 3

1 +1 +1 . 2 +1 +32 2 2 2 2 2±±±±

= 2+1+3

1+1+1. 4 +1+9±±±±

= 6

3. 14±±±±

∴∴∴∴6

cos =7

θ ±θ ±θ ±θ ±

∴∴∴∴

6= cos

7

–1θθθθ or

6= cos –

7

–1θθθθ .


through the point A(4, –2, 1) and parallel

to the line �� (((( )))) �� (((( ))))= + – + –2 +3 – 5r i j k t i j k� �� .

Ans. The position vector a of the point A(4, –2, 1)

is � ��= 4 – 2 + .a i j k

Since the required line is parallel to the line

� �( ) � �( )� �= + – + –2 + 3 – 5 ,r i j k t i j k the l ine is

parallel to the vector � ��= –2 + 3 – 5 .b i j k

The vector equation of the line passing

through ( )A a and parallel to b is = +r a tb

∴∴∴∴ the vector equation of the required line is

� �( ) � �( )� �= 4 – 2 + + –2 + 3 – 5r i j k t i j k


through A(2, –5, 7) and parallel to

�� – +2 – 6i j k�� in nonparametric form and

symmetric form.

Ans. i) The vector equation of the line in non-

parametric form passing through the

Mahesh Tutorials Science 51

3D-Line

Q-9) Show that the point lies on the line passing

through the point A(9, –9, 5), B(–7, 11, 3)

and C(–3, 6, –1) by vector method..

Ans. Let , ,a b c be the positive vectors of the points

A, B, C respectively w.r.t. the origin.

∴∴∴∴ � ��= 9 – 9 +5 ,a i j k � ��= –7 +11 – 3 ,b i j k

� ��= –3 + 6 – .c i j k

The vector eqution of the line AB is

( )= + – ,r a t b a where

� �( ) � �( )� �– = –7 +11 – 3 – –9 – 9 +b a i j k i j k

= � ��–16 + 20 – 8i j k

∴∴∴∴ the vector equation of the line AB is

� �( ) � �( )� �= 9 – 9 +5 + –16 +20 – 8r i j k t i j k ...(I)

where t is a parameter.

To show that this line passses through C, we

have to show that the position vector c of C

must satisfy the equation (I) for some scalar t.

Replacing r by c in equation (I), we get,

� � � �( ) � �( )� � �–3 + 6 – = 9 – 9 +5 + –16 +20 – 8i j k i j k t i j k

∴∴∴∴� �( ) � �( )� �–3 +6 – – 9 – 9 +5i j k i j k

=� �( )�–16 +20 – 8t i j k

∴∴∴∴� � � �( )� �–12 +15 – 6 = –16 +20 – 8i j k t i j k

∴∴∴∴ = � �( )�3–16 + 20 – 8

4i j k

=� �( )�–16 +20 – 8t i j k

∴∴∴∴ t = 3

4

∴∴∴∴ the line passes through the point C.

∴∴∴∴ the line passing through the point A, B, C is

� �( ) � �( )� �= 9 – 9 +5 + –16 +20 – 8r i j k t i j k

Let p be the position vector of the point P

w.r.t. the origin.

∴∴∴∴ � ��= 5 – 4 +3p i j k

Replacing r by p in the equation of the line,

we get,

� � � �( ) � �( )� � �5 – 4 + 3 = 9 – 9 +5 –16 +20 – 8i j k i j k t i j k

∴∴∴∴ � �( ) � �( )� �5 – 4 +3 – 9 – 9 +5i j k i j k

=� �( )�–16 +20 – 8t i j k

∴∴∴∴ � � � �( )� �–4 +5 – 2 = –16 +20 – 8i j k t i j k

∴∴∴∴ � �( )�1–16 +20 – 8

4i j k

=� �( )�–16 +20 – 8t i j k

∴∴∴∴1

=4

t

∴∴∴∴ the position vectors p of P satisfy equation

of the line for1

=4

t .

Hence the point P lies on the line passing

through the points A, B, C.

point ( )A a and parallel to b is =r b a b× ×× ×× ×× ×

Here � ��= 2 – 5 +7a i j k and � ��= – +2 – 6b i j k

∴∴∴∴

� ��

= 2 –5 7

–1 2 –6

i j k

a a××××

= ( ) � ( ) � ( )� 30 –14 – –12+ 7 + 4 – 5i j k

= � ��16 +5 –i j k

∴∴∴∴ the vector equation of the required line in

non-parametric form is

� �( ) � �� – +2 – 6 =16 +5 –r i j k i j k××××

Q-10)Find vector equation of the line passing

through A(1, 2, 3) and perpendicular to

�� 2 + –i j k�� and �� + 3 +2i j k�� in parametric

form.

Ans. Let � ��= 2 + –b i j k and � ��= +3 +2c i j k .

The vector perpendicular to the vectors b and

c is given by

3D-Line


Q-11)If the points A(5, 5, λλλλ), B(–1, 3, 2) and

C(–4, 2, –2) are collinear, find the value of

λλλλ.

Ans. The direction ratios of AB are –1–5, 3–5, 2–λλλλ

i.e, –6, –2, 2–λλλλ

The direction ratios as AC are –4–5, 2–5,

–2–λλλλ i.e. –9, –3, –2–λλλλ

Since the point A,B, C are collinear, the

direction ratios of AB and AC are in same

proportion

–6 –2 2 –= =

–9 –3 –2 –

λλλλ

λλλλ

2 – 2=

3 + 2

λλλλ

λλλλ

2λλλλ + 4 = 3λλλλ – 6

λλλλ – 10

Q-12)Show that the lines

(((( )))) (((( ))))= 18 – 9 +10 + 3 –16 + 7r a b c a b cλλλλ and

(((( )))) (((( ))))= 15 +29 +5 + 3 +8 – 5r a b c a b cµµµµ

are non-coplanar.

� ��

= 2 1 –1

1 3 2

i j k

b c××××

= ( ) � ( ) � ( )� 2+ 3 – 4 +1 + 6 –1i j k

= � ��5 – 5 +5i j k

Since the line is perpendicular to the vectors

b and c , it is parallel to b c×××× .

The equation of the line in parametric form

passing through ( )A a and parallel to b c×××× is

( )= +r a t b a×××× , Where t is a parameter

Here, � ��= + 2 + 3a i j k

� �( ) � �( )� �= +2 +3 + 5 – 5 +5r i j k t i j k

= � �( ) � �( )� �+2 +3 +5 – +i j k t i j k

i.e.,� �( ) � �( )� �= +2 +3 + – + ,r i j k s i j k

(where s = 5t, is a parameter).

Ans. The equations of the given lines are

( ) ( )= 18 – 9 +10 + 3 –16 + 7r a b c a b cλλλλ ...(i)

and ( ) ( )= 15 + 29 +5 + 3 + 8 – 5r a b c a b cµµµµ

...(ii)

The line (i) is parallel to the vector and

3 –16 + 7a b c and line (ii) is parallel to the

vector 3 + 8 – 5a b c .

Since the vectors 3 –16 + 7a b c and

3 + 8 – 5a b c are not parallel because

3 –16 7

3 8 –5≠ ≠≠ ≠≠ ≠≠ ≠ .

