Shortest Distance Between Two Lines in 3D Geometry Class 12 OP Malhotra Exe-23F ISC Maths Solutions Ch-23. In this article you would learn how to calculate shortest distance between two lines in Three Dimensional Geometry using formula. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Shortest Distance Between Two Lines in 3D Geometry Class 12 OP Malhotra Exe-23F ISC Maths Solutions Ch-23

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-23	Three Dimensional Geometry
Writer	OP Malhotra
Exe-23(f)	Shortest Distance Between Two Lines

Practice Questions / Problems on Shortest Distance Between Two Lines in 3D Geometry Using Formula

Three Dimensional Geometry Class 12 OP Malhotra Exe-23F Solutions

Que-1: x−3/1 = y−5/−2 = z−7/1 and x+1/7 = y+1/−6 = z+1/1

Sol: In Cartesian form, these eqns. can be written as
x−3/1 = y−5/−2 = z−7/1
and
x+1/7 = y+1/−6 = z+1/1
Any point on line (1) is P(t + 3, -2 t + 5, t + 7)
and any point on line (2) is Q(7 s – 1, -6 s – 1, s – 1)
∴ D’ratios of line PQ are < 7 s – t – 4, -6 s + 2 t – 6, s – t – 8 >
Now line PQ is ⊥ to line (1).
∴ (7 s – t – 4) 1 + (-6 s + 2 t – 6)(-2) + (s – t + 8) 1 = 0
⇒ 20 s – 6 t = 0
also line PQ is ⊥ to line (2)
∴ (7 s – t – 4) + 7 + (-6 s + 2 t – 6)(-6) + (s – t – 8) 1 = 0
⇒ 86 s – 20 t = 0
On solving (3) and (4); s = t = 0
∴ points on line (1) is P(3, 5, 7) and on line (2) be, Q}
(-1, -1, -1)
Thus |P Q| =√(3+1)²+(5+1)²+(7+1)²
= √16+36+64 = √116 = 2√29

Que-2: x+3/−4 = y−6/3 = z/2 and x+2/−4 = y/1 = z−7/1

Sol: Given eqns of lines are
x+3/−4 = y−6/3 = z/2
and
x+2/−4 = y/1 = z−7/1
Let < l, m, n> be the direction cosines of the line of shortest distance. Now line of shortest distance P Q is ⊥ to both given lines.
-4 l + 3 m + 2 n = 0
-4 l + m + n = 0
Using cross-multiplication method, we have
l/3−2 = m/−8+4 = n/−4+12
∴ l/1 = m/−4 = n/8
= 1/9
i.e., l/1 = m/−4
= n/8 = √l²+m²+n²/√1+16+81
∴ length of S.D. = projection of line joining A(-3, 6, 0) and B(-2, 0, 7) on line PQ
= |l(x₂−x₁)+m(y₂−y₁)+n(z₂−z₁)|
= |1/9(−2+3)−4/9(0−6)+8/9(7−0)|
= |1/9+24/9+56/9|
= 9 units
Aliter: We know that shortest distance between lines
x−x₁/l₁
= y−y₁/m₁
= z−z₁/n₁ and x−x₂/l₂
= y−y₂/m₂
= z−z₂/n₂
is given by |x₂−x₁y₂−y₁z₂−z₁l₁m₁n₁l₂m₂n₂|/√(l₁m₂−l₂m₁)²+(m₁n₂−m₂n₁)²+(n₁l₂-l₁n₂)
Here,
x₁ = -3 ; y1 = 6 ; z1 = 0 ;
l₁ = -4 ; m1 = 3 ; n1 = 2
x₂ = -2 ; y2 = 0 ; z2 = 7 ;
l₂ = -4 ; m2 = 1 ; n2 = 1
∴ Required S.D. =(1/√(−4+12)²+(3−2)²+(−8+4)²)|−2+30−67−0−432−411|
= 1/√64+1+16
= |1−67−432−411|;
Expanding along R1
= 1/9[1(3 – 2) + 6(-4 + 8) + 7(-4 + 12)]
= 1/9[1 + 24 + 56]
= 9 units
Aliter : any point on line (1) be
x+3/−4
= y−6/3
= z/2 = t (say)
i.e., P(-4 t – 3, 3 t + 6, 2 t)
and any point on line (2) be
x+2/−4
= y/1
= z−7/1 = r (say)
i.e., Q(-4 r – 2, r, r + 7)
Then line PQ be the line of shortest distance
∴ direction ratios of line PQ are
< -4 r – 2 + 4 t + 3, r – 3 t – 6, r + 7 – 2 t >
i.e., < -4 r + 4 t + 1, r – 3 t – 6, r – 2 t + 7 >
Now, line PQ be the line of S.D. and it is ⊥ to line (1) and (2).
∴ (-4 r + 4 t + 1)(-4) + (r – 3 t – 6) 3 + (r – 2 t + 7) 2 = 0
i.e., 16 r – 16 t – 4 + 3 r – 9 t – 18 + 2 r – 4 t + 14 = 0
⇒ 21 r – 29 t – 8 = 0
and (-4 r + 4 t + 1)(-4) + (r – 3 t – 6)1 + (r – 2 t + 7) 1 = 0
i.e., 16 r – 16 t – 4 + r – 3 t – 6 + r – 2 t + 7 = 0
⇒ 18 r – 21 t – 3 = 0
⇒ 6 r – 7 t – 1 = 0
On solving eqn. (1) and eqn. (2); we have
r/29−56
= t/−48+21 = 1/−147+174
i.e., r/−27 = t/−27
= 1/+27
i .e., r = -1 ; t = -1
Thus, coordinates of P and Q becomes (1, 3, -2) and (2, -1, 6)
∴ |PQ| = √(2−1)²+(−1−3)²+(6+2)²
= √1+16+64 =9 units

