Question

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

• A. $3\hat{i}+2\hat{j}+\hat{k}$
• B. $\hat{i}-\hat{j}-\hat{k}$
• C. $4\hat{i}+2\hat{j}-\hat{k}$
• D. $3\hat{i}+\hat{j}-\hat{k}$

Hint:

Whenever you see a system of symmetric cross product equations like $\vec{r}\times\vec{a}=\vec{b}\times\vec{a}$ and $\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$, the intersection point is always simply the vector sum of the two constant baseline vectors: $\vec{r} = \vec{a} + \vec{b}$.

Answer 1

The Correct Option is D

Concept: Vector cross-product equations can be simplified by collecting all terms on one side and factoring out the common vector multipliers. For any two vectors where $\vec{u} \times \vec{v} = 0$, the vectors must be collinear or parallel, allowing us to represent them via a scalar parameter equations system. Step 1: Factor and rewrite the first vector equation.
The first vector intersection condition is given as: $$\vec{r}\times\vec{a} = \vec{b}\times\vec{a}$$ Bring all terms to the left side and factor out the common cross multiplier $\times\vec{a}$: $$\vec{r}\times\vec{a} - \vec{b}\times\vec{a} = 0 \quad \Rightarrow \quad (\vec{r} - \vec{b})\times\vec{a} = 0$$ Since the cross product is zero, the vector $(\vec{r} - \vec{b})$ must be parallel to $\vec{a}$. We can express this using a scalar parameter $\lambda$: $$\vec{r} - \vec{b} = \lambda\vec{a} \quad \Rightarrow \quad \vec{r} = \vec{b} + \lambda\vec{a} \quad \cdots (1)$$

Step 2: Factor and rewrite the second vector equation.
The second vector intersection condition is given as: $$\vec{r}\times\vec{b} = \vec{a}\times\vec{b}$$ Similarly, group the cross terms on one side and factor out $\times\vec{b}$: $$\vec{r}\times\vec{b} - \vec{a}\times\vec{b} = 0 \quad \Rightarrow \quad (\vec{r} - \vec{a})\times\vec{b} = 0$$ This implies that $(\vec{r} - \vec{a})$ must be parallel to $\vec{b}$. Expressing this via a second scalar parameter $\mu$: $$\vec{r} - \vec{a} = \mu\vec{b} \quad \Rightarrow \quad \vec{r} = \vec{a} + \mu\vec{b} \quad \cdots (2)$$

Step 3: Equate expressions to solve for the parameters.
Equating the two parametric representations for the shared intersection point $\vec{r}$ from equation (1) and equation (2): $$\vec{b} + \lambda\vec{a} = \vec{a} + \mu\vec{b}$$ Rearranging the terms to group parameters on one side: $$\lambda\vec{a} - \mu\vec{b} = \vec{a} - \vec{b} \quad \Rightarrow \quad (\lambda - 1)\vec{a} = (\mu - 1)\vec{b}$$ We are given that $\vec{a} = \hat{i}+\hat{j}$ and $\vec{b} = 2\hat{i}-\hat{k}$. Since these two vectors point in completely independent directions, they are non-collinear. A linear combination of non-parallel vectors can only equal zero if both coefficients vanish simultaneously: $$\lambda - 1 = 0 \implies \lambda = 1$$ $$\mu - 1 = 0 \implies \mu = 1$$

Step 4: Calculate the explicit coordinates of the intersection point.
Substitute $\lambda = 1$ back into our first parametric vector line equation (1): $$\vec{r} = \vec{b} + 1\vec{a} = \vec{a} + \vec{b}$$ Substitute the given unit component values: $$\vec{r} = (\hat{i} + \hat{j}) + (2\hat{i} - \hat{k}) = 3\hat{i} + \hat{j} - \hat{k}$$ This matches option (D) perfectly.