∴∴∴∴ the lines (i) and (ii) are not parallel.

if = + +r xa yb zc then equation (i) becomes

+ +xa yb zc

= (8 + 3λλλλ)a + (–9 – 16λλλλ) b + (10 + 7λλλλ)c

∴∴∴∴ x = 8 + 3λλλλ, y = –9 – 16λ λ λ λ , z = 10 + 7λλλλ

∴∴∴∴ the coordinates of any point on the line (i)

are (8 + 3λλλλ, –9 – 16λλλλ, 10 + 7λλλλ)

Also, equation (ii) becomes,

+ +xa yb zc = (15 + 3µµµµ)a + (29 + 8µµµµ)b

+ (5 –5µµµµ)c

∴∴∴∴ x = 15 + 3µµµµ, y = 29 + 8µµµµ, z = 5 –5µµµµ

∴∴∴∴ the coordinates of any point on the line

(ii) are (15 + 3µµµµ, y = 29 + 8µµµµ, z = 5 – 5µµµµ)

If the lines (i) and (ii) intersect, then

8 + 3λλλλ = 15 + 3λλλλ ...(iii)

–9 – 16λλλλ = 29 + 8µµµµ ...(iv)

and 10 + 7λ λ λ λ = 5 – 5µµµµ ...(v)

are simultaneously true.

Solving equations (iii) and (iv), we get,

λ λ λ λ = –29

36, µ µ µ µ =

–113

36

These values of λ λ λ λ and µ µ µ µ do not satisfy equation

(iv).

∴∴∴∴ the lines (i) and (ii) do not intersect.

Hence, the given lies are neither parallel nor

intersecting

∴∴∴∴ the lines are non-coplanar.


3D-Line

Q-13)Find the equation of the line equally

inclined to co-ordinate axes and passes

through (–5, 1, –2).

Ans. The equation of the line passing through the

point (x1, y

1, z

1) and having direction cosines

l, m, n is – – –

= =x x y y z z

l m n

1 1 1

Since the required line is equally inclined to

the coord nate axes, α α α α = β β β β = γγγγ are the angles

made to the line with the coordinate axes

∴∴∴∴ cosα α α α = cosβ β β β = cosγγγγ

∴∴∴∴ l = m = n

∴∴∴∴ the equation of the required line passing

through the point (–5, 1, –2) is

+ 5 –1 + 2= =

x y z

l l l

∴∴∴∴ x + 5 = y – 1 = z + 2

Q-14)Find the vector and the cartesian equations

of the line passes through

i) the origin and (5, –2, 3)

ii) the points (3, –2, –5) and (3, –2, 6).

Ans. Let a be the position vector of the point A(5,

–2, 3) w.r.t. the origin

∴∴∴∴ � ��= 5 – 2 +3a i j k

The vector equations of the line passing

through ( )A a and ( )B b is ( )= + –r a b aλλλλ ,

where λλλλ is a scalar.

∴∴∴∴ the vector equation of the line passing

through the origin and ( )A a is

( )= 0 + – 0r aλλλλ i.e. =r aλλλλ

= � �( )�= 5 – 2 +3r i j kλλλλ

The cartesian equation of the line passing

through the point A(x1, y

1, z

1) and B(x

2, y

2, z

2)

is

– – –= =

– – –

x x y y z z

x x y y z z

1 1 1

2 1 2 1 2 1

∴∴∴∴ the cartesian equation of the line passing

through the origin O(0, 0, 0) and A(5, –2, 3) is

– 0 – 0 – 0= =

5 – 0 –2 – 0 3 – 0

x y z

i.e., = =5 –2 3

x y z

GROUP (B) – HOME WORK PROBLEMS

Q-1) Find the coordinates of the foot of

perpendicular drawn the point ˆˆ ˆ2 – + 5i j k

to the line

(((( )))) (((( ))))ˆ ˆˆ ˆ ˆ ˆ= 11 – 2 – 8 + 10 – 4 – 11r i j k i j k . Also,

find the length of perpendicular.

Ans. Let M be the foot of perpendicular drawn from

the point ˆ ˆ ˆ2 – +5i j k on the line

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= 2 – – 8 + 10 – 4 –11r i j k i j k .