Find the magnitude and the equations of the line of shortest distance between the two lines:

Que-3: x−3/−1 = y−4/2 = z+2/1 and x−1/1 = y+7/2 = z+2/1

Sol: The equations of given lines are
and
x−3/−1 = y−4/2
= z+2/1 = t (say)
x−1/1 = y+7/3 = z+2/2 = r (say)
So, any points on lines (1) and (2) are P(-t + 3, 2 t + 4, t – 2) and Q(r + 1, 3 r – 7, 2 r – 2)
Then PQ be the line of S.D. if it is ⊥ to both line (1) and (2).
∴ D ratios of line PQ are < r + 1 + t – 3, 3 r – 7 – 2 t – 4, 2 r – 2 – t + 2)
i.e., < r + t – 2, 3 r – 2 t – 11, 2 r – t >
Also, line PQ is ⊥ to line (1) and line (2).
∴ -1(r + t – 2) + 2(3 r – 2 t – 11) + 1(2 r – t) = 0
⇒ 7 r – 6 t – 20 = 0
and 1(r + t – 2) + 3(3 r – 2 t – 11) + 2(2 r – t) = 0
⇒ 14 r – 7 t – 35 = 0
i.e., 2 r – t – 5 = 0
On solving eqn. (3) and eqn. (4); we have
∴ r = 2, t = -1
r/30−20 = t/−40+35
= 1/−7+12
i.e., r/10
= t/−5
= 1/5
Hence, the coordinates of P and Q are
(1 + 3, -2 + 4, -1 – 2)
i.e., (4, 2, -3) and (2 + 1, 6 – 7, 4 – 2)
i.e., (3, -1, 2)
∴ required S.D. between lines = |P Q| = √(3−4)²+(−1−2)²+(2+3)²
= √1+9+25
= √35 units
∴ D ratios of line PQ be < 3 – 4, -1 – 2, 2 + 3 >
i.e., < -1 , -3, 5 >
i.e., < 1, 3, -5 >
Hence the required eqnn. of line of S.D. which pass through (4, 2, -3) and having direction ratios < 1, 3, -5 > is given by
x−4/1
= y−2/3
= z+3/−5

Que-4: Obtain the coordinates of the points where the shortest distance between the following lines meets them:
x−23/−6 = y−19/−4 = z−25/3;
x−12/−9 = y−1/4 = z−5/2

