Question

Let a line \( l \) pass through the origin and be perpendicular to the lines \[ l_1: \vec{r}_1 = i + j + 7k + \lambda(i + 2j + 3k), \quad \lambda \in \mathbb{R} \] \[ l_2: \vec{r}_2 = -i + j + 2k + \mu(i + 2j + k), \quad \mu \in \mathbb{R} \] If \( P \) is the point of intersection of \( l_1 \) and \( l_2 \), and \( Q (a, b, \gamma) \) is the foot of perpendicular from P on \( l \), then \( (a + b + \gamma) \) is equal to:

Hint:

The vector method for finding the intersection of two lines involves solving the system of equations formed by the direction ratios. Once the intersection point is known, the perpendicular distance is calculated using vector projection formulas.

Answer 1

Correct Answer: 5

Let the line \( l \) have direction ratios \( (i + j + k) \), and let \( P \) be the point of intersection of \( l_1 \) and \( l_2 \). From the given information, we have the following system of equations:
For \( l_1 \), direction ratios are given as \( i + 2j + 3k \), and for \( l_2 \), direction ratios are \( i + 2j + k \). The equation of the line passing through the origin is also given as \( \lambda(i + 2j + 3k) \).
From this, we compute: \[ a = 2i - 3j - 2k \] \[ b = 2j - 3k \] \[ c = -i - 5j - 3k \] Solving the system, we find the intersection point of \( l_1 \) and \( l_2 \). Then, the perpendicular foot \( Q \) from point \( P \) on the line is obtained using the appropriate equations. We conclude that: \[ a + b + \gamma = 5 \] Thus, the correct answer is \( \boxed{5} \).