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The shortest distance between the line $\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})$ and the plane $\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ is
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2931 Upvotes
Verified Answer
The correct answer is:
$\frac{10}{3 \sqrt{3}}$
We have,
$\begin{aligned}
& L: \mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\
& P: \mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5
\end{aligned}$
$\quad$ Now, $\quad(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=1-5+4=0$
$\therefore$ Line is parallel to plane.
$\begin{aligned}
& L: \mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\
& P: \mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5
\end{aligned}$
$\quad$ Now, $\quad(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=1-5+4=0$
$\therefore$ Line is parallel to plane.
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