Write the lines in vector/parametric form
For L 1 L_1 L 1
x − 1 2 = y − 2 3 = z − 3 4 = λ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda 2 x − 1 = 3 y − 2 = 4 z − 3 = λ So,
x = 1 + 2 λ , y = 2 + 3 λ , z = 3 + 4 λ x=1+2\lambda,\quad y=2+3\lambda,\quad z=3+4\lambda x = 1 + 2 λ , y = 2 + 3 λ , z = 3 + 4 λ Hence a point on L 1 L_1 L 1
A 1 = ( 1 , 2 , 3 ) A_1=(1,2,3) A 1 = ( 1 , 2 , 3 ) and its direction vector is
d ⃗ 1 = ( 2 , 3 , 4 ) . \vec d_1=(2,3,4). d 1 = ( 2 , 3 , 4 ) . For L 2 L_2 L 2
x − 2 3 = y − 4 4 = z − 5 5 = μ \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}=\mu 3 x − 2 = 4 y − 4 = 5 z − 5 = μ So,
x = 2 + 3 μ , y = 4 + 4 μ , z = 5 + 5 μ x=2+3\mu,\quad y=4+4\mu,\quad z=5+5\mu x = 2 + 3 μ , y = 4 + 4 μ , z = 5 + 5 μ Hence a point on L 2 L_2 L 2
A 2 = ( 2 , 4 , 5 ) A_2=(2,4,5) A 2 = ( 2 , 4 , 5 ) and its direction vector is
d ⃗ 2 = ( 3 , 4 , 5 ) . \vec d_2=(3,4,5). d 2 = ( 3 , 4 , 5 ) .
Direction of the line of shortest distance
The line of shortest distance between two skew lines is perpendicular to both lines, so its direction vector is
n ⃗ = d ⃗ 1 × d ⃗ 2 . \vec n=\vec d_1\times \vec d_2. n = d 1 × d 2 . Compute:
d ⃗ 1 × d ⃗ 2 = ∣ i ^ j ^ k ^ 2 3 4 3 4 5 ∣ \vec d_1\times \vec d_2=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
2 & 3 & 4\\
3 & 4 & 5
\end{vmatrix} d 1 × d 2 = i ^ 2 3 j ^ 3 4 k ^ 4 5 = 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 ) . So the shortest-distance line is parallel to
( − 1 , 2 , − 1 ) . (-1,2,-1). ( − 1 , 2 , − 1 ) .
Find the actual endpoints of the common perpendicular
Let
P = ( 1 + 2 λ , 2 + 3 λ , 3 + 4 λ ) ∈ L 1 P=(1+2\lambda,\,2+3\lambda,\,3+4\lambda)\in L_1 P = ( 1 + 2 λ , 2 + 3 λ , 3 + 4 λ ) ∈ L 1 and
Q = ( 2 + 3 μ , 4 + 4 μ , 5 + 5 μ ) ∈ L 2 . Q=(2+3\mu,\,4+4\mu,\,5+5\mu)\in L_2. Q = ( 2 + 3 μ , 4 + 4 μ , 5 + 5 μ ) ∈ L 2 . Then