Let the position vector of the point M be

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ11 – 2 – 8 + 10 – 4 –11i j k i j kλλλλ

= ( ) ( ) ( )ˆ ˆ ˆ11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ

Then

PM = Position vector of M – Position vector of P

= ( ) ( ) ( ){ }ˆ ˆ ˆ11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ

– ( )ˆ ˆ ˆ2 – +5i j k

= ( ) ( ) ( )ˆ ˆ ˆ9 +10 + –1 – 4 + –13 –11i j kλ λ λλ λ λλ λ λλ λ λ

Since PM is perpendicular to the given line

which parallel to ˆ ˆ ˆ=10 – 4 –11b i j k ,

PM b⊥ ∴∴∴∴ = 0PM b⋅

∴ ∴ ∴ ∴ ( ) ( ) ( )ˆ ˆ ˆ9 +10 + –1 – 4 + –13 –11i j k λ λ λλ λ λλ λ λλ λ λ

( )ˆ ˆ ˆ10 – 4 –11i j k =

∴ ∴ ∴ ∴ 10(9 + 10λλλλ) – 4(–1– 4λλλλ) – 11(–13 – 11λλλλ) = 0

∴ ∴ ∴ ∴ 90 + 100λλλλ + 4 + 16λ λ λ λ + 143 + 121λ λ λ λ = 0

∴ ∴ ∴ ∴ 237λλλλ + 237 = 0

∴ λ∴ λ∴ λ∴ λ = –1

Putting this value of λλλλ, we get the position

vector of M as ˆ ˆ ˆ+2 +3i j k .

∴ ∴ ∴ ∴ coordinates of the foot of perpendicular Mare (1,2,3).

Now, ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= + 2 + 3 – 2 – + 5PM i j k i j k

= ˆ ˆ ˆ– +3 – 2i j k

3D-Line


∴∴∴∴ ( ) ( ) ( )= –1 + 3 + –2PM2 2 2

= 1+9+ 4 = 14

Hence, the coordinates of the foot of

perpendicular are (1,2,3) and length of

perpendicular = 14 units.

Q-2) Find the perpendicular distance of the

point (1, 0, 0) from the line

– 1 – 1 +10= =

2 –3 8

x y z. Also, find the co-

ordinates of the foot of the perpendicular.

Ans. Let P ≡ ≡ ≡ ≡ (1, 0, 0)

Equation of line is –1 +1 +10

= =2 –3 8

x y z

∴∴∴∴ x = 2αααα + 1, y = –3α α α α –1, z = 8α α α α – 10

Let Q ≡≡≡≡ (2αααα + 1, –3α α α α –1,8α α α α – 10)

Direction ratios of line PQ are 2αααα , –3α α α α –1,

8α α α α – 10

Direction ratios of line given line are 2, –3, 8.

∴∴∴∴ 2(2αααα) – 3(–3α α α α –1) + 8(8α α α α – 10) = 0

∴∴∴∴ 77αααα = 77

∴∴∴∴ α α α α = 1

∴∴∴∴ Q ≡ ≡ ≡ ≡ (3, –4, –2)

l (perpendicular) = l(PQ)

= ( ) ( ) ( )3 –1 + –4 – 0 + –2 – 02 2 2

= 4 +16 + 4

= 2 6

Q

P(1, 0, 0)

Q-3) By computing the shortest distance,

determine whether the following lines

intersect or not:

