If the shortest distance between the lines (x- )/(-2)=(y-2)/(1)=(z-1)/(1) and x-√31=(y-1)/…

Question

If the shortest distance between the lines

$\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda $ is :

Accepted Answer

Given the two lines:

$$ L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1} $$

$$ L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1} $$

We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$. The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.

The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$, and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$. The cross product of $d_1$ and $d_2$, which will be perpendicular to both lines, is given by:

$$ d = d_1 \times d_2 = \left| \begin{matrix}

\mathbf{i} & \mathbf{j} & \mathbf{k} \

-2 & 1 & 1 \
1 & -2 & 1 \

\end{matrix} \right| $$

Expanding the determinate gives:

$$ d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1)) $$

$$ d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1) $$

$$ d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} $$

$$ d = \langle 3, 3, 3 \rangle $$

The shortest distance $D$ between the two lines can then be given by the formula:

$$ D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|} $$

Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$, respectively, and '$\cdot$' denotes the dot product.

From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$, let's choose the point $A(\lambda, 2, 1)$, and for $L_2$, let's choose the point $B(\sqrt{3}, 1, 2)$. These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$, respectively.

The vector $\mathbf{a_2} - \mathbf{a_1}$ is:

$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle $$

$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle $$

$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle $$

We can now substitute this, along with $d$, into the distance formula:

$$ D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|} $$

$$ D = \frac{\left| 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) \right|}{\sqrt{3^2 + 3^2 + 3^2}} $$

$$ D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}} $$

$$ D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}} $$

$$ D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} $$

Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$:

$$ \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1 $$

$$ \left| \sqrt{3} - \lambda \right| = \sqrt{3} $$

This absolute value equation gives us two possible cases:

Case 1: $\sqrt{3} - \lambda = \sqrt{3}$, which gives $\lambda = 0$.

Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$, which gives $\lambda = 2\sqrt{3}$.

Therefore, the sum of all possible values of $\lambda$ is:

$$ \lambda_{sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3} $$

Hence, option B ($2 \sqrt{3}$) is the correct answer.

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