Question

A line from origin meets the lines $\frac{x-1}{1}=\frac{y}{2}=\frac{z+1}{2}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z}{1}$ at A and B, then length of AB is equal to 6.

Answer 1

6

Answer 2

Let the line from the origin be $L_0$. $L_0$ passes through $O(0,0,0)$. Let $L_1$ be the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z+1}{2}$ and $L_2$ be the line $\frac{x}{2}=\frac{y-1}{3}=\frac{z}{1}$. $L_0$ intersects $L_1$ at point A and $L_2$ at point B. Since A and B lie on $L_0$, they are collinear with the origin O. Thus, the vector $\vec{OA}$ is parallel to $\vec{OB}$.

A general point on $L_1$ is $A = (1+t_1, 2t_1, -1+2t_1)$ for some scalar $t_1$. A general point on $L_2$ is $B = (2t_2, 1+3t_2, t_2)$ for some scalar $t_2$.

Since O, A, and B are collinear, $\vec{OA}$ must be parallel to $\vec{OB}$. This means the coordinates of A and B are proportional. So, $(1+t_1, 2t_1, -1+2t_1) = k(2t_2, 1+3t_2, t_2)$ for some scalar $k$. This gives the system of equations:

1. $1+t_1 = 2kt_2$
2. $2t_1 = k(1+3t_2)$
3. $-1+2t_1 = kt_2$

Substitute $kt_2$ from equation (3) into equation (1): $1+t_1 = 2(-1+2t_1)$ $1+t_1 = -2+4t_1$ $3 = 3t_1 \implies t_1 = 1$.

Now substitute $t_1=1$ back into the equations: From (3): $-1+2(1) = kt_2 \implies 1 = kt_2$. From (1): $1+1 = 2kt_2 \implies 2 = 2kt_2 \implies kt_2 = 1$. (Consistent) From (2): $2(1) = k(1+3t_2) \implies 2 = k+3kt_2$. Substitute $kt_2=1$ into this equation: $2 = k+3(1) \implies 2 = k+3 \implies k = -1$.

Now we have $k=-1$ and $kt_2=1$. $(-1)t_2 = 1 \implies t_2 = -1$.

Using $t_1=1$, the coordinates of point A are: $A = (1+1, 2(1), -1+2(1)) = (2, 2, 1)$.

Using $t_2=-1$, the coordinates of point B are: $B = (2(-1), 1+3(-1), -1) = (-2, 1-3, -1) = (-2, -2, -1)$.

The length of the segment AB is the distance between points A and B: $AB = \sqrt{(-2-2)^2 + (-2-2)^2 + (-1-1)^2}$ $AB = \sqrt{(-4)^2 + (-4)^2 + (-2)^2}$ $AB = \sqrt{16 + 16 + 4}$ $AB = \sqrt{36}$ $AB = 6$.

The problem statement asserts that "A line from origin meets the lines ... at A and B, then length of AB is equal to 6." Our calculation confirms this statement by finding the points A and B and verifying that the distance between them is indeed 6.