Question

Consider the lines $L_1: \frac{x-7}{3} = \frac{y-5}{2} = \frac{z-3}{1}$ and $L_2: \frac{x-1}{2} = \frac{y+1}{4} = \frac{z+1}{3}$. If a line $L$ whose direction ratios are (2, 2, 1) intersects $L_1$ and $L_2$ at $A$ and $B$. Then the distance from the point (3, 3, 2) to the line $L$ is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

Let the line $L_1$ be given by the parametric equation: $L_1: (x, y, z) = (7 + 3t_1, 5 + 2t_1, 3 + t_1)$ A point on $L_1$ is $P_1(7, 5, 3)$ and its direction vector is $\vec{d_1} = (3, 2, 1)$.

Let the line $L_2$ be given by the parametric equation: $L_2: (x, y, z) = (1 + 2t_2, -1 + 4t_2, -1 + 3t_2)$ A point on $L_2$ is $P_2(1, -1, -1)$ and its direction vector is $\vec{d_2} = (2, 4, 3)$.

The line $L$ has direction ratios $(2, 2, 1)$, so its direction vector is $\vec{d} = (2, 2, 1)$. Line $L$ intersects $L_1$ at point $A$ and $L_2$ at point $B$. Thus, point $A$ lies on $L_1$ and point $B$ lies on $L_2$. $A = (7 + 3t_1, 5 + 2t_1, 3 + t_1)$ for some $t_1$. $B = (1 + 2t_2, -1 + 4t_2, -1 + 3t_2)$ for some $t_2$.

Since $A$ and $B$ both lie on line $L$, the vector $\vec{AB}$ must be parallel to the direction vector $\vec{d}$ of line $L$. $\vec{AB} = B - A = (1 + 2t_2 - (7 + 3t_1), -1 + 4t_2 - (5 + 2t_1), -1 + 3t_2 - (3 + t_1))$ $\vec{AB} = (2t_2 - 3t_1 - 6, 4t_2 - 2t_1 - 6, 3t_2 - t_1 - 4)$

Since $\vec{AB} \parallel \vec{d} = (2, 2, 1)$, their components are proportional: $\frac{2t_2 - 3t_1 - 6}{2} = \frac{4t_2 - 2t_1 - 6}{2} = \frac{3t_2 - t_1 - 4}{1}$ From the first equality: $2t_2 - 3t_1 - 6 = 4t_2 - 2t_1 - 6$ $0 = 2t_2 + t_1 \implies t_1 = -2t_2$

From the second equality: $4t_2 - 2t_1 - 6 = 2(3t_2 - t_1 - 4)$ $4t_2 - 2t_1 - 6 = 6t_2 - 2t_1 - 8$ $4t_2 - 6 = 6t_2 - 8$ $2t_2 = 2 \implies t_2 = 1$

Substituting $t_2 = 1$ into $t_1 = -2t_2$, we get $t_1 = -2(1) = -2$.

Now we find the coordinates of points $A$ and $B$: For point $A$ (using $t_1 = -2$): $A = (7 + 3(-2), 5 + 2(-2), 3 + (-2)) = (7 - 6, 5 - 4, 3 - 2) = (1, 1, 1)$

For point $B$ (using $t_2 = 1$): $B = (1 + 2(1), -1 + 4(1), -1 + 3(1)) = (1 + 2, -1 + 4, -1 + 3) = (3, 3, 2)$

The line $L$ passes through point $A(1, 1, 1)$ and has direction vector $\vec{d} = (2, 2, 1)$. The parametric equation of line $L$ is: $L: (x, y, z) = (1 + 2\lambda, 1 + 2\lambda, 1 + \lambda)$, where $\lambda$ is a parameter.

We need to find the distance from the point $P(3, 3, 2)$ to the line $L$. We substitute the coordinates of $P$ into the parametric equation of $L$: $3 = 1 + 2\lambda \implies 2\lambda = 2 \implies \lambda = 1$ $3 = 1 + 2\lambda \implies 2\lambda = 2 \implies \lambda = 1$ $2 = 1 + \lambda \implies \lambda = 1$

Since we obtain a consistent value of $\lambda = 1$, the point $P(3, 3, 2)$ lies on the line $L$. The distance from a point to a line on which it lies is 0.

Alternatively, we found that point $B$ is $(3, 3, 2)$. Since $B$ is the intersection of $L$ and $L_2$, $B$ must lie on $L$. Therefore, the given point $(3, 3, 2)$ lies on $L$, and the distance is 0.