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

Last updated on Jul 17, 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:

Shortest distance between the lines 

x12=y+87=z45 and x12=y21=z63 is

  1. 2√3 
  2. 4√3 
  3. 3√3 
  4. 5√3 

Answer (Detailed Solution Below)

Option 2 : 4√3 

Skew Lines Question 3 Detailed Solution

Calculation: 

x12=y+87=z45a=i^8j^+4k^

x12=y21=z63b=i^+2j^+6k^

⇒ p=2i^7j^+5k^,q=2i^+j^3k^

⇒ p×q=|i^j^k^275213|

i^(16)j^(16)+k^(16)

16(i^+j^+k^)

⇒ d = |(ab)(p×q)|p×q||=|(10j^2k^)16(i^+j^+k^)163|

|123|=43

Hence, the correct answer is Option 2.

Skew Lines Question 4:

Line L1 passes through the point (1, 2, 3) and is parallel to z-axis. Line L2 passes through the point (λ , 5, 6) and is parallel to y-axis. Let for λ = λ1 , λ ,  λ21 , the shortest distance between the two lines be 3. Then the square of the distance of the point (λ1, λ2, 7) from the line L1 is  

  1. 40
  2. 32
  3. 25
  4. 37

Answer (Detailed Solution Below)

Option 3 : 25

Skew Lines Question 4 Detailed Solution

Calculation:

We have two Lines

⇒ L1:x10=y20=z31,

⇒  L2:xλ0=y51=z60.

To find their shortest distance, form the determinant

⇒ det(λ133001010)=(λ1)det(0110)3det(0100)+3det(0001) 

=(λ1).

⇒ d1×d2=det(ijk001010)=i,i=1.

SD=|(λ1)|=|λ1|=3λ=1±3λ=4 or 2.

Foot of the perpendicular from P  to L1

Take P(λ1,λ2,7)=(4,2,7)A general point on L1 ay be written as Q ( (1,2,3+t).

Then, PQ=(14,2(2),(3+t)7)=(3,4,t4).

Perpendicularity to L1 direction d1 = (0, 0, 1) gives 

⇒ PQd1=0(3,4,t4)(0,0,1)=0t4=0t=4.

Thus, Q=(1,2,3+4)=(1,2,7).

⇒ PQ2=PQ2=(41)2+(22)2+(77)2=32+(4)2+02

=9+16=25.

Hence, the correct answer is Option 3.

Skew Lines Question 5:

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 5 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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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