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$L_1$ is a line passing through the points with position vectors $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}-3 \hat{\mathbf{k}}$. $L_2$ is a line passing through the points with position vectors $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$. Then the distance between $L_1$ and $L_2$ is
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Verified Answer
The correct answer is:
$\frac{4}{3}$
Equation of line $L_1$ is
$\mathbf{r}=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
Equation of line $L_2$ is
$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$
Distance between $L_1$ and $L_2$
$\begin{array}{r}
D=\frac{(4 \hat{\mathbf{j}}) \cdot[(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})]}{\mid(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-4 \hat{\mathbf{k}}) \mid} \\
(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 2 & -2 \\
1 & -6 & -4
\end{array}\right| \\
=-(20 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+20 \hat{\mathbf{k}}) \\
D=\left|\frac{4 \hat{\mathbf{j}} \cdot(-20 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}-20 \hat{\mathbf{k}})}{\sqrt{(20)^2+(10)^2+(20)^2}}\right|=\frac{4}{3}
\end{array}$
$\mathbf{r}=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
Equation of line $L_2$ is
$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$
Distance between $L_1$ and $L_2$
$\begin{array}{r}
D=\frac{(4 \hat{\mathbf{j}}) \cdot[(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})]}{\mid(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-4 \hat{\mathbf{k}}) \mid} \\
(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 2 & -2 \\
1 & -6 & -4
\end{array}\right| \\
=-(20 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+20 \hat{\mathbf{k}}) \\
D=\left|\frac{4 \hat{\mathbf{j}} \cdot(-20 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}-20 \hat{\mathbf{k}})}{\sqrt{(20)^2+(10)^2+(20)^2}}\right|=\frac{4}{3}
\end{array}$
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