If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$, then the vector $\bar{b}$ satisfying $\bar{a} imes \bar{b}+\bar{c}=\overline{0}$ and $\bar{a} \cdot \bar{b}=3$ is

Question

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$, then the vector $\bar{b}$ satisfying $\bar{a} 	imes \bar{b}+\bar{c}=\overline{0}$ and $\bar{a} \cdot \bar{b}=3$ is

Aicharya · Accepted Answer

Given: - $\mathbf{a}=\mathbf{j}-\mathbf{k}$, - $\mathbf{c}=\mathbf{i}-\mathbf{j}-\mathbf{k}$, - Find $\mathbf{b}$ such that $(\mathbf{a} \times \mathbf{b})+\mathbf{c}=0$ and $\mathbf{a} \cdot \mathbf{b}=\mathbf{3}$.
Step 1: Write b in component form: $\mathbf{b}=p \mathbf{i}+q \mathbf{j}+r \mathbf{k}$
Step 2: Compute $\mathbf{a} \times \mathbf{b}$ : $\begin{aligned} \mathbf{a} & =(0,1,-1), \quad \mathbf{b}=(p, q, r) \\ \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -1 \\ p & q & r \end{array}\right| & =\mathbf{i}(1 r-(-1) q)-\mathbf{j}(0 r-(-1) p)+\mathbf{k}(0 q-1 p) \\ \mathbf{a} & \times \mathbf{b}=(r+q) \mathbf{i}-(-p) \mathbf{j}-p \mathbf{k} \\ \mathbf{a} & \times \mathbf{b}=(r+q) \mathbf{i}+p \mathbf{j}-p \mathbf{k} \end{aligned}$
Step 3: Use $(\mathbf{a} \times \mathbf{b})+\mathbf{c}=0$ : $(r+q) \mathbf{i}+p \mathbf{j}-p \mathbf{k}+(\mathbf{i}-\mathbf{j}-\mathbf{k})=0$
Equating components: - $i$-component: $r+q+1=0 \Rightarrow r+q=-1$, - $j$-component: $p-1=0 \quad \Rightarrow \quad p=1$, - $k$-component: $-p-1=0 \Rightarrow p=-1$.
Step 4: Solve $\mathbf{a} \cdot \mathbf{b}=3$ : $\mathbf{a} \cdot \mathbf{b}=(0) p+(1) q+(-1) r=q-r=3$
From $r+q=-1$ and $q-r=3$, solve for $q$ and $r$ : - Add equations: $2 q=2 \Rightarrow q=1$, - Substitute: $r+1=-1 \Rightarrow r=-2$.
Final b: $\mathbf{b}=-\mathbf{i}+\mathbf{j}-\mathbf{k}$
Answer: $\mathbf{- i}+\mathbf{j}-2\mathbf{k}$.

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$, then the vector $\bar{b}$ satisfying…

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