Question

The lines \( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \) and \( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \) intersect at a point, where \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = \hat{i} - \hat{k} \). Find the point of intersection.

• (A) \( (2,1,-1) \)
• (B) \( (1,1,-1) \)
• (C) \( (2,0,-1) \)
• (D) \( (1,0,-1) \)

Hint:

Whenever a vector equation is of the form \[ (\vec{r}-\vec{a})\times\vec{b}=0, \] the position vector \( \vec{r}-\vec{a} \) is parallel to \( \vec{b} \). Directly convert it into the parametric form \[ \vec{r}=\vec{a}+\lambda\vec{b}. \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Vector product equations can often be simplified into standard line equations in parametric form.
The equation \( \vec{r} \times \vec{m} = \vec{c} \times \vec{m} \) can be rewritten as \( (\vec{r} - \vec{c}) \times \vec{m} = 0 \).
This implies that the vector \( (\vec{r} - \vec{c}) \) is parallel to the vector \( \vec{m} \).
Therefore, \( \vec{r} - \vec{c} = \lambda \vec{m} \), or \( \vec{r} = \vec{c} + \lambda \vec{m} \).
This is the parametric form of a line passing through point \( \vec{c} \) and parallel to direction vector \( \vec{m} \).
We have two such lines, and we need to find the specific values of the parameters for each line that yield the same position vector \( \vec{r} \).
Step 2: Key Formula or Approach:
For Line 1 (\( L_1 \)): \( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \implies \vec{r} = \vec{b} + \lambda \vec{a} \).
For Line 2 (\( L_2 \)): \( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \implies \vec{r} = \vec{a} + \mu \vec{b} \).
We convert vectors \( \vec{a} \) and \( \vec{b} \) into coordinate form:
\( \vec{a} = (1, 1, 0) \)
\( \vec{b} = (1, 0, -1) \)
Step 3: Detailed Explanation:
Write the parametric equations for both lines.
For \( L_1 \):
\( \vec{r} = (1, 0, -1) + \lambda(1, 1, 0) \)
\( \vec{r} = (1 + \lambda, \lambda, -1) \)
For \( L_2 \):
\( \vec{r} = (1, 1, 0) + \mu(1, 0, -1) \)
\( \vec{r} = (1 + \mu, 1, -\mu) \)
At the point of intersection, the coordinates of \( L_1 \) and \( L_2 \) must be identical.
Comparing corresponding components:
x-components: \( 1 + \lambda = 1 + \mu \implies \lambda = \mu \)
y-components: \( \lambda = 1 \)
z-components: \( -1 = -\mu \implies \mu = 1 \)
All three equations lead to a consistent solution: \( \lambda = 1 \) and \( \mu = 1 \).
Now, substitute \( \lambda = 1 \) into the equation for \( L_1 \) (or \( \mu = 1 \) into \( L_2 \)) to find the intersection point:
\( \vec{r} = (1 + 1, 1, -1) = (2, 1, -1) \)
The point of intersection is \( (2, 1, -1) \).
This matches Option (A).
Step 4: Final Answer:
By converting the cross-product line equations into parametric form, we found that both lines contain the point \( (2, 1, -1) \) when their respective parameters are set to 1. This corresponds to Option (A).