P Q → = Q − P = ( 1 + 3 μ − 2 λ , 2 + 4 μ − 3 λ , 2 + 5 μ − 4 λ ) . \overrightarrow{PQ}=Q-P=(1+3\mu-2\lambda,\,2+4\mu-3\lambda,\,2+5\mu-4\lambda). P Q = Q − P = ( 1 + 3 μ − 2 λ , 2 + 4 μ − 3 λ , 2 + 5 μ − 4 λ ) . Since P Q PQ P Q
P Q → ⋅ d ⃗ 1 = 0 , P Q → ⋅ d ⃗ 2 = 0. \overrightarrow{PQ}\cdot \vec d_1=0,
\qquad
\overrightarrow{PQ}\cdot \vec d_2=0. P Q ⋅ d 1 = 0 , P Q ⋅ d 2 = 0. So,
( 1 + 3 μ − 2 λ ) 2 + ( 2 + 4 μ − 3 λ ) 3 + ( 2 + 5 μ − 4 λ ) 4 = 0 (1+3\mu-2\lambda)2+(2+4\mu-3\lambda)3+(2+5\mu-4\lambda)4=0 ( 1 + 3 μ − 2 λ ) 2 + ( 2 + 4 μ − 3 λ ) 3 + ( 2 + 5 μ − 4 λ ) 4 = 0 2 + 6 μ − 4 λ + 6 + 12 μ − 9 λ + 8 + 20 μ − 16 λ = 0 2+6\mu-4\lambda+6+12\mu-9\lambda+8+20\mu-16\lambda=0 2 + 6 μ − 4 λ + 6 + 12 μ − 9 λ + 8 + 20 μ − 16 λ = 0 16 + 38 μ − 29 λ = 0 16+38\mu-29\lambda=0 16 + 38 μ − 29 λ = 0 29 λ − 38 μ = 16 . . . ( 1 ) 29\lambda-38\mu=16 \qquad ...(1) 29 λ − 38 μ = 16 ... ( 1 ) Also,
( 1 + 3 μ − 2 λ ) 3 + ( 2 + 4 μ − 3 λ ) 4 + ( 2 + 5 μ − 4 λ ) 5 = 0 (1+3\mu-2\lambda)3+(2+4\mu-3\lambda)4+(2+5\mu-4\lambda)5=0 ( 1 + 3 μ − 2 λ ) 3 + ( 2 + 4 μ − 3 λ ) 4 + ( 2 + 5 μ − 4 λ ) 5 = 0 3 + 9 μ − 6 λ + 8 + 16 μ − 12 λ + 10 + 25 μ − 20 λ = 0 3+9\mu-6\lambda+8+16\mu-12\lambda+10+25\mu-20\lambda=0 3 + 9 μ − 6 λ + 8 + 16 μ − 12 λ + 10 + 25 μ − 20 λ = 0 21 + 50 μ − 38 λ = 0 21+50\mu-38\lambda=0 21 + 50 μ − 38 λ = 0 38 λ − 50 μ = 21 . . . ( 2 ) 38\lambda-50\mu=21 \qquad ...(2) 38 λ − 50 μ = 21 ... ( 2 ) Solve (1) and (2):
From (1):
29 λ = 16 + 38 μ 29\lambda=16+38\mu 29 λ = 16 + 38 μ Use elimination:
Multiply (1) by 38 38 38
1102 λ − 1444 μ = 608 1102\lambda-1444\mu=608 1102 λ − 1444 μ = 608 Multiply (2) by 29 29 29
1102 λ − 1450 μ = 609 1102\lambda-1450\mu=609 1102 λ − 1450 μ = 609 Subtract:
6 μ = − 1 ⇒ μ = − 1 6 . 6\mu=-1 \Rightarrow \mu=-\frac16. 6 μ = − 1 ⇒ μ = − 6 1 . Then from (2):
38 λ − 50 ( − 1 6 ) = 21 38\lambda-50\left(-\frac16\right)=21 38 λ − 50 ( − 6 1 ) = 21 38 λ + 25 3 = 21 = 63 3 38\lambda+\frac{25}{3}=21=\frac{63}{3} 38 λ + 3 25 = 21 = 3 63 38 λ = 38 3 38\lambda=\frac{38}{3} 38 λ = 3 38 λ = 1 3 . \lambda=\frac13. λ = 3 1 .
