The lines $ \frac{x-1}{2} = \frac{y-4}{4} = \frac{z-2}{3} \quad \text{and} \quad \frac{1-x}{1} = \frac{y-2}{5} = \frac{3-z}{a} \quad \text{are perpendicular to each other, then} \ a \ \text{equals to} $
The lines $ \frac{x-1}{2} = \frac{y-4}{4} = \frac{z-2}{3} \quad \text{and} \quad \frac{1-x}{1} = \frac{y-2}{5} = \frac{3-z}{a} \quad \text{are perpendicular to each other, then} \ a \ \text{equals to} $
Show Hint
- -6
- 6
- \( \frac{22}{3} \)
- \( -\frac{22}{3} \)
The Correct Option is B
Solution and Explanation
The condition for two lines to be perpendicular is that the dot product of their direction ratios must be zero.
For the first line, we have the direction ratios as \( \mathbf{l_1} = (2, 4, 3) \). For the second line, we have the direction ratios as \( \mathbf{l_2} = (-1, 5, -a) \). The condition for perpendicularity is: \[ 2 \cdot (-1) + 4 \cdot 5 + 3 \cdot (-a) = 0 \] Simplifying this: \[ -2 + 20 - 3a = 0 \] \[ 18 - 3a = 0 \] \[ 3a = 18 \] \[ a = 6 \] Thus, the value of \( a \) is 6.
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