Sol: Given eqns. of lines are
x−23/−6 = y−19/−4 = z−25/3 = t (say)
and
x-12/-9 = y-1/4 = z-25/2 = r(say)
So, any point on lines (1) and (2) are
P(-6 t + 23, -4 t + 19, 3 t + 25) and Q(-9 + 12, 4 r + 1, 2 r + 5)
Then PQ be the line of shortest distance if PQ be perpendicular to both lines (1) and (2).
∴ direction ratios of line PQ are
i.e., < -9 r + 12 + 6 t – 23, 4 r + 1 + 4 t – 19, 2 r + 5 – 3 t – 25 >
< -9 r + 6 t – 11, 4 r + 4 t – 18, 2 r – 3 t – 20 >
Since, PQ is ⊥ to both lines (1) and (2).
∴ (-9 r + 6 t – 11)(-6) + (4 r + 4 t – 18)(-4) + (2 r – 3 t – 20) 3 = 0
⇒ 54 r – 36 t + 66 – 16 r – 16 t + 72 + 6 r – 9 t – 60 = 0
⇒ 44 r – 61 t + 78 = 0
and (-9 r + 6 t – 11)(-9) + (4 r + 4t – 18) 4 + (2 r – 3 t – 20) 2 = 0
101 r – 44 t – 13 = 0
On solving eqn. (3) and (4); we have
r = 1, t = 2
Thus, required coordinates of points be
(-12 + 23, -8 + 19, 6 + 35)
i.e., (11, 11, 3) and (-9 + 12, 4 + 1, 2 + 5)
i.e., (3, 5, 7).

Find the shortest distance between the pairs of lines whose equations are

Que-5: r→ = 6î + 2ĵ + 2k̂ + λ(î – 2ĵ + 2k̂) and r →= -4î – k̂ + µ(3 j^ – 2ĵ – 2k̂)

Sol: Comparing given eqns. with  → r₁  = → a₁ + λ→ b₁ and → r₂ = → a₂ + λ→ b₂; we have
→ a₁ = 6î + 2ĵ + 2k̂;
→ a₂ = -4î – k̂
→ b₁ = î + 2ĵ + 2k̂;
→ b₂ = 3î – 2ĵ – 2k̂

Que-6: r→ = 2î – 5ĵ + k̂ + µ(3î + 2ĵ + 6k̂) and r →= 7î – 6k̂ + µ(î + 2ĵ + 2k̂)

Sol: Given equations of lines are
r → = 2î– 5ĵ +k̂ + λ(3î + 2ĵ + 6k̂) and r →
= 7î – 6k̂ + µ(î + 2ĵ + 2k̂)
On comparing with r⃗ = →a₁ + λ→b₁ and r →
= →a₂ + λ→b₂
Here,
Here, →a₁ = 2î – 5ĵ + k̂;
→b₁ = 3î + 2ĵ + 6k̂
→a₂ = 7î – 6k̂;
→b₂ = î + 2ĵ + 2k̂
∴→a₂ – →a₁
= (7î – 6k̂) – (2î – 5ĵ + k̂)
= 5î + 5ĵ – 7k̂
→b₁ × →b₂ = |îĵk̂326122|;
Expanding along R1 = î(4 – 12) – ĵ(6 – 6) + k̂(6 – 2)
= -8î + 0ĵ + 4k̂
∴ |→b₁ × →b₂| = √(−8)²+0+4²
= √64+16 = √80 = 4√5
Now,
(→a₂ – →a₁) (→b₁ × →b₂)
= (5î + 5ĵ – 7k̂) (-8î + 0ĵ+ 4k̂)
= 5(-8) + 5(0) – 7(4) = -68
Thus required S.D. between them = |(→a₂−→a₁)(→b₁ × →b₂)/|→b₁ × →b₂|
= |−68|/4√5
= 17/√5 units

Que-7: r → = (8 + 3λ)î – (9 + 16λ)ĵ + (10 + 7λ)k̂ and r → = 15î + 29ĵ + 5k̂ + µ(3î + 8ĵ – 5k̂)