i) (((( )))) (((( ))))= + + ˆˆ ˆ ˆ– 2r i j i kλλλλ

(((( )))) (((( ))))= + + ˆˆ ˆ ˆ ˆ2 – –r i j i j kµµµµ

ii)+

= =– 5 – 7 3

4 –5 –5

x y z,

–= =

– 8 – 7 5

7 1 3

x y z

Ans. i) The shortest distance between the lines

= +r a b1 1λλλλ and = +r a b2 2µµµµ is given by

( ) ( )–=a b a a

db b

⋅1 2 2 1

1 2

××××

××××

Here, ˆ ˆ= –a i j1, ˆ ˆ= 2 –a i j2

, ˆ ˆ= 2 +b i k1,

ˆ ˆ ˆ= + –b i j k2

∴∴∴∴

ˆ ˆ ˆ

= 2 0 1

1 1 –1

i j k

b b1 2××××

( ) ( ) ( )ˆ ˆ ˆ= 0 –1 – –2 –1 + 2 – 0i i k

ˆ ˆ ˆ= – + 3 + 2i j k

and ( ) ( )ˆ ˆ ˆ ˆ ˆ– = 2 – – – =a a i j i j i2 1

∴∴∴∴ ( ) ( ) ( )ˆ ˆ ˆ ˆ– = – + 3 + 2b b a a i j k i⋅ ⋅1 2 2 1××××

= –1(1) + 3(0) + 2(0) = –1

and b b1 2×××× ( )= –1 + 3 + 22 2 2

= 1+9+ 4

= 14

∴∴∴∴ the shortest distance between the given lines

–1=

14

1=

14 unit

Hence, the given lines do not intersect.

ii) Refer to the solution of Q.No.4

The Shortest distance between the lines

282=

3830 units.

Hence, the given lines do not intersect.


3D-Line

Q-4) Find the foot of the perpendicular from the

point (0, 2, 3) on the line

+ += =

3 –1 4

5 2 3

x y z. Also, find the length

of the perpendicular.

Ans. Let M be the perpendicular drawn from the

point P(0, 2, 3) to the given line. The

coordinates of any point on

+ 3 –1 + 4= =

5 2 3

x y z are

+ 3 –1 + 4= =

5 2 3

x y z λλλλ

x = 5λ λ λ λ –3, y = 2λ λ λ λ + 1, z = 3λ λ λ λ – 4

Therefore, coordinates of M are

(5λ λ λ λ –3, 2λ λ λ λ + 1, 3λ λ λ λ – 4) The direction ratios of

PM are 5λ λ λ λ –3 – 0, 2λ λ λ λ + 1 – 2, 3λ λ λ λ – 4 – 3

i.e. 5λ λ λ λ –3, 2λ λ λ λ – 1, 3λ λ λ λ – 7

Direction ratio of the given line are 5, 2, 3.


∴∴∴∴ 2(5λ λ λ λ –3) + 2(2λ λ λ λ –1) + 3(3λ λ λ λ –7) = 0

∴∴∴∴ λ λ λ λ = 1

Putting λ λ λ λ = 1 in e.q.1, the co-ordination of M

are (2, 3, –1) length of the perpendicular from

P on the given line

( ) ( ) ( )PM = 2 – 0 + 3 – 2 + –1, –32 2 2

PM = 21 units.

Q-5) Find the length of the perpendicular from

the point (5, 4, –1) to the line

(((( ))))+ + + ˆˆ ˆ ˆ= 2 9 5r i i j kλλλλ

Ans. Equation of line is ( )ˆ ˆ ˆ ˆ= + 2 +9 +5r i i j kλλλλ

∴∴∴∴ ( ) ˆ ˆ ˆ= 1+ 2 + 9 +5r i j kλ λ λλ λ λλ λ λλ λ λ

Let M be the foot of the perpendicular as ‘M’

lies on the line also.

( ) ˆ ˆ ˆ= 1+2 +9 +5m i j kλ λ λλ λ λλ λ λλ λ λ

Let ˆ ˆ ˆ= 2 +9 +5b i j k

( ) ( ) ( )ˆ ˆ ˆ= – = 2 – 4 + 9 – 4 + 5 +1PM m p i j kλ λ λλ λ λλ λ λλ λ λ

PM b⊥ ∴ ∴ ∴ ∴ = 0PM b⋅

∴∴∴∴ 2(2λ λ λ λ – 4) + 9(9λ λ λ λ – 4) + 5(5λ λ λ λ –1) = 0

110λ λ λ λ – 39

39=110

λλλλ

∴∴∴∴188 351 195

= , ,110 110 110

M

Length of perpendicular =l(PM)

