If the shortest distance between the lines and is , then the sum of all possible values of is [2025]

Question

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

Smart Achievers · Accepted Answer

(1)

The given lines are $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ ,

$\vec{n} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & α & 1 \end{matrix} |$

$= \hat{i} (3 - 4 α) - \hat{j} (2 - 4) + \hat{k} (2 α - 3)$

$= \hat{i} (3 - 4 α) - \hat{j} (- 2) + \hat{k} (2 α - 3)$

Shortest distance = $| \frac{\vec{B A} \cdot \vec{n}}{| \vec{n} |} |$ $=$ $| \frac{(\hat{i} + 2 \hat{j} - 2 \hat{k}) \cdot \vec{n}}{| \vec{n} |} |$ $= \frac{5}{\sqrt{6}}$ [Given]

$=$ $| \frac{13 - 8 α}{\sqrt{{(3 - 4 α)}^{2} + 4 + {(2 α - 3)}^{2}}} |$ $= \frac{5}{\sqrt{6}}$

$\Rightarrow 6 {(13 - 8 α)}^{2} = 25 ({(4 α - 3)}^{2} + {(2 α - 3)}^{2} + 4)$

$\Rightarrow 6 (64 α^{2} - 208 α + 169) = (25 (20 α^{2} - 36 α + 22))$

$\Rightarrow 116 α^{2} + 348 α - 464 = 0$

$∴$ Sum of roots $α_{1}$ and $α_{2} = \frac{- 348}{116} = - 3$ .

Solution

Q.
If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

1 –3

2 $- \frac{3}{2}$

3 $\frac{3}{2}$

4 3

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Solution

Q. If the shortest distance between the lines x–12=y–23=z–34 and x1=yα=z–51 is 56, then the sum of all possible values of α is [2025]

1 –3

2 –32

3 32

4 3

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Q.
If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

2 $- \frac{3}{2}$

3 $\frac{3}{2}$