Sol: r → = (8î – 9ĵ + 10k̂) + λ(3î – 16ĵ + 7k̂) and r⃗ = (15î + 29ĵ + 5 k^) + µ(3î + 8ĵ – 5k̂)
On comparing given eqns. with
r → = →a₁ + λb1→ and r⃗ = →a₂ + λ→b₂
∴ →a₁ = 8î – 9ĵ + 10k̂;
→a₂ = 15î + 29ĵ + 5k̂;
→b₁ = 3î – 16ĵ + 7 k̂;
→b₂ = 3î + 8ĵ – 5k̂
∴ →a₂ – →a₁ = 7î + 38ĵ – 5k̂
→b₁ × →b₂ = |îĵk̂3−16738−5|
= 24î + 36ĵ+ 72k̂
= 12(2î + 3ĵ + 6k̂)
∴ |→b₁ × →b₂|
= 12√4+9+36 = 12 × 7 = 84
and

= 1/7 [2(7) + 3(38) + 6(-5)]
= 1/7[14 + 114 – 30]
= 98/7 = 14

Que-8: Show that the lines r → = (î – ĵ) + λ(2î + k̂) and r → =(2î – ĵ) + µ(î + ĵ – k̂) do not intersect.

Sol: Given lines are
r → = (î – ĵ) + λ(2 i^ + k̂)
and
r → = (2î – ĵ) + µ(î + ĵ – k̂)
Comparing eqn. (1) and (2) with r → = →a₁ + λ→b₁;
r → = →a₂ + λ→b₂
where, a1→ = î – ĵ;
→b₁ = 2î + k̂;
→a₂ = 2î – ĵ;
→b₂ =î + ĵ – k̂
Here →a₂ – →a₁ = 2î – ĵ – î + ĵ = î + 0ĵ
and
→b₁ ×→ b₂= |îĵk̂20111−1| = î(0 – 1) – ĵ(-2 – 1) + k̂(2 – 1)
= −î + 3ĵ + 2k̂
Here, (→a₂ – →a₁) (→b₁ × →b₂)
= (î + 0k̂) (−î + 3ĵ + 2k̂) = -1 ≠ 0
Thus given lines do not intersects.

Que-9: By computing the shortest distance determine whether the following pairs of lines intersect or not.
(i) r →= (î + ĵ – k̂) + λ(3î – ĵ) and r →
= (4î – k̂) + µ(2î + 3ĵ)
(ii) x−5/4 = y−7/−5 = z+3/−5, x−8/7 = y−71 = z−53
(iii) r → = (3 – t)î + (4 + 2 t)ĵ + (t – 2)k̂,
r → = (1 + s)î + (3s – 7)ĵ + (2s – 2)k̂

Sol: (i) Given lines are, r → = (î + ĵ – k̂) + λ(3î – ĵ)
and
r → = (4î – k̂) + (2î + 3k̂)
Comparing eqn. (1) and eqn. (2) with r →
= →a₁ + λ→b₁;
r → = →a₂ + λ→b₂
Here
→a₁ = î + ĵ – k̂;
→b₁ = 3î – ĵ;
→a₂ = 4î + 0ĵ – k̂;
→b₂ = 2î + 0ĵ + 3k̂
→a₂ – →a₁
= (4î + 0ĵ – k̂) – (î + ĵ – k̂)
= 3î – ĵ + 0k̂
→b₁ × →b₂ = |îĵk̂3−10203|
= î(-3 – 0) – ĵ(9 – 0) + k̂(0 + 2)
= -3î + 9ĵ + 2k̂
∴ (→a₂ – →a₁) (→b₁ × →b₂)
= (3î – ĵ + 0k̂) (-3î – 9ĵ + 2k̂)
= 3(-3) – 1(-9) + 0(2) = -9 + 9 = 0
Thus S.D. = |(→a₂−→a₁)⋅(→b₁×→b₂)/|→b₁×→b₂|| = 0
Thus the given lines are intersecting.

(iii) Eqn’s of given lines are r → = 3î + 4ĵ – 2k̂ + t(−î + 2ĵ + k̂) and r → = î – 7ĵ – 2k̂ + r(î + 3ĵ + 2k̂)
On comparing with
→r₁ = →a₁ + λ→b₁ and →r₂ = →a₂ + µ→b₂
Here,
→a₁ = 3î + 4ĵ– 2k̂;
→b₁ = −î + 2ĵ + k̂
→a₂ = î – 7ĵ – 2k̂;
→b₂ = î + 3ĵ + 2k̂
∴ →a₂ – →a₁ = -2î – 11ĵ + 0k̂
→b₁ ×→b₂
= |îĵk̂−121132|
= î(4 – 3) – ĵ(-2 – 1) + k̂(-3 – 2)
= î + 3ĵ – 5k̂
∴ |→b₁ × →b₂| = √1+9+25
= √35
∴ (→a₂ – →a₁) (→b₁ × →b₂)
= (-2î – 11ĵ + 0k̂) (î + 3ĵ – 5k̂)
= 2(1) – 11(3) + 0(-5) = -35
∴ S.D. between given lines = |((→a₂ – →a₁)⋅(→b₁ × →b₂)|/|→b₁ × →b₂|
= |−35|/√35 = √35 ≠ 0
Hence, the given lines do not intersects.