362 89 305= + +

110 110 110

2 2 2

231990=

110

Q-6) Find the coordinates of the foot of

perpendicular drawn from the point

+ ˆˆ ˆ2 – 5i j k to the line

(((( )))) (((( ))))– – + – –ˆ ˆˆ ˆ ˆ ˆ= 11 2 8 10 4 11r i j k i j kλλλλ . Also,

find the length of perpendicular.


the point ( )ˆ ˆ ˆ2 – +5P i j k on the line

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= 11 – 2 – 8 + 10 – 4 –11r i j k i j kλλλλ .


( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ11 – 2 – 8 + 10 – 4 –11i j k i j kλλλλ

( ) ( ) ( )ˆ ˆ ˆ= 11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ

Then

PM = Position vector of M – Position vector of P

= ( ) ( ) ( ){ }ˆ ˆ ˆ11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ

– ( )ˆ ˆ ˆ2 – +5i j k

= ( ) ( ) ( )ˆ ˆ ˆ9 +10 + –1 – 4 + –13 –11i j kλ λ λλ λ λλ λ λλ λ λ


which parallel to ˆ ˆ ˆ=10 – 4 –11b i j k ,

PM b⊥ ∴∴∴∴ = 0PM b⋅

∴ ∴ ∴ ∴ ( ) ( ) ( )ˆ ˆ ˆ9 +10 + –1 – 4 + –13 –11i j k λ λ λλ λ λλ λ λλ λ λ

( )ˆ ˆ ˆ10 – 4 –11i j k =

∴ ∴ ∴ ∴ 10(9 + 10λλλλ) – 4(–1– 4λλλλ) – 11(–13 – 11λλλλ) = 0

∴ ∴ ∴ ∴ 90 + 100λλλλ + 4 + 16λ λ λ λ + 143 + 121λ λ λ λ = 0

∴ ∴ ∴ ∴ 237λλλλ + 237 = 0 ∴ λ∴ λ∴ λ∴ λ = –1

Putting this value of λλλλ, we get the position

vector of M as ˆ ˆ ˆ+ 2 + 3i j k .

3D-Line


∴ ∴ ∴ ∴ coordinates of the foot of perpendicular Mare (1,2,3).

Now, ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= + 2 + 3 – 2 – +5PM i j k i j k

= ˆ ˆ ˆ– +3 – 2i j k

∴∴∴∴ ( ) ( ) ( )= –1 + 3 + –2PM2 2 2

= 1+9+ 4 = 14

Hence, the coordinates of the foot of

perpendicular are (1,2,3) and length of

perpendicular = 14 units.

BASIC ASSIGNMNTS (BA) :

BA–1

Q-1) Find the equation of the line passing

through the point (5, 4, 3) and having ratios

–3, 4, 2.

Ans. Let a be the position vector of the point A(5,

4, 3) w.r.t. the origin.

∴∴∴∴ ˆ ˆ ˆ= 5 + 4 +3a i j k

Let b be the vector parallel of the line whose

direction ratios are –3, 4, 2.

Then ˆ ˆ ˆ= –3 + 4 +2b i j k


through A ( )a and parallel to ( )b is = +r a bλλλλ ,

where λλλλ is a scalar.


( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= 5 + 4 + 3 + –3 + 4 + 2r i j k i j kλλλλ


through the point (–1, –1, 2) and parallel

to the line 2x – 2 = 3y + 1 = 6z – 2.

Ans. Let a be the position vector of the point

A(–1, –1, 2) w.r.t. the origin.

