Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

3D Geometry question

2021 · 16 Mar · Shift 2 · Q27
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /3D Geometry
  5. /2021 · 16 Mar · Shift 2 · Q27

3D Geometry question

2021 · 16 Mar · Shift 2 · Q27

JEE MainMathematics3D GeometryMCQ+4 / −1
If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4{L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}L1​:lx−a​=3y−2​=4z−b​, l eee 0 is (3, 5, 7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55{L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}L2​:3x−2​=4y−4​=5z−5​ is equal to :
  1. A
    16{1 \over {\sqrt 6 }}6​1​
  2. B
    12{1 \over 2}21​
  3. C
    13{1 \over {\sqrt 3 }}3​1​
  4. D
    23\sqrt {{2 \over 3}}32​​
View written solutionFree

Correct answer: A

  1. Use the foot of perpendicular information to determine line L1L_1L1​.

The line is

L1:x−al=y−23=z−b4L_1: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4}L1​:lx−a​=3y−2​=4z−b​

So a point on L1L_1L1​ is (a,2,b)(a,2,b)(a,2,b) and its direction vector is

d⃗1=(l,3,4).\vec d_1=(l,3,4).d1​=(l,3,4).

We are given:

  • external point P=(4,3,8)P=(4,3,8)P=(4,3,8)
  • foot of perpendicular on L1L_1L1​ is Q=(3,5,7)Q=(3,5,7)Q=(3,5,7)

Since QQQ lies on L1L_1L1​, and PQPQPQ is perpendicular to L1L_1L1​, we use both conditions.

  1. Direction of the perpendicular segment.
PQ⃗=Q−P=(3−4, 5−3, 7−8)=(−1,2,−1).\vec{PQ}=Q-P=(3-4,\,5-3,\,7-8)=(-1,2,-1).PQ​=Q−P=(3−4,5−3,7−8)=(−1,2,−1).

Because PQ⊥L1PQ \perp L_1PQ⊥L1​, we must have

PQ⃗⋅d⃗1=0.\vec{PQ}\cdot \vec d_1=0.PQ​⋅d1​=0.

That is,

(−1,2,−1)⋅(l,3,4)=0(-1,2,-1)\cdot (l,3,4)=0(−1,2,−1)⋅(l,3,4)=0 −l+6−4=0-l+6-4=0−l+6−4=0 −l+2=0-l+2=0−l+2=0 l=2.\boxed{l=2}.l=2​.

Hence the direction vector of L1L_1L1​ is

d⃗1=(2,3,4).\vec d_1=(2,3,4).d1​=(2,3,4).
  1. Find a point on L1L_1L1​.

Since Q=(3,5,7)Q=(3,5,7)Q=(3,5,7) lies on L1L_1L1​,

3−a2=5−23=7−b4.\frac{3-a}{2}=\frac{5-2}{3}=\frac{7-b}{4}.23−a​=35−2​=47−b​.

Now

5−23=1.\frac{5-2}{3}=1.35−2​=1.

So

3−a2=1⇒3−a=2⇒a=1,\frac{3-a}{2}=1 \Rightarrow 3-a=2 \Rightarrow a=1,23−a​=1⇒3−a=2⇒a=1,

and

7−b4=1⇒7−b=4⇒b=3.\frac{7-b}{4}=1 \Rightarrow 7-b=4 \Rightarrow b=3.47−b​=1⇒7−b=4⇒b=3.

Therefore,

L1:x−12=y−23=z−34.L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}.L1​:2x−1​=3y−2​=4z−3​.

A point on L1L_1L1​ is A=(1,2,3)A=(1,2,3)A=(1,2,3).

  1. Write line L2L_2L2​ in vector form.

Given

L2:x−23=y−44=z−55.L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}.L2​:3x−2​=4y−4​=5z−5​.

