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Question: Let $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = 2\hat{i} - \hat{k}$, $\vec{r}$ is a vector satisfying ...

Let a⃗=i^+j^\vec{a} = \hat{i} + \hat{j}, b⃗=2i^−k^\vec{b} = 2\hat{i} - \hat{k}, r⃗\vec{r} is a vector satisfying r⃗×a⃗=b⃗×a⃗\vec{r} \times \vec{a} = \vec{b} \times \vec{a} and r⃗×b⃗=a⃗×b⃗\vec{r} \times \vec{b} = \vec{a} \times \vec{b}, then r⃗\vec{r} is

A

i^+j^−k^\hat{i} + \hat{j} - \hat{k}

B

3i^+j^−k^3\hat{i} + \hat{j} - \hat{k}

C

3i^+j^+k^3\hat{i} + \hat{j} + \hat{k}

D

−i^+j^−k^-\hat{i} + \hat{j} - \hat{k}

Answer

3\hat{i} + \hat{j} - \hat{k}

Explanation

Solution

The given equations are r⃗×a⃗=b⃗×a⃗\vec{r} \times \vec{a} = \vec{b} \times \vec{a} and r⃗×b⃗=a⃗×b⃗\vec{r} \times \vec{b} = \vec{a} \times \vec{b}. Rearranging the first equation: r⃗×a⃗−b⃗×a⃗=0⃗\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0} (r⃗−b⃗)×a⃗=0⃗(\vec{r} - \vec{b}) \times \vec{a} = \vec{0} This implies that (r⃗−b⃗)(\vec{r} - \vec{b}) is parallel to a⃗\vec{a}, so r⃗−b⃗=ka⃗\vec{r} - \vec{b} = k\vec{a} for some scalar kk. Thus, r⃗=b⃗+ka⃗\vec{r} = \vec{b} + k\vec{a}.

Rearranging the second equation: r⃗×b⃗−a⃗×b⃗=0⃗\vec{r} \times \vec{b} - \vec{a} \times \vec{b} = \vec{0} (r⃗−a⃗)×b⃗=0⃗(\vec{r} - \vec{a}) \times \vec{b} = \vec{0} This implies that (r⃗−a⃗)(\vec{r} - \vec{a}) is parallel to b⃗\vec{b}, so r⃗−a⃗=mb⃗\vec{r} - \vec{a} = m\vec{b} for some scalar mm. Thus, r⃗=a⃗+mb⃗\vec{r} = \vec{a} + m\vec{b}.

Equating the two expressions for r⃗\vec{r}: b⃗+ka⃗=a⃗+mb⃗\vec{b} + k\vec{a} = \vec{a} + m\vec{b} ka⃗−a⃗=mb⃗−b⃗k\vec{a} - \vec{a} = m\vec{b} - \vec{b} (k−1)a⃗=(m−1)b⃗(k-1)\vec{a} = (m-1)\vec{b}

Given a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=2i^−k^\vec{b} = 2\hat{i} - \hat{k}. Since a⃗\vec{a} and b⃗\vec{b} are not parallel, they are linearly independent. For the equation (k−1)a⃗=(m−1)b⃗(k-1)\vec{a} = (m-1)\vec{b} to hold, the coefficients must be zero: k−1=0  ⟹  k=1k-1 = 0 \implies k = 1 m−1=0  ⟹  m=1m-1 = 0 \implies m = 1

Substitute k=1k=1 into r⃗=b⃗+ka⃗\vec{r} = \vec{b} + k\vec{a}: r⃗=b⃗+1a⃗=a⃗+b⃗\vec{r} = \vec{b} + 1\vec{a} = \vec{a} + \vec{b}

Now, calculate r⃗\vec{r}: r⃗=(i^+j^)+(2i^−k^)\vec{r} = (\hat{i} + \hat{j}) + (2\hat{i} - \hat{k}) r⃗=(1+2)i^+1j^−1k^\vec{r} = (1+2)\hat{i} + 1\hat{j} - 1\hat{k} r⃗=3i^+j^−k^\vec{r} = 3\hat{i} + \hat{j} - \hat{k}