∴∴∴∴ ˆ ˆ ˆ= – – + 2a i j k

The equation of given line is

2x – 2 = 3y + 1 = 6z – 2

∴∴∴∴ 2(x – 1) = 31 1

+ = 6 –3 3

y z

∴∴∴∴+ –

–1= =y z

x1 1

3 3

1 1 1

2 3 6

The direction ratios of this line are

1

2, 1

3, 1

6 i.e. 3, 2, 1

Let b be the vector parallel to this line.

Then ˆ ˆ ˆ= 3 + 2 +b i j k


through A ( )a and parallel to b is

= +r a bλλλλ where λλλλ is a scalar.


( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= – – + 2 + 3 + 2 +r i j k i j kλλλλ

Q-3) Show that the point whose position vectors

are + ˆˆ5 5i k , + – ˆˆ ˆ– 4 3i j k and + + ˆˆ ˆ2 3i j k

are collinear.

Ans. Let ˆ ˆ= 5 +5a i k

ˆ ˆ ˆ= 2 + +3b i j k

ˆ ˆ ˆ= –4 +3 –c i j k

∴∴∴∴ ˆ ˆ ˆ= –3 + – 2AB i j k

ˆ ˆ ˆ= 6 + 2 – 4BC i j k

Direction ratios of line AB are –3, 1, –2 and

direction ratios of line BC are –6, 2, –4 i.e. –3,

1, –2 Which are proportional to each other

Hence, lines are parallel But they have a

common point ie. point B.

Hence points A,B,C are collinear.


through the point with position vector

+ – ˆˆ ˆ2i j k and parallel to the line joining

the points + + ˆˆ ˆ– 2 4i j k . Also, find the

Cartesian from of this equation.

Ans. Let A,B,C be the points with position vector

ˆ ˆ ˆ= 2 + +a i j k , ˆ ˆ ˆ= – + + 4b i j k and

ˆ ˆ ˆ= +2 +2c i j k respectively.


3D-Line

Now, ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= – + 2 + 2 – – + + 4BC c b i j k i j k

ˆ ˆ ˆ ˆ ˆ ˆ= +2 +2 + – – 4i j k i j k

ˆ ˆ ˆ= 2 + – 2i j k

BC is parallel to the line passing through A.

We know the equation of the line passing

through a and parallel to b is = +r a bλλλλ

∴∴∴∴ The equation of the required l ine is

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= 2 + – + 2 + – 2r i j k i j kλλλλ ...(1)

Cartesian from:

Putting ˆ ˆ ˆ= + +r xi yj zk in equation (1), we get

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ+ + = 2+2 + 1+ + –1 – 2xi yj zk i j kλ λ λλ λ λλ λ λλ λ λ

x = 2 + 2λλλλ, y = 1 + λλλλ, z = –1 – 2λλλλ

∴∴∴∴– 2 –1 +1

= =2 1 –2

x y z

BA-2

Q-1) Find the angle between the lines

+= =

–1 1 – 4

2 2 1

x y z and

+= =

– 2 1 – 4

2 2 1

x y z

Ans. Direction ratio’s of given lines are 2, 2, 1 and

2, 2, 1

Let a and b be the vectors along these lines

respectively.

∴∴∴∴ a = ˆ ˆ ˆ2 2i j k+ +

b = ˆ ˆ ˆ2 2i j k+ +

Let θ be the ∠∠∠∠ between the lines.

Then cos θ = .a b

ab

where

.a b = ( ) ( )ˆ ˆ ˆ ˆ ˆ2 2 . 2 2i j k i j k+ + + +

= (2) + (2) + (2)(2) + (1)(1)

= 9

a = ( ) ( ) ( )2 2 22 2 1+ +

a = 9 = 3

b = ( ) ( ) ( )2 2 22 2 1+ +

b = 9 = 3

cos θ = .a b

ab =

9

3 3× = 1

Q-2) Find the foot of the perpendicular drawn

from the point �2 – +5i j k� � to the line

�(((( )))) �(((( ))))λλλλ= 11 – 2 – 8 + 10 – 4 – 11r i j k i j k� � � � . Also,

find the length of the perpendicular.


the point �( )2 – +5P i j k� � on the line

�( ) �( )= 11 – 2 – 8 + 10 – 4 –11r i j k i j kλλλλ� � � �.


