Question

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a\hat{i} - \hat{j})$, $a \neq 0$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + a\hat{k})$ from the origin is:

• A. 5
• B. 10
• C. 17
• D. 26

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two lines in 3D space in vector form. We need to find their point of intersection, and then calculate the square of the distance of this point from the origin.
Step 2: Key Formula or Approach:
1. Write the parametric equations for x, y, and z for each line.
2. To find the point of intersection, equate the corresponding coordinate expressions. This will give a system of three equations in two unknowns, $\lambda$ and $\mu$.
3. Solve any two equations to find the values of $\lambda$ and $\mu$. We will also need to find the value of the unknown constant 'a'.
4. Substitute the values of $\lambda$ and $\mu$ into the third equation to ensure consistency (i.e., that the lines actually intersect).
5. Once $\lambda$ or $\mu$ is found, substitute it back into the corresponding line's vector equation to find the position vector of the intersection point.
6. Calculate the square of the distance from the origin, which is simply the square of the magnitude of this position vector ($x^2+y^2+z^2$).
Step 3: Detailed Explanation:
Let the two lines be $L_1$ and $L_2$.
For $L_1$: $\vec{r} = (1, 1, -1) + \lambda(a, -1, 0)$.
The parametric equations are:
$x = 1 + a\lambda$
$y = 1 - \lambda$
$z = -1$
For $L_2$: $\vec{r} = (4, 0, -1) + \mu(2, 0, a)$.
The parametric equations are:
$x = 4 + 2\mu$
$y = 0$
$z = -1 + a\mu$
At the point of intersection, the coordinates must be the same. Let's equate them:
1) $1 + a\lambda = 4 + 2\mu$
2) $1 - \lambda = 0$
3) $-1 = -1 + a\mu$
From equation (2), we can directly find $\lambda$:
$1 - \lambda = 0 \implies \lambda = 1$.
From equation (3), we have:
$-1 = -1 + a\mu \implies a\mu = 0$.
We are given that $a \neq 0$, so we must have $\mu = 0$.
Now we use equation (1) to find the value of $a$. Substitute $\lambda=1$ and $\mu=0$:
$1 + a(1) = 4 + 2(0)$
$1 + a = 4$
$a = 3$.
Since we found a consistent value for $a$, the lines do intersect.
To find the point of intersection, we can substitute $\lambda=1$ into the equations for $L_1$ or $\mu=0$ into the equations for $L_2$.
Using $L_1$ with $\lambda=1$ and $a=3$:
$x = 1 + (3)(1) = 4$
$y = 1 - 1 = 0$
$z = -1$
The point of intersection is P(4, 0, -1).
The distance of this point from the origin O(0, 0, 0) is $|\vec{OP}|$.
$|\vec{OP}| = \sqrt{(4-0)^2 + (0-0)^2 + (-1-0)^2} = \sqrt{4^2 + 0^2 + (-1)^2} = \sqrt{16 + 0 + 1} = \sqrt{17}$.
The square of the distance is:
$|\vec{OP}|^2 = (\sqrt{17})^2 = 17$.
Step 4: Final Answer:
The square of the distance from the origin is 17.