So a point on L2L_2L2​ is

B=(2,4,5),B=(2,4,5),B=(2,4,5),

and direction vector is

d⃗2=(3,4,5).\vec d_2=(3,4,5).d2​=(3,4,5).
  1. Shortest distance between two skew lines.

For lines through points A,BA,BA,B with direction vectors d⃗1,d⃗2\vec d_1,\vec d_2d1​,d2​, the shortest distance is

D=∣(AB→)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\overrightarrow{AB})\cdot (\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(AB)⋅(d1​×d2​)∣​.

Here,

AB→=B−A=(2−1,4−2,5−3)=(1,2,2).\overrightarrow{AB}=B-A=(2-1,4-2,5-3)=(1,2,2).AB=B−A=(2−1,4−2,5−3)=(1,2,2).

Compute the cross product:

d⃗1×d⃗2=∣i^j^k^234345∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & 4\\ 3 & 4 & 5 \end{vmatrix}d1​×d2​=​i^23​j^​34​k^45​​ =i^(15−16)−j^(10−12)+k^(8−9)=\hat i(15-16)-\hat j(10-12)+\hat k(8-9)=i^(15−16)−j^​(10−12)+k^(8−9) =(−1,2,−1).=(-1,2,-1).=(−1,2,−1).

Its magnitude is

∣d⃗1×d⃗2∣=(−1)2+22+(−1)2=6.|\vec d_1\times \vec d_2|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt{6}.∣d1​×d2​∣=(−1)2+22+(−1)2​=6​.

Now

(AB→)⋅(d⃗1×d⃗2)=(1,2,2)⋅(−1,2,−1)(\overrightarrow{AB})\cdot (\vec d_1\times \vec d_2) =(1,2,2)\cdot(-1,2,-1)(AB)⋅(d1​×d2​)=(1,2,2)⋅(−1,2,−1) =−1+4−2=1.=-1+4-2=1.=−1+4−2=1.

Therefore,

D=∣1∣6=16.D=\frac{|1|}{\sqrt{6}}=\boxed{\frac{1}{\sqrt{6}}}.D=6​∣1∣​=6​1​​.
  1. Check options.
  • A: 16\dfrac{1}{\sqrt6}6​1​ ✅
  • B: 12\dfrac1221​ ❌
  • C: 13\dfrac{1}{\sqrt3}3​1​ ❌
  • D: 23\sqrt{\dfrac23}32​​ ❌

So the correct option is

A.\boxed{\text{A}}.A​.
PreviousNext

More from 3D Geometry

  • The lines x = ay − 1 = z − 2 and x = 3y − 2 = bz − 2, (ab e 0) are coplanar, if :2021 · MCQ
  • If the shortest distance between the straight lines 3(x−1)=6(y−2)=2(z−1) and 4(x−2)=2(y−λ)=(z−3),λ∈R is 38​1​, then the integral value of λ is equal to :2021 · MCQ
  • Let a, b ∈ R. If the mirror image of the point P(a, 6, 9) with respect to the line 7x−3​=5y−2​=−9z−1​ is (20, b, − a − 9), then | a + b |, is equal to :2021 · MCQ
  • Let λ be an integer. If the shortest distance between the lines x −λ= 2y − 1 =− 2z and x = y + 2 λ= z −λ is 22​7​​, then the value of | λ | is ​.2021 · Numerical
  • The equation of the line through the point (0, 1, 2) and perpendicular to the line 2x−1​=3y+1​=−2z−1​ is :2021 · MCQ
  • Let α be the angle between the lines whose direction cosines satisfy the equations l + m − n = 0 and l2 + m2 − n2 = 0. Then the value of sin4 α + cos4 α is :2021 · MCQ
  • A line 'l' passing through origin is perpendicular to the lines l1​:r=(3+t)i+(−1+2t)j​+(4+2t)kl2​:r=(3+2s)i+(3+2s)j​+(2+s)k…2021 · Numerical
  • The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m − n = 0 and mn + nl + lm = 0, is :2021 · MCQ