Question

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{c} = \hat{j} - \hat{k}$, $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$, then $\vec{b} = $

• A. $\hat{i}$
• B. $-\hat{i}$
• C. $\hat{j}$
• D. $\hat{k}$

Hint:

To find the answer rapidly without algebra, test the options against the simple dot product condition $\vec{a} \cdot \vec{b} = 1$. For option (A), $(\hat{i}+\hat{j}+\hat{k})\cdot\hat{i} = 1$. For option (C), $(\hat{i}+\hat{j}+\hat{k})\cdot\hat{j} = 1$. To break the tie, test the cross product condition with option (A): $(\hat{i}+\hat{j}+\hat{k})\times\hat{i} = 0 - \hat{k} + \hat{j} = \hat{j} - \hat{k} = \vec{c}$. This confirms it immediately!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{j} - \hat{k}$. We need to solve for an unknown vector $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ that simultaneously satisfies the conditions $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$.

Step 2: Key Formula or Approach:
We set up algebraic component systems using the definition of dot and cross products: $$\vec{a} \cdot \vec{b} = x + y + z = 1$$ The cross product condition $\vec{a} \times \vec{b} = \vec{c}$ can be expanded using a matrix determinant format: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$$

Step 3: Detailed Explanation:
Expanding the vector cross product determinant along its first row: $$\hat{i}(z - y) - \hat{j}(z - x) + \hat{k}(y - x) = 0\hat{i} + 1\hat{j} - 1\hat{k}$$ Equating matching components from both sides gives the following relationships: align z - y &= 0 y = z
-(z - x) &= 1 x - z = 1 z = x - 1
y - x &= -1 y = x - 1 align Now, substitute these expressions for $y$ and $z$ into the dot product condition equation ($x + y + z = 1$): $$x + (x - 1) + (x - 1) = 1$$ $$3x - 2 = 1 \implies 3x = 3 \implies x = 1$$ Substitute $x = 1$ back to evaluate $y$ and $z$: $$y = 1 - 1 = 0$$ $$z = 1 - 1 = 0$$ Putting these coordinates back into our definition for vector $\vec{b}$: $$\vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$$

Step 4: Final Answer:
The vector $\vec{b}$ is equal to $\hat{i}$, which matches option (A).