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The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is
$\begin{array}{ll}\text { a. } \sqrt{30} \quad \text { b. } 2 \sqrt{30} & \text { c. } 5 \\ \lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) & =\end{array}$
Options:
$\begin{array}{ll}\text { a. } \sqrt{30} \quad \text { b. } 2 \sqrt{30} & \text { c. } 5 \\ \lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) & =\end{array}$
Solution:
2538 Upvotes
Verified Answer
The correct answer is:
$\frac{2 b}{a}$
$L_{1}: \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$
$L_{2}: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$
Shortest distance between two lines
$$
=\left|\frac{\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}\right|
$$
$$
=\left|\frac{\left|\begin{array}{ccc}
-3-3 & -7-8 & 6-3 \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right|}{\sqrt{\left(-4-2\right)^{2}+\left(-3-12\right)^{2}+\left(6-3\right)^{2}}}\right|
$$
$=\left|\frac{\left|\begin{array}{ccc}-6 & -15 & 3 \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{(-6)^{2}+(-15)^{2}+(3)^{2}}}\right|$ On expanding the
$$
\begin{aligned}
&=\frac{3(6-3)-1(-12-45)+4(6+45)}{\sqrt{36+225+9}} \\
&=\frac{9+57+204}{\sqrt{270}}=\frac{270}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30}
\end{aligned}
$$
$L_{2}: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$
Shortest distance between two lines
$$
=\left|\frac{\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}\right|
$$
$$
=\left|\frac{\left|\begin{array}{ccc}
-3-3 & -7-8 & 6-3 \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right|}{\sqrt{\left(-4-2\right)^{2}+\left(-3-12\right)^{2}+\left(6-3\right)^{2}}}\right|
$$
$=\left|\frac{\left|\begin{array}{ccc}-6 & -15 & 3 \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{(-6)^{2}+(-15)^{2}+(3)^{2}}}\right|$ On expanding the
$$
\begin{aligned}
&=\frac{3(6-3)-1(-12-45)+4(6+45)}{\sqrt{36+225+9}} \\
&=\frac{9+57+204}{\sqrt{270}}=\frac{270}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30}
\end{aligned}
$$
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