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Shortest Distance between Two Lines
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the shortest distance between the lines x 3 1 y 2
Question:
The shortest distance between the lines
\(\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3} \space and \space \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\)
is: =
CUET (UG) - 2023
CUET (UG)
Updated On:
Oct 10, 2024
\(\frac{197}{\sqrt{179}}\)
\(\frac{197}{\sqrt{191}}\)
\(\frac{197}{\sqrt{189}}\)
\(\frac{197}{\sqrt{237}}\)
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The Correct Option is
D
Solution and Explanation
The correct option is (D):
\(\frac{197}{\sqrt{237}}\)
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Top Questions on Shortest Distance between Two Lines
The shortest distance (in units) between the lines
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Let the line of the shortest distance between the lines
\(L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})\)
and
\(L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})\)
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L2: $\vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k}$, $\mu \in \mathbb{R}$ is $\frac{m}{\sqrt{n}}$, where gcd(m, n) = 1, then the value of m + n equals.
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\[\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}\]and
\[\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}\] is:
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