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 vector form that are known to intersect. These lines contain an unknown constant 'a'. The task is to first find the coordinates of their intersection point, and then find the square of its distance from the origin.
Step 2: Key Formula or Approach:
1. Convert to Parametric Form: Express the x, y, and z coordinates for each line in terms of its parameter ($\lambda$ or $\mu$).
2. Equate Coordinates: Set the corresponding coordinate expressions equal to each other. This creates a system of three linear equations with the variables $\lambda, \mu,$ and $a$.
3. Solve the System: Solve the system of equations to find the specific values of the parameters $\lambda$ and $\mu$, and the constant $a$.
4. Determine Intersection Point: Substitute the found value of $\lambda$ (or $\mu$) back into the parametric equations of its line to get the (x, y, z) coordinates of the intersection point.
5. Calculate Squared Distance: For a point $(x, y, z)$, the square of its distance from the origin is $x^2 + y^2 + z^2$.
Step 3: Detailed Explanation:
Let's write the parametric equations for both lines.
Line 1: $\vec{r} = (1, 1, -1) + \lambda(a, -1, 0)$
$x = 1 + a\lambda$
$y = 1 - \lambda$
$z = -1$
Line 2: $\vec{r} = (4, 0, -1) + \mu(2, 0, a)$
$x = 4 + 2\mu$
$y = 0$
$z = -1 + a\mu$
At the point of intersection, the coordinates from both lines must be equal. This gives us a system of three equations:
(i) $1 + a\lambda = 4 + 2\mu$
(ii) $1 - \lambda = 0$
(iii) $-1 = -1 + a\mu$
Let's solve the system, starting with the simplest equations.
From equation (ii), we immediately get $\lambda = 1$.
From equation (iii), we get $a\mu = 0$. Since we are given that $a \neq 0$, it must be that $\mu = 0$.
Now, we substitute $\lambda = 1$ and $\mu = 0$ into equation (i) to find $a$:
$1 + a(1) = 4 + 2(0)$
$1 + a = 4 \implies a = 3$.
The values are consistent, confirming the lines intersect.
To find the intersection point, substitute $\lambda = 1$ into the equations for Line 1:
$x = 1 + (3)(1) = 4$
$y = 1 - 1 = 0$
$z = -1$
The point of intersection is P(4, 0, -1).
Finally, we calculate the square of the distance from the origin O(0,0,0) to P(4,0,-1):
Distance$^2 = (4-0)^2 + (0-0)^2 + (-1-0)^2 = 16 + 0 + 1 = 17$.
Step 4: Final Answer:
The square of the distance from the origin is 17.