Question:
Let \( L_1 : \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 1}{-1} \) and \( L_2 : \frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 2}{2} \) 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:
Let \( L_1 : \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 1}{-1} \) and \( L_2 : \frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 2}{2} \) 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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Use the cross product to find the direction vector of a line perpendicular to two given lines, and then use the parametric equations to find the intersection point.
Updated On: Mar 20, 2025
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The Correct Option is D
Solution and Explanation
- First, find the direction vectors of \( L_1 \) and \( L_2 \), and the point of intersection.
- The direction vector of \( L_3 \) will be the cross product of the direction vectors of \( L_1 \) and \( L_2 \).
- Use this to calculate the coordinates of the intersection point and the required value. Thus, the value of \( 5x - 11y - 8z = 3 \).
- The direction vector of \( L_3 \) will be the cross product of the direction vectors of \( L_1 \) and \( L_2 \).
- Use this to calculate the coordinates of the intersection point and the required value. Thus, the value of \( 5x - 11y - 8z = 3 \).
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