Question 61

Mathematics Scalar and Vector Products Hard

Let \(\vec{a}=\hat{i}+\hat{j}\) and \(\vec{b}=2\hat{i}-\hat{k}\), the point of intersection of the lines \(\vec{r}\times\vec{a}=\vec{b}\times\vec{a}\) and \(\vec{r}\times\vec{b}=\vec{a}\times\vec{b}\) is

(A) \(-\hat{i}+\hat{j}+\hat{k}\)
(B) \(3\hat{i}-\hat{j}+\hat{k}\)
(C) \(\hat{i}-\hat{j}-\hat{k}\)
(D) \(3\hat{i}+\hat{j}-\hat{k}\)
View Dynamic Solution & Explanation
Correct Solution: Option D

Step-by-step Solution:

Given that \[ \overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{a} \] or \[ \overrightarrow{r} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{a} = 0 \Rightarrow (\vec{r} - \vec{b}) \times \vec{a} = 0 \] So, the line passes through \(\vec{b}\) and is parallel to \(\vec{a}\): \[ \Rightarrow \vec{r} = \vec{b} + \alpha \vec{a} \] Again, given that \[ \overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{a} \times \overrightarrow{b} \] \[ \Rightarrow \vec{r} = \vec{a} + \beta \vec{b} \] For the intersection of both the lines, \[ \vec{b} + \alpha \vec{a} = \vec{a} + \beta \vec{b} \] By comparing coefficients, \[ \alpha = \beta = 1 \] Hence, the point of intersection is \[ \vec{a} + \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \]