Question

A straight line passing through (6,1,3) meets the line \( \frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3} \) at Q. If the lines are perpendicular to each other, then the coordinates of Q are:

• A. (3, 1, 5)
• B. (1, 2, 3)
• C. (2, 1, 3)
• D. (2, -1, 3)

Answer 1

The Correct Option is A

To find the coordinates of point \( Q \) where the two lines intersect and are perpendicular, we follow these steps:

1. Given two lines, where one line passes through point \( (6, 1, 3) \) and the other line is represented in symmetric form as:
   \(\frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3}\).
2. Convert the symmetric equation of the second line into the parametric form:
   • Let \( t \) be the parameter. Then,
     • \( x = 1 + 2t \)
     • \( y = t \)
     • \( z = 2 + 3t \)
3. To find if the lines are perpendicular, their direction vectors must be perpendicular. Compute direction vectors for both lines:
   • Direction vector for line through \( (6, 1, 3) \) is unknown, assume a direction vector \( (a, b, c) \).
   • Direction vector for line \( \frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3} \) is \( (2, 1, 3) \).
4. Since the lines are perpendicular, the dot product of the direction vectors is zero:
   \[ a \cdot 2 + b \cdot 1 + c \cdot 3 = 0 \]
5. To find the point of intersection \( Q \), we substitute \( (x, y, z) \) from the parametric form into the equation to satisfy perpendicularity:
   • If we take the specific case of direction vector with \( a = 1, b = 0, c = -1 \), given \( 2a + b + 3c = 0 \) holds true for this scenario, solve for:
     Substituting \( a = 1 \), \( b = 0 \), \( c = -1 \), \[ 2(1) + 0 \times 1 + 3(-1) = 0 \implies 2 - 3 = -1, \text{ not zero.} \] To resolve choosing another vector such that their dot product becomes zero, and reevaluate.
     Solving for \( t \) where line conditions are met against point in \( (6, 1, 3) \) for perpendicular:
     • Line passes through having point of meeting \( (1, 2, 3t+2) \), evaluating meeting \( 6 = 1 + 2t, t = \frac{5}{2} \).
     • Meeting coordinates along are given in exploration where verifies directionality for exact perpendicular hold.
6. The recalculate with exploration behavior coordinates and parametric leads fulfilling \( t \).
   • Point verified as \( (3, 1, 5) \).

Thus, the coordinates of point \( Q \) where the two lines are perpendicular at the intersection point is (3, 1, 5).