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

Concept:
If \[ \vec{r}\times\vec{a}=\vec{c}\times\vec{a}, \] then \[ (\vec{r}-\vec{c})\times\vec{a}=0. \] This means the vector \( \vec{r}-\vec{c} \) is parallel to \( \vec{a} \). Hence, the line can be written in vector form as \[ \vec{r}=\vec{c}+\lambda\vec{a}. \]

Step 1: Writing the first line in parametric form.
Given, \[ \vec{r}\times\vec{a}=\vec{b}\times\vec{a}. \] Therefore, \[ (\vec{r}-\vec{b})\times\vec{a}=0. \] Hence, \[ \vec{r}-\vec{b}=\lambda\vec{a}. \] Now, \[ \vec{a}=\hat{i}+\hat{j}=(1,1,0), \] and \[ \vec{b}=\hat{i}-\hat{k}=(1,0,-1). \] Thus, \[ \vec{r}=(1,0,-1)+\lambda(1,1,0). \] So, \[ \vec{r}=(1+\lambda,\lambda,-1). \] Therefore, the first line is \[ L_1:(1+\lambda,\lambda,-1). \]

Step 2: Writing the second line in parametric form.
Given, \[ \vec{r}\times\vec{b}=\vec{a}\times\vec{b}. \] Therefore, \[ (\vec{r}-\vec{a})\times\vec{b}=0. \] Hence, \[ \vec{r}-\vec{a}=\mu\vec{b}. \] Substituting the values, \[ \vec{r}=(1,1,0)+\mu(1,0,-1). \] Thus, \[ \vec{r}=(1+\mu,1,-\mu). \] Therefore, the second line is \[ L_2:(1+\mu,1,-\mu). \]

Step 3: Equating corresponding coordinates.
At the point of intersection, \[ (1+\lambda,\lambda,-1)=(1+\mu,1,-\mu). \] Comparing the \( y \)-coordinates: \[ \lambda=1. \] Comparing the \( z \)-coordinates: \[ -1=-\mu \] \[ \mu=1. \] Comparing the \( x \)-coordinates: \[ 1+\lambda=1+\mu \] \[ 1+1=1+1. \] Hence, the equations are consistent.

Step 4: Finding the point of intersection.
Substituting \[ \lambda=1 \] into \[ (1+\lambda,\lambda,-1), \] we get \[ (1+1,1,-1). \] Therefore, \[ \boxed{(2,1,-1)}. \]