�( ) �( )11 – 2 – 8 + 10 – 4 –11i j k i j kλλλλ� � � �

= ( ) ( ) ( ) �11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ� � .

Then PM = Position vector of M – Position

vector of P

= ( ) ( ) ( ) �11+10 + –2 – 4 + –8 –11i j kλ λ λλ λ λλ λ λλ λ λ

� �

�( )– 2 – +5i j k� �

= ( ) ( ) ( ) �9 +10 + –1 – 4 + –13 –11i j kλ λ λλ λ λλ λ λλ λ λ� �


which is parallel to �=10 – 4 –11b i j k� � ,

PM b⊥⊥⊥⊥′ . = 0PM b∴

∴∴∴∴ ( ) ( ) ( ) �9+10 + –1 – 4 + –13 –11i j kλ λ λλ λ λλ λ λλ λ λ

� �

�( )10 – 4 –11 = 0i j k� �

∴∴∴∴ 10(9 + 10λλλλ) –4(–1 –4λλλλ) –11(–13 – 11λλλλ) = 0

∴∴∴∴ 90 + 100λλλλ + 4 + 16λλλλ + 143 + 121λλλλ = 0

∴∴∴∴ 237λλλλ + 237 = 0 ∴∴∴∴ λλλλ = –1

3D-Line


Q-3) Find the shortest distance between the

lines = =–1 – 2 – 3

2 3 4

x y z and

= =– 2 – 4 – 5

3 4 5

x y z

Ans. The lines are –1 – 2 – 3

= =2 3 4

x y z...(i)

and– 2 – 4 – 5

= =3 4 5

x y z...(ii)

Here, x1 = 1, y

1 = 2, z

1 = 3

a1 = 2, b

1 = 3, c

1 = 4

x2 = 2, y

2 = 4, z

2 = 5

a2 = 3, b

2 = 4, c

2 = 5

Shortest distance between the lines is

( ) ( ) ( )

– – –

– + – –

x x y y z z

a b c

a b c

b c b c c a c a a b a b+

2 1 2 1 2 1

1 1 1

2 2 2

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

Now

– – –x x y y z z

a b c

a b c

2 1 2 1 2 1

1 1 1

2 2 2

=

1 2 2

2 3 4

3 4 5

= 1 (15 – 16) –2 (10 – 12) + 2(8 – 9)

= –1 + 4 – 2

= 1

and (b1c2 – b

2c1)2 + (c

1a

2 – c

2a

1)2 + (a

1b

2 – a

2b

1)2

= (15 – 16)2 + (12 – 10)2 + (8 – 9)2

= 1 + 4 + 1

= 6

Hence, the shortest distance between the

lines (i) and (ii) is

=1

6

=1

6 units.

Q-4) Find the equation of line equally inclined

to co-ordinate axes and passes through

(–5, 1, –2).

Ans. The equation of the line passing through the

point (x1, y

1, z

1) and having direction cosines

l, m, n is – – –

= =x x y y z z

l m n

1 1 1

Since the required line is equally inclined to

the coordinate axes, αααα = ββββ = γγγγ, where αααα, β β β β, γγγγ

are the angles made by the line with the

coordinate axes

∴∴∴∴ cos αααα = cos ββββ = cos γγγγ

∴∴∴∴ l = m = n

∴∴∴∴ the equation of the required line passing

through the point (–5, 1, –2) is

+5 –1 + 2= =

x y z

l l l

∴∴∴∴ x + 5 = y – 1 = z + 2

1 3D-LINE - CBSE Coaching Classes for Std. IX & X...

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Transcript of 1 3D-LINE - CBSE Coaching Classes for Std. IX & X...