If the shortest distance between the lines $L_1:\vec{r}=(2+\lambda)\hat{i}+(1-3\lambda)\hat{j}+(3+4\lambda)\hat{k}$, $\lambda\in\mathbb{R}$ and $L_2:\vec{r}=2(1+\mu)\hat{i}+3(1+\mu)\hat{j}+(5+\mu)\hat{k}$, $\mu\in\mathbb{R}$ is $\dfrac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals:
Step-by-Step Solution
Key Concept: $L_1$ through $A(2,1,3)$ with direction $\vec{p}=(1,-3,4)$. $L_2$ through $B(2,3,5)$ with direction $\vec{q}=(2,3,1)$. $\overrightarrow{AB}=(0,2,2)$. $\vec{p}\times\vec{q}=(-3\cdot1-4\cdot3)\hat{i}-(1\cdot1-4\cdot2)\hat{j}+(1\cdot3-(-3)\cdot2)\hat{k}=(-15,7,9)$.
To find the shortest distance between two lines $L_1$ and $L_2$, we first need to identify a point on each line and the direction vectors of the lines.
Step 1:
Identify the direction vectors $\vec{d_1}$ and $\vec{d_2}$ of $L_1$ and $L_2$ respectively.
For $L_1$, the direction vector $\vec{d_1}$ can be extracted from the coefficients of $\lambda$ in its equation, yielding $\vec{d_1} = \hat{i} - 3\hat{j} + 4\hat{k}$.
For $L_2$, the direction vector $\vec{d_2}$ can be extracted from the coefficients of $\mu$ in its equation, yielding $\vec{d_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Step 2:
Find a point on each line.
For $L_1$, setting $\lambda = 0$ gives us a point $P_1 = (2, 1, 3)$.
For $L_2$, setting $\mu = 0$ gives us a point $P_2 = (2, 3, 5)$.
Step 3:
Calculate the vector $\vec{P_2P_1}$ from $P_2$ to $P_1$.
$\vec{P_2P_1} = (2 - 2)\hat{i} + (1 - 3)\hat{j} + (3 - 5)\hat{k} = 0\hat{i} - 2\hat{j} - 2\hat{k}$.
Step 4:
The shortest distance between two lines can be found using the formula $SD = \frac{|\vec{d_1} \times \vec{d_2} \cdot \vec{P_2P_1}|}{|\vec{d_1} \times \vec{d_2}|}$, where $\times$ denotes the cross product and $\cdot$ denotes the dot product.
First, calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$$
\begin{aligned}
\vec{d_1} \times \vec{d_2} &=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 4 \\
2 & 3 & 1 \\
\end{vmatrix} \\
&= \hat{i}((-3)(1) - (4)(3)) - \hat{j}((1)(1) - (4)(2)) + \hat{k}((1)(3) - (-3)(2)) \\
&= \hat{i}(-3 - 12) - \hat{j}(1 - 8) + \hat{k}(3 + 6) \\
&= \hat{i}(-15) - \hat{j}(-7) + \hat{k}(9) \\
&= -15\hat{i} + 7\hat{j} + 9\hat{k}
\end{aligned}
$$
Step 5:
Next, calculate the dot product $(\vec{d_1} \times \vec{d_2}) \cdot \vec{P_2P_1}$:
$$
\begin{aligned}
(\vec{d_1} \times \vec{d_2}) \cdot \vec{P_2P_1} &= (-15\hat{i} + 7\hat{j} + 9\hat{k}) \cdot (0\hat{i} - 2\hat{j} - 2\hat{k}) \\
&= (-15)(0) + (7)(-2) + (9)(-2) \\
&= 0 - 14 - 18 \\
&= -32
\end{aligned}
$$
Step 6:
Then, calculate the magnitude of $\vec{d_1} \times \vec{d_2}$:
$$
\begin{aligned}
|\vec{d_1} \times \vec{d_2}| &= \sqrt{(-15)^2 + 7^2 + 9^2} \\
&= \sqrt{225 + 49 + 81} \\
&= \sqrt{355}
\end{aligned}
$$
Step 7:
Finally, substitute these values into the formula for the shortest distance:
$$
\begin{aligned}
SD &= \frac{|\vec{d_1} \times \vec{d_2} \cdot \vec{P_2P_1}|}{|\vec{d_1} \times \vec{d_2}|} \\
&= \frac{|-32|}{\sqrt{355}} \\
&= \frac{32}{\sqrt{355}}
\end{aligned}
$$
Given that the shortest distance is $\frac{m}{\sqrt{n}}$, we have $m = 32$ and $n = 355$, so $m + n = 32 + 355 = 387$.
Therefore: $m + n = 387$ corresponds to option 4.
<div class="key-concept"><strong>Key Concept:</strong> $L_1$ through $A(2,1,3)$ with direction $\vec{p}=(1,-3,4)$. $L_2$ through $B(2,3,5)$ with direction $\vec{q}=(2,3,1)$. $\overrightarrow{AB}=(0,2,2)$. $\vec{p}\times\vec{q}=(-3\cdot1-4\cdot3)\hat{i}-(1\cdot1-4\cdot2)\hat{j}+(1\cdot3-(-3)\cdot2)\hat{k}=(-15,7,9)$.</div>
<div class="trap-box"><strong>Trap:</strong> $|\vec{p}\times\vec{q}|=\sqrt{225+49+81}=\sqrt{355}$. SD $=\frac{|(0,2,2)\cdot(-15,7,9)|}{\sqrt{355}}=\frac{|0+14+18|}{\sqrt{355}}=\frac{32}{\sqrt{355}}$. $m=32,n=355,\gcd=1$. $m+n=387$.</div>
Correct Answer: 4