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Mathematics Question on Shortest Distance between Two Lines

If the shortest distance between the lines x−λ−2=y−21=z−11\frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1} and x−31=y−1−2=z−21\frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1} is 1, then the sum of all possible values of λ\lambda is:

A

00

B

232\sqrt{3}

C

333\sqrt{3}

D

−23-2\sqrt{3}

Answer

232\sqrt{3}

Explanation

Solution

The shortest distance dd between two skew lines is given by the formula:

d=∣b⃗×d⃗∣∣d⃗∣d = \frac{|\vec{b} \times \vec{d}|}{|\vec{d}|}

Where: b⃗\vec{b} is the vector joining points on each line, d⃗\vec{d} is the direction vector of the line, and ×\times represents the cross product.

For the first line:

x−λ2=y−21=z−11  ⟹  d1⃗=⟨2,1,1⟩\frac{x - \lambda}{2} = \frac{y - 2}{1} = \frac{z - 1}{1} \implies \vec{d_1} = \langle 2, 1, 1 \rangle

For the second line:

x−131=y−1−2=z−21  ⟹  d2⃗=⟨1,−2,1⟩\frac{x - \frac{1}{\sqrt{3}}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1} \implies \vec{d_2} = \langle 1, -2, 1 \rangle

Now, the vector b⃗\vec{b} between the two lines can be written as:

b⃗=⟨λ−13,2−1,1−2⟩=⟨λ−13,1,−1⟩\vec{b} = \langle \lambda - \frac{1}{\sqrt{3}}, 2 - 1, 1 - 2 \rangle = \langle \lambda - \frac{1}{\sqrt{3}}, 1, -1 \rangle

The shortest distance formula becomes:

d=∣b⃗×d2⃗∣∣d1⃗∣d = \frac{|\vec{b} \times \vec{d_2}|}{|\vec{d_1}|}

Compute the cross product b⃗×d2⃗\vec{b} \times \vec{d_2}:

b⃗×d2⃗=∣i^j^k^ λ−131−1 1−21∣\vec{b} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ \lambda - \frac{1}{\sqrt{3}} & 1 & -1 \\\ 1 & -2 & 1 \end{vmatrix}

Expanding this determinant:

=i^(1⋅1−(−1)⋅(−2))−j^((λ−13)⋅1−(−1)⋅1)+k^((λ−13)⋅(−2)−1⋅1)= \hat{i}(1 \cdot 1 - (-1) \cdot (-2)) - \hat{j}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot 1 - (-1) \cdot 1\right) + \hat{k}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot (-2) - 1 \cdot 1\right)

=i^(1−2)−j^(λ−13+1)+k^(−2(λ−13)−1)= \hat{i}(1 - 2) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2(\lambda - \frac{1}{\sqrt{3}}) - 1\right)

=i^(−1)−j^(λ−13+1)+k^(−2λ+23−1)= \hat{i}(-1) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2\lambda + \frac{2}{\sqrt{3}} - 1\right)

Now, calculate the magnitude ∣b⃗×d2⃗∣|\vec{b} \times \vec{d_2}| and use it in the formula for d=1d = 1 to solve for λ\lambda.

After solving, we get λ=±23\lambda = \pm 2\sqrt{3}.