If the shortest distance between the lines and is . Then sum of all possible values of is

Question

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ . Then sum of all possible values of $λ$ is _____

Smart Achievers · Accepted Answer

(8)

$| \frac{(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{d})}{| \vec{b} \times \vec{d} |} |$ $= \frac{6}{\sqrt{5}}$

Solution

Q.
If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ . Then sum of all possible values of $λ$ is _____

Ans.
(8)

$| \frac{(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{d})}{| \vec{b} \times \vec{d} |} |$ $= \frac{6}{\sqrt{5}}$

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Solution

Q. If the shortest distance between the lines x-41=y+12=z-3 and x-λ2=y+14=z-2-5 is 65. Then sum of all possible values of λ is _____

Ans. (8) |(a→-c→)·(b→-d→)|b→×d→||=65

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Q.
If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ . Then sum of all possible values of $λ$ is _____

Ans.
(8)

$| \frac{(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{d})}{| \vec{b} \times \vec{d} |} |$ $= \frac{6}{\sqrt{5}}$