Let
\[
L_1 : \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{-1} \quad {and} \quad L_2 : \frac{x+1}{1} = \frac{y-2}{2} = \frac{z-2}{2}
\]
be two lines. Let \( L_3 \) be a line passing through the point \( (\alpha, \beta, \gamma) \) and be perpendicular to both \( L_1 \) and \( L_2 \). If \( L_3 \) intersects \( L_1 \) where \( 5x - 11y - 8z = 1 \), then \( 5x - 11y - 8z \) equals:
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Solution and Explanation
Step 1: The direction ratios of the lines \( L_1 \) and \( L_2 \) can be extracted from their parametric equations. For \( L_1 \), the direction ratios are \( (1, -1, -1) \), and for \( L_2 \), the direction ratios are \( (1, 2, 2) \).
Step 2: The direction vector of the line \( L_3 \) is perpendicular to both \( L_1 \) and \( L_2 \), so we compute the cross product of the direction vectors of \( L_1 \) and \( L_2 \). \[ \mathbf{d_3} = \mathbf{d_1} \times \mathbf{d_2} \] After computing the cross product, we get the direction vector of \( L_3 \).
Step 3: The equation of the line \( L_3 \) is now known, and we substitute the parametric equations of \( L_3 \) into the plane equation \( 5x - 11y - 8z = 1 \) to find the value of \( 5x - 11y - 8z \). Thus, the final value of \( 5x - 11y - 8z \) is found.
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