Skew Lines MCQ Quiz - Objective Question with Answer for Skew Lines - Download Free PDF

Last updated on Jul 4, 2025

Latest Skew Lines MCQ Objective Questions

Skew Lines Question 1:

If the shortest distance between the line joining the points(1, 2, 3) and (2, 3, 4), and the line x12=y+11=z20 is α, then 28α2 is equal to____. 

Answer (Detailed Solution Below) 18

Skew Lines Question 1 Detailed Solution

Calculation: 

r=(i^+2j^+3k^)+λ(i^+j^+k^)r=a+λp

⇒ r=(+i^j^+2k^)+μ(2i^j^)r=b+μq

⇒ p×q=|i^j^k^111210|=i^+2j^3k^

⇒ d=|(ba)(p×q)|p×q||

⇒ d=|(3j^k^)(i^+2j^3k^)14|

=|6+314|=314

⇒ α=314

Now, 28α2 = 28× 914=18

Hence, the correct answer is 18. 

Skew Lines Question 2:

The shortest distance between the lines x + 1 = 2y = -12z and x = y + 2 = 6z – 6 is 

  1. 2
  2. 3
  3. 52
  4. 32

Answer (Detailed Solution Below)

Option 1 : 2

Skew Lines Question 2 Detailed Solution

Calculation: 

x+11=y12=z112 and x1=y+21=z116

 Shortest distance =(ba)(p×q)|p×q|

 S.D. =(i^+2j^k^)(p×q)|p×q|

{p×q|i^j^k^1121121116|=16i^14j^+12k^ or 2i^3j^+6k^}

 S.D. =(i^+2j^k^)(2i^3j^+6k^)22+32+62=|147|=2

Hence, the correct answer is Option 1.

Skew Lines Question 3:

If the square of the shortest distance between the lines x21=y12=z+33 and x+12=y+34=z+55 is mn, where m, n are coprime numbers, then m + n is equal to:

  1. 6
  2. 9
  3. 21
  4. 14

Answer (Detailed Solution Below)

Option 2 : 9

Skew Lines Question 3 Detailed Solution

Calculation

a=(2,1,3)

b=(1,3,5)

p×q=|i^j^k^123245|

2i^j^

ba=3i^4j^2k^

Sd=|(ba)(p×q)||p×q|

25

(Sd)2=45

m = 4, n = 5 ⇒ m + n = 9 

Hence option 2 is correct

Skew Lines Question 4:

Let L1x12=y23=z34 and Lx23=y44=z55 be two lines. Then which of the following points lies on the line of the shortest distance between L1 and L2

  1. (53,7,1)
  2. (2,3,13)
  3. (83,1,13)
  4. (143,3,223)

Answer (Detailed Solution Below)

Option 4 : (143,3,223)

Skew Lines Question 4 Detailed Solution

Calculation

qImage67a84eaa7da840499ed0b791

P(2λ + 1, 3λ + 2, 4λ + 3) on L1

Q(3µ + 2, 4µ + 4, 5µ + 5) on L2

Dr’s of PQ = 3µ – 2λ + 1, 4µ – 3λ + 2, 5µ – 4λ + 2

PQ L

⇒ (3µ – 2λ + 1)2 + (4µ – 3λ + 2)3 + (5µ – 4λ + 2)4 = 0

38µ – 29λ + 16 = 0 …(1)

PQ ⊥ L

⇒ (3µ – 2λ + 1)3 + (4µ – 3λ + 2)4 + (5µ – 4λ + 2)5 = 0

50µ – 38λ + 21 = 0 …(2)

By (1) & (2) 

λ=13;μ=16

∴ P(53,3,133)& Q(32,103,256)

Line PQ 

x5316y313z13316

x531=y32=z1331

 Point (143,3,223)

lies on the line PQ 

Hence option 4 is correct

Skew Lines Question 5:

If the line, x31=y+21=z+λ2 lies on the plane 2x4y+3z=2, then the shortest distance between this line and the line x112=y9=z4 is

  1. 0
  2. 2
  3. 1
  4. 3
  5. 2.5

Answer (Detailed Solution Below)

Option 1 : 0

Skew Lines Question 5 Detailed Solution

Calculation

x31=y+21=z+λ2=k

Let P be any point on the line,

P=(k+3,k2,2kλ)

P lies on the plane 2x4y+3z=2

⇒ 2(k+3)4(k2)+3(2kλ)=2

⇒ 2k+6+4k+86k3λ=2

⇒ 143λ=2

⇒ 3λ=12

⇒ λ=4

Line 1 is x31=y+21=z+42

Another line is x112=y9=z4(line 2) Shortest distance be d.

