If the lines x - 1 -3 = y - 2 -2k = z - 3 2 and x - 1 k = y - 2 1… - Vidyadip

Question

If the lines $\frac{\text{x - 1}}{-3}=\frac{\text{y - 2}}{\text{-2k}}=\frac{\text{z - 3}}{2}$ and $\frac{\text{x - 1}}{\text{k}}=\frac{\text{y - 2}}{\text{1}}=\frac{\text{z - 3}}{5}$are perpendicular, find the value of k and hence find the equation of plane containing these lines.

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Answer

The lines are $\frac{\text{x - 1}}{-3}=\frac{\text{y - 2}}{\text{-2k}}=\frac{\text{z - 3}}{2}$ and $\frac{\text{x - 1}}{\text{k}}=\frac{\text{y - 2}}{\text{1}}=\frac{\text{z - 3}}{5}$ the lines are perpendicular $\Rightarrow$ ( – 3) k + (–2 k) 1 + 2 (5) = 0– 3 k – 2 k + 10 = 0 $\Rightarrow$ k = 2
$\therefore$ The lines become $\frac{\text{x - 1}}{-3}=\frac{\text{y - 2}}{-4}=\frac{\text{z - 3}}{2},\frac{\text{x - 1}}{\text{2}}=\frac{\text{y - 2}}{\text{1}}=\frac{\text{z - 3}}{5}$ The equation of plane containing the lines is $ \begin{vmatrix} \text{x - 1} & \text{y - 2} & \text{z - 3} \\ -3 & -4 & 2 \\ 2 & 1 & 5 \end{vmatrix}=0$
$\Rightarrow$ – 22 (x – 1) – (y – 2) (– 19) + (z – 3) 5 = 0
$\Rightarrow$ – 22x + 19y + 5z = 31
OR 22x – 19y – 5z + 31 = 0.

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