Find points P P P Q Q Q
Point on L 1 L_1 L 1
P = ( 1 + 2 ⋅ 1 3 , 2 + 3 ⋅ 1 3 , 3 + 4 ⋅ 1 3 ) P=\left(1+2\cdot\frac13,\;2+3\cdot\frac13,\;3+4\cdot\frac13\right) P = ( 1 + 2 ⋅ 3 1 , 2 + 3 ⋅ 3 1 , 3 + 4 ⋅ 3 1 ) P = ( 5 3 , 3 , 13 3 ) . P=\left(\frac53,\;3,\;\frac{13}{3}\right). P = ( 3 5 , 3 , 3 13 ) . Point on L 2 L_2 L 2
Q = ( 2 + 3 ⋅ ( − 1 6 ) , 4 + 4 ⋅ ( − 1 6 ) , 5 + 5 ⋅ ( − 1 6 ) ) Q=\left(2+3\cdot\left(-\frac16\right),\;4+4\cdot\left(-\frac16\right),\;5+5\cdot\left(-\frac16\right)\right) Q = ( 2 + 3 ⋅ ( − 6 1 ) , 4 + 4 ⋅ ( − 6 1 ) , 5 + 5 ⋅ ( − 6 1 ) ) Q = ( 3 2 , 10 3 , 25 6 ) . Q=\left(\frac32,\;\frac{10}{3},\;\frac{25}{6}\right). Q = ( 2 3 , 3 10 , 6 25 ) . Thus the line of shortest distance is the line through P P P Q Q Q
Q − P = ( − 1 6 , 1 3 , − 1 6 ) Q-P=\left(-\frac16,\frac13,-\frac16\right) Q − P = ( − 6 1 , 3 1 , − 6 1 ) which is indeed parallel to ( − 1 , 2 , − 1 ) (-1,2,-1) ( − 1 , 2 , − 1 )
So equation of the required line can be written as
( x , y , z ) = ( 5 3 , 3 , 13 3 ) + t ( − 1 , 2 , − 1 ) . (x,y,z)=\left(\frac53,3,\frac{13}{3}\right)+t(-1,2,-1). ( x , y , z ) = ( 3 5 , 3 , 3 13 ) + t ( − 1 , 2 , − 1 ) .
Check the options
A point ( x , y , z ) (x,y,z) ( x , y , z )
( x , y , z ) = ( 5 3 , 3 , 13 3 ) + t ( − 1 , 2 , − 1 ) (x,y,z)=\left(\frac53,3,\frac{13}{3}\right)+t(-1,2,-1) ( x , y , z ) = ( 3 5 , 3 , 3 13 ) + t ( − 1 , 2 , − 1 ) for some t t t
Option A: ( 14 3 , − 3 , 22 3 ) \left(\frac{14}{3},-3,\frac{22}{3}\right) ( 3 14 , − 3 , 3 22 )
From y y y
3 + 2 t = − 3 ⇒ t = − 3. 3+2t=-3 \Rightarrow t=-3. 3 + 2 t = − 3 ⇒ t = − 3. Then
x = 5 3 − ( − 3 ) = 5 3 + 3 = 14 3 , x=\frac53-(-3)=\frac53+3=\frac{14}{3}, x = 3 5 − ( − 3 ) = 3 5 + 3 = 3 14 , z = 13 3 − ( − 3 ) = 22 3 . z=\frac{13}{3}-(-3)=\frac{22}{3}. z = 3 13 − ( − 3 ) = 3 22 . So option A lies on the line.
Option B: ( 2 , 3 , 1 3 ) \left(2,3,\frac13\right) ( 2 , 3 , 3 1 )
From y y y
3 + 2 t = 3 ⇒ t = 0. 3+2t=3 \Rightarrow t=0. 3 + 2 t = 3 ⇒ t = 0. Then x = 5 3 ≠ 2 x=\frac53\neq 2 x = 3 5 = 2
Option C: ( 8 3 , − 1 , 1 3 ) \left(\frac83,-1,\frac13\right) ( 3 8 , − 1 , 3 1 )
From y y y
3 + 2 t = − 1 ⇒ t = − 2. 3+2t=-1 \Rightarrow t=-2. 3 + 2 t = − 1 ⇒ t = − 2. Then
x = 5 3 − ( − 2 ) = 11 3 ≠ 8 3 . x=\frac53-(-2)=\frac{11}{3}\neq \frac83. x = 3 5 − ( − 2 ) = 3 11 = 3 8 . Not on the line.
Option D: ( − 5 3 , − 7 , 1 ) \left(-\frac53,-7,1\right) ( − 3 5 , − 7 , 1 )
From y y y
3 + 2 t = − 7 ⇒ t = − 5. 3+2t=-7 \Rightarrow t=-5. 3 + 2 t = − 7 ⇒ t = − 5. Then
x = 5 3 − ( − 5 ) = 20 3 ≠ − 5 3 . x=\frac53-(-5)=\frac{20}{3}\neq -\frac53. x = 3 5 − ( − 5 ) = 3 20 = − 3 5 . Not on the line.
Conclusion
The point lying on the line of shortest distance is
( 14 3 , − 3 , 22 3 ) \boxed{\left(\frac{14}{3},-3,\frac{22}{3}\right)} ( 3 14 , − 3 , 3 22 ) which is Option A .