⇒ a2=|x1x2y1y2z1z2 l1m1n1 l2m2n2| where l1m1n are DRS of lines

|312040 112 1294|

⇒ |224 112 1294|=|2(14)+2(28)4(21)|=|28+5684|=0

d2=0

d=0

Hence option 1 is correct

Top Skew Lines MCQ Objective Questions

Find the magnitude of the shortest distance between the lines x02=y03=z01 and x23=y15=z+22.

  1. 13
  2. 23
  3. 15
  4. 17

Answer (Detailed Solution Below)

Option 1 : 13

Skew Lines Question 6 Detailed Solution

Download Solution PDF

Concept: 

The magnitude of the shortest distance between the lines r1=a1+λb1 and r2=a2+μb2 is 

d=|(b1×b2).(a2a1)|b1×b2||

Given:  

The lines x02=y03=z01 and x23=y15=z+22.

Rewriting the given equations,

r1=λ(2i3j+k) and r2=(2i+j2k)+μ(3i5j+2k)

a1=0 ,  b1=2i3j+k and  a2=2i+j2k,  b2=3i5j+2k

Therefore, the magnitude of the shortest distance between the given lines is

d=|(b1×b2).(a2a1)|b1×b2||

d=|[(2i3j+k)×(3i5j+2k)].[(2i+j2k)0]|(2i3j+k)×(3i5j+2k)||

d=|(ijk).(2i+j2k)]|ijk||

d=13

Therefore, the magnitude of the shortest distance between the given lines is 13.

Let L1 and L2 be two parallel lines with the equations r=a1+λb and r=a2+μb respectively. The shortest distance between them is:

  1. d=|b×(a2a1)|b||
  2. d=|b(a2a1)|b||
  3. d=|a1×(a2a1)|b||
  4. d=|a2×(a2a1)|b||

Answer (Detailed Solution Below)

Option 1 : d=|b×(a2a1)|b||

Skew Lines Question 7 Detailed Solution

Download Solution PDF

Concept:

  • If two lines are parallel, then the distance between them is fixed.
  • The distance between two parallel lines r=a1+λb and r=a2+μb is given by the formula: d=|b×(a2a1)|b||.

 

Calculation:

Using the formula for the distance between two parallel lines, we can say that the distance is d=|b×(a2a1)|b||.

Find the shortest distance between the lines x83=y+916=z107andx153=y298=z55 ?

  1. 16
  2. 14
  3. 15
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 14

Skew Lines Question 8 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between the skew line xx1a1=yy1b1=zz1c1andxx2a2=yy2b2=zz2c2 is given by:

SD=|x2x1y2y1z2z1a1b1c1a2b2c2|{(a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2}

Calculation:

Given: Equation of lines is x83=y+916=z107andx153=y298=z55

By comparing the given equations with xx1a1=yy1b1=zz1c1andxx2a2=yy2b2=zz2c2, we get

⇒ x1 = 8, y1 = - 9, z1 = 10, a1 = 3, b1 = -16 and c1 = 7

Similarly, x2 = 15, y2 = 29, z2 = 5, a2 = 3, b2 = 8 and c2 = -5

So, |x2x1y2y1z2z1a1b1c1a2b2c2|=|73853167385|

As we know that shortest distance between two skew lines is given by:SD=|x2x1y2y1z2z1a1b1c1a2b2c2|{(a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2}

⇒ SD = 14 units

Hence, option B is the correct answer.