Que-10: Find the shortest distance between the following pairs of parallel lines:
(i) r → = (î + 2ĵ + 3k̂) + λ(î – ĵ +k̂) and r → = (2î – ĵ – k̂) + µ(−î + ĵ – k̂)
(ii) r → = (î + ĵ) + λ(2î – ĵ + k̂) and r → = (2î + ĵ – k̂) + µ(4î – 2ĵ + 2k̂)

Sol: (i) Given lines are
r → = (î + 2ĵ + 3k̂) + λ(î – ĵ + k̂)
r → = (2î – ĵ – k̂) + µ(−î + ĵ – k̂)
⇒ r → = (2î –ĵ – k̂) + µ(î – ĵ + k̂)
where µ = -µ
Hence both given lines are || to b⃗ = î – ĵ + k̂
Here →a₁ = î + 2 ĵ + 3k̂;
→a₂= 2î – ĵ – k̂
Now →a₂ – →a₁
= (2î – ĵ – k̂) – (î + 2ĵ + 3k̂)
= î – 3ĵ – 4k̂
∴ (→a₂ – →a₁) × → b
= î(-3 – 4) – ĵ(1 + 4) + k̂(-1 + 3)
= -7î – 5ĵ + 2k̂
|(→a₂ – →a₁) × →b| = √(−7)²+(−5)²+2²
= √49+25+4
= √78
|→b| = √1²+1²+1²
= √3
∴ required S.D. = |(→a₂ – →a₁)×→ b|/|→ b|
= √78/√3 = √26

(ii) The equation of given lines are;
r → = (î + ĵ) + λ(2î – ĵ + k̂)
r → = (2î + ĵ – k^) + µ(4î – 2ĵ + 2k̂)
⇒ r → = ( 2î + ĵ – ĵ) + µ(2î – ĵ +k̂)
where µ = 2µ
Clearly given lines (1) and (2) are parallel and || to vector → b= 2î – ĵ + k̂
Here →a₁ = î + ĵ;
→a₂ = 2î + ĵ – k̂
∴ →a₂ – →a₁ = î + 0ĵ – k̂;
(→a₂ – →a₁) × → b =
= î(0 – 1) – ĵ(1 + 2) + k̂(-1 – 0)
= −î – 3ĵ – k̂
∴ |(→a₂ – →a₁) × → b | = √(−1)²+(−3)²+(−1)²
= √11
|→ b| = √2²+(−1)²+(1)² = √6
and
|→ b| = √2²+(−1)²+(1)² = √6
Thus required S.D. = ∣(→a₂ – →a₁)×→ b∣/∣→b∣
= √11/√6

Que-11: Define the line of shortest distance between two skew lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
(i) r → = (8 + 3λ)î – (9 + 16λ)ĵ + (10 + 7λ)k̂ and r → = 15 î + 29ĵ + 5k̂ + µ(3î + 8ĵ – 5k̂)
(ii) r → = (3î + 8ĵ + 3k̂) + λ(3î – ĵ + k̂) and r →
= (-3µ – 3)î + (2µ – 7)ĵ + (4µ + 6)k̂

Sol: Line of shortest distance : Let l and m be two skew lines then there is one and only one line p which is ⊥ to l and m. Then line p is called line of shortest distance.

