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Question: The shortest distance between the lines $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z-0}{1}$ and $\frac{x+4}{...

The shortest distance between the lines x−22=y−32=z−01\frac{x-2}{2}=\frac{y-3}{2}=\frac{z-0}{1} and x+4−1=y−78=z−54\frac{x+4}{-1}=\frac{y-7}{8}=\frac{z-5}{4} lies in the interval

A

[0, 1]

B

[1, 2]

C

[2, 3]

D

[3, 4]

Answer

[2, 3]

Explanation

Solution

The shortest distance between two skew lines is given by the formula: d=∣(a2⃗−a1⃗)⋅(v1⃗×v2⃗)∣∣∣v1⃗×v2⃗∣∣d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})|}{||\vec{v_1} \times \vec{v_2}||}

For the first line x−22=y−32=z−01\frac{x-2}{2}=\frac{y-3}{2}=\frac{z-0}{1}, a point on the line is P1=(2,3,0)P_1 = (2, 3, 0) and its direction vector is v1⃗=⟨2,2,1⟩\vec{v_1} = \langle 2, 2, 1 \rangle. Thus, a1⃗=⟨2,3,0⟩\vec{a_1} = \langle 2, 3, 0 \rangle.

For the second line x+4−1=y−78=z−54\frac{x+4}{-1}=\frac{y-7}{8}=\frac{z-5}{4}, a point on the line is P2=(−4,7,5)P_2 = (-4, 7, 5) and its direction vector is v2⃗=⟨−1,8,4⟩\vec{v_2} = \langle -1, 8, 4 \rangle. Thus, a2⃗=⟨−4,7,5⟩\vec{a_2} = \langle -4, 7, 5 \rangle.

The vector connecting the two points is: a2⃗−a1⃗=⟨−4−2,7−3,5−0⟩=⟨−6,4,5⟩\vec{a_2} - \vec{a_1} = \langle -4-2, 7-3, 5-0 \rangle = \langle -6, 4, 5 \rangle.

The cross product of the direction vectors is: v1⃗×v2⃗=∣ijk221−184∣=(8−8)i−(8−(−1))j+(16−(−2))k=0i−9j+18k=⟨0,−9,18⟩\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ -1 & 8 & 4 \end{vmatrix} = (8 - 8)\mathbf{i} - (8 - (-1))\mathbf{j} + (16 - (-2))\mathbf{k} = 0\mathbf{i} - 9\mathbf{j} + 18\mathbf{k} = \langle 0, -9, 18 \rangle.

The magnitude of the cross product is: ∣∣v1⃗×v2⃗∣∣=02+(−9)2+182=81+324=405=95||\vec{v_1} \times \vec{v_2}|| = \sqrt{0^2 + (-9)^2 + 18^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}.

The scalar triple product is: (a2⃗−a1⃗)⋅(v1⃗×v2⃗)=⟨−6,4,5⟩⋅⟨0,−9,18⟩=(−6)(0)+(4)(−9)+(5)(18)=0−36+90=54(\vec{a_2}-\vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2}) = \langle -6, 4, 5 \rangle \cdot \langle 0, -9, 18 \rangle = (-6)(0) + (4)(-9) + (5)(18) = 0 - 36 + 90 = 54.

The shortest distance is: d=∣54∣95=65=655d = \frac{|54|}{9\sqrt{5}} = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}.

To determine the interval, we can square the distance: d2=(655)2=36×525=365=7.2d^2 = \left(\frac{6\sqrt{5}}{5}\right)^2 = \frac{36 \times 5}{25} = \frac{36}{5} = 7.2.

We check which interval contains dd.

  • [0,1]2=[0,1][0, 1]^2 = [0, 1]
  • [1,2]2=[1,4][1, 2]^2 = [1, 4]
  • [2,3]2=[4,9][2, 3]^2 = [4, 9]
  • [3,4]2=[9,16][3, 4]^2 = [9, 16]

Since 4≤7.2≤94 \le 7.2 \le 9, the distance dd lies in the interval [2, 3].