Find the shortest distance between the lines whose vector equations are r=(3s+2)i^3j^+(4s+4)k^ and r=(3t+2)i^3j^+(4t)k^

  1. 2.4
  2. 2
  3. 1.4
  4. 1.8
  5. 0

Answer (Detailed Solution Below)

Option 1 : 2.4

Skew Lines Question 9 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between parallel lines r=a1+λb and r=a2+μb is given by: d=|b×(a2a1)|b||

Calculation:

L1r=(3s+2)i^3j^+(4s+4)k^ can be written as r=2i^3j^+4k^+s(3i^+4k^).

L2r=(3t+2)i^3j^+(4t)k^ can be written as r=2i^3j^+t(3i^+4k^).

Here, we see both lines are parallel and a1=2i^3j^+4k^ , a2=2i^3j^ and b=3i^+4k^.

 The shortest distance between parallel lines L1 and L2

d=|(3i^+4k^)×[(2i^3j^)(2i^3j^+4k^)]|3i^+4k^||

⇒ d=|(3i^+4k^)×(4k^)9+16|

⇒ d=|12j^5| ⇒ d=125=2.4 unit.

Hence, option 1 is correct.

Find the shortest distance between the lines whose vector equations are r=(3s+2)i^3j^+(4s+4)k^ and r=(3t+2)i^3j^+(4t)k^

  1. 2.4
  2. 2
  3. 1.4
  4. 0

Answer (Detailed Solution Below)

Option 1 : 2.4

Skew Lines Question 10 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between parallel lines r=a1+λb and r=a2+μb is given by: d=|b×(a2a1)|b||

Calculation:

L1r=(3s+2)i^3j^+(4s+4)k^ can be written as r=2i^3j^+4k^+s(3i^+4k^).

L2r=(3t+2)i^3j^+(4t)k^ can be written as r=2i^3j^+t(3i^+4k^).

Here, we see both lines are parallel and a1=2i^3j^+4k^ , a2=2i^3j^ and b=3i^+4k^.

 The shortest distance between parallel lines L1 and L2

d=|(3i^+4k^)×[(2i^3j^)(2i^3j^+4k^)]|3i^+4k^||

⇒ d=|(3i^+4k^)×(4k^)9+16|

⇒ d=|12j^5| ⇒ d=125=2.4 unit.

Hence, option 1 is correct.

Find the shortest distance between the lines x1=y20=z1 and x+21=y1=z0

  1. 1
  2. 3
  3. 2
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Skew Lines Question 11 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between the lines xx1a1=yy1b1=zz1c1  and xx2a2=yy2b2=zz2c2 is given by:d=|x2x1y2y1z2z1a1b1c1a2b2c2|(b1c2b2c1)2+(c1a2c2a1)2+(a1b2a2b1)2

Calculation:

Here we have to find the shortest distance between the lines ​​x1=y20=z1 and x+21=y1=z0

Let line L1 be represented by the equation x1=y20=z1 and line L2 be represented by the equation x+21=y1=z0

⇒ x1 = 0, y1 = 2, z1 = 0  and a1 = -1, b1 = 0, c1 = 1.

⇒ x2 = -2, y2 = 0, z2 = 0  and a2 = 1, b2 = 1, c2 = 0.

∵ The shortest distance between the lines is given by:  d=|x2x1y2y1z2z1a1b1c1a2b2c2|(b1c2b2c1)2+(c1a2c2a1)2+(a1b2a2b1)2

⇒ d=|200200101110|(01)2+(01)2+(10)2    

⇒ d=|220101110|1+1+1

⇒ d = 0

Hence, option 4 is correct.

If the shortest distance between parallel lines r=i^+2k^+λ(i^+2j^+3k^) and r=i^+2j^+2k^+λ(i^+2j^+3k^). is k7 then k?

  1. 8
  2. 40
  3. 10
  4. 20

Answer (Detailed Solution Below)

Option 4 : 20

Skew Lines Question 12 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between parallel lines r=a1+λb and r=a2+μb is given by: d=|b×(a2a1)|b||

Calculation:

Given: Equation of lines r=i^+2k^+λ(i^+2j^+3k^)  and r=i^+2j^+2k^+λ(i^+2j^+3k^).

So, by comparing the above equations with r=a1+λb and r=a2+μb we get

⇒ a1=i^+2k^ , a2=i^+2j^+2k^  and b=i^+2j^+3k^.