Aliter
(i) Given equs. can be written as
r → = 8î – 9ĵ + 10k̂ + λ(3î – 16ĵ + 7k̂) and r⃗
= (15î + 29ĵ + 5k̂) + µ(3î + 8ĵ -5k̂)
and in Cartesian form are;
x−8/3 = y+9/−16
= z−10/7 = t (say)
x−15/3 = y−29/8
= z−5/−5 = r (say)
Any point on lines (1) and (2) are given by
P(3 t + 8, -16 t – 9,7 t + 10)
Q(3 r + 15, 8 r + 29), -5 r + 5)
and
Q(3 r + 15, 8 r + 29), -5 r + 5)
Then PQ be the line of S.D. if it is ⊥ to both lines (1) and (2)
Now direction ratios of P Q are
< 3 r + 15 – 3 t – 8, 8 r + 29 + 16 t + 9, -5 r + 5 – 7 t – 10 >
i.e. < 3 r – 3 t + 7, 8 r + 16 t + 38, -5 r – 7 t – 5 >
Since P Q is ⊥ to both lines (1) and (2).
(3 r – 3 t + 7) 3 + (8 r + 16 t + 38)(-16) + (-5 r – 7 t – 5) 7 = 0
⇒ -154 r – 314 t – 622 = 0
⇒ 77 r + 157 t + 311 = 0
Also, (3 r – 3 t + 7) 3 + (8 r + 16 t + 38) 8 + (-5 r – 7 t – 5)(-5) = 0
⇒ 98 r + 154 t + 350 = 0
⇒ 49 r + 77 t + 175 = 0
On solving eqn. (3) and (4); we have
r = -2 ; t = -1
Hence the coordinates of P and Q are:
P(-3 + 8, 16 – 9, -7 + 10)
i.e. P(5, 7, 3)
and Q(-6 + 15, -16 + 29, 10 + 5)
i.e. Q(9, 13, 15)
∴ required S.D. between lines = |P Q|
= √(9−5)²+(13−7)²+(15−3)²
= √16+36+144
= √196 = 14 units
∴ direction ratios of line PQ are
< 9 – 5, 13 – 7, 15 – 3 >
i.e. < 4, 6, 12 >
i.e. < 2, 3, 6 >
Hence, required vector eqn. of line of S.D. which pass through (5, 7, 3) and having direction ratio < 2, 3, 6 > is given by r → = (5î + 7ĵ + 3k̂) + λ(2î + 3ĵ + 6k̂)

(ii) Given equations in cartesian form are;
x−3/3 = y−8/−1 = z−31 = t (say)
and x+3/−3 = y+7/2 = z−6/4 = r (say)
So any points on line (1) and (2) are
P(3 t + 3, -t + 8, t + 3)
and Q(-3 r – 3, 2 r – 7, 4 r + 6)
Then line PQ be the line of S.D. if it is ⊥ to both lines (1) and (2);
Now direction ratios of line P Q are
< -3 r – 3 – 3 t – 3, 2 r – 7 + t – 8, 4 r + 6 – t – 3 >
i.e. < -3 r – 3 t – 6, 2 r + t – 15, 4 r – t + 3 >
Since line PQ is ⊥ to line (1)
(-3 r – 3 t – 6) 3 + (2 r + t – 15)(-1) + (4 r – t + 3) 1 = 0
⇒ -7 r – 11 t = 0
Also line PQ is ⊥ to line (2)
(-3 r – 3 t – 6)(-3) + (2 r + t – 15) 2 + (4 r – t + 3) 4 = 0
⇒ +29 r + 7 t = 0
On solving eqn. (3) and eqn. (4); we have
r = t = 0
Hence coordinates of P and Q are P(3, 8, 3) and Q(-3, -7, 6)
∴ required S.D. between lines = |P Q| = √(−3−3)²+(−7−8)²+(6−3)²
= √36+225+9
= √270
= √30×9
= 3√30 units
and vector eqn. of S.D. which pass through the point whose P.V. be 3î + 8ĵ + 3k̂ and || to vector b → = (-3 – 3)î + (-7 – 8)ĵ + (6 – 3)k̂ is given by
= -6î – 15ĵ + 3k̂
r → = (3î + 8ĵ + 3k̂) + λ(-6î – 15ĵ + 3k̂)
⇒ r → = (3î + 8ĵ + 3k̂) + µ(2î + 5ĵ – k̂)
where µ = -2λ

–: End of Shortest Distance Between Two Lines in 3D Geometry Class 12 OP Malhotra Exe-23F ISC Maths Solutions :–

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