Question

If \(\overset{̄}{a} = \hat{i}+\hat{j}\) and \(\overset{̄}{b} = \hat{i}-\hat{k}\), then the point of intersection of the lines \(\overset{̄}{r}\times \overset{̄}{a} = \overset{̄}{b}\times \overset{̄}{a}\) and \(\overset{̄}{r}\times \overset{̄}{b} = \overset{̄}{a}\times \overset{̄}{b}\) is

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

Hint:

Each equation says r minus a fixed vector is parallel to another vector, so each is a line.

Answer 1

The Correct Option is A

Step 1: Rewrite each line in a simple form:
The first equation is \(\vec{r}\times\vec{a} = \vec{b}\times\vec{a}\), so \((\vec{r}-\vec{b})\times\vec{a} = \vec{0}\). A zero cross product means \(\vec{r}-\vec{b}\) is parallel to \(\vec{a}\). So the first line is \(\vec{r} = \vec{b} + \lambda\vec{a}\).

Step 2: Write the first line in components:
With \(\vec{a} = (1,1,0)\) and \(\vec{b} = (1,0,-1)\):
\[ \vec{r} = (1+\lambda,\ \lambda,\ -1) \]

Step 3: Write the second line:
The second equation \(\vec{r}\times\vec{b} = \vec{a}\times\vec{b}\) gives \((\vec{r}-\vec{a})\times\vec{b} = \vec{0}\). So \(\vec{r} = \vec{a} + \mu\vec{b} = (1+\mu,\ 1,\ -\mu)\).

Step 4: Find the point of intersection:
Equate the components of the two lines. From the \(y\) coordinates, \(\lambda = 1\). From the \(z\) coordinates, \(-1 = -\mu\), so \(\mu = 1\). The \(x\) coordinates are \(1+\lambda = 2\) and \(1+\mu = 2\), which agree. So the point is \((2, 1, -1)\).

Step 5: Why the other options are wrong:
The first line always has \(z = -1\), and the second always has \(y = 1\). Options (B), (C) and (D) have \(z = 1\), which breaks the first line, so none of them can be the common point.

Final Answer:
The lines meet at \((2, 1, -1)\), which is option (A). \[ \boxed{(2,\,1,\,-1)} \]