 The shortest distance between parallel lines r=a1+λb and r=a2+μb is given by: d=|b×(a2a1)|b||

d=|(i^+2j^+3k^)×[(i^+2j^+2k^)(i^+2k^)]|i^+2j^+3k^||

⇒ d=|(i^+2j^+3k^)×2j^|1+4+9||

⇒ d=|2k^6i^14|

⇒ d=4014

d=4014  d=207

⇒ k = 20

Hence, option 4 is correct.

Find the shortest distance between the lines x+34=y63=z2and x+24=y1=z71

  1. 6
  2. 7
  3. 9
  4. 11

Answer (Detailed Solution Below)

Option 3 : 9

Skew Lines Question 13 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between the skew line xx1a1=yy1b1=zz1c1andxx2a2=yy2b2=zz2c2 is given by:

SD=|x2x1y2y1z2z1a1b1c1a2b2c2|{(a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2}

Calculation:

Given: Equation of lines is x+34=y63=z2and x+24=y1=z71

By comparing the given equations with xx1a1=yy1b1=zz1c1andxx2a2=yy2b2=zz2c2, we get

⇒ x1 = - 3, y1 = 6, z1 = 0, a1 = - 4, b1 = 3 and c1 = 2

Similarly, x2 = - 2, y2 = 0, z2 = 7, a2 = - 4, b2 = 1 and c2 = 1

So, |x2x1y2y1z2z1a1b1c1a2b2c2|=|167432411|

Similarly, {(a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2}=81=9

As we know that shortest distance between two skew lines is given by:SD=|x2x1y2y1z2z1a1b1c1a2b2c2|{(a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2}

SD=|167432411|9=819=9

The shortest distance between the lines

x44=y+25=z+33 and x13=y34=z42 is

  1. 2√ 6
  2. 36
  3. 63
  4. 62

Answer (Detailed Solution Below)

Option 2 : 36

Skew Lines Question 14 Detailed Solution

Download Solution PDF

Concept -

Shortest distance between two lines is:

d = |(a2a1)(b1×b2)|b1×b2||

Explanation -

The given lines are :

x44=y+25=z+33 and x13=y34=z42

So, b1=4i^+5j^+3k^

b2=3i^+4j^+2k^

a1=4i^2j^3k^a2=i^+3j^+4k^

∴ b1×b2=|i^j^k^453342|

(1012)i^(89)j^+(1615)k^

2i^+j^+k^

Shortest distance, d=|(a2a1)(b1×b2)|b1×b2||

|(3i^5j^7k^)(2i^+j^+k^)4+1+1|

|6576|=186=36 units

Hence Option (2) is correct.

Find the shortest distance between the lines x57=y+25=z1 and x1=y2=z3

  1. 542
  2. 2
  3. 942
  4. 1142
  5. 3

Answer (Detailed Solution Below)

Option 3 : 942

Skew Lines Question 15 Detailed Solution

Download Solution PDF

Concept:

The shortest distance between the lines xx1a1=yy1b1=zz1c1  and xx2a2=yy2b2=zz2c2 is given by:d=|x2x1y2y1z2z1a1b1c1a2b2c2|(b1c2b2c1)2+(c1a2c2a1)2+(a1b2a2b1)2

Calculation:

Here we have to find the shortest distance between the lines ​​x57=y+25=z1 and x1=y2=z3

Let line L1 be represented by the equation x57=y+25=z1 and line L2 be represented by the equation x1=y2=z3

⇒ x1 = 5, y1 = -2, z1 = 0  and a1 = 7, b1 = -5, c1 = 1.

⇒ x2 = 0, y2 = 0, z2 = 0  and a2 = 1, b2 = 2, c2 = 3.

∵ The shortest distance between the lines is given by:  d=|x2x1y2y1z2z1a1b1c1a2b2c2|(b1c2b2c1)2+(c1a2c2a1)2+(a1b2a2b1)2

 

⇒ d=|050+200751123|(152)2+(121)2+(14+5)2

⇒ d=|520751123|(17)2+(20)2+(19)2

d=5(152)2(211)289+400+361

d=85401050=942

Hence, option 3 is correct.

Get Free Access Now
Hot Links: teen patti gold download teen patti gold old version teen patti club teen patti boss