Question

Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \left((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}\right) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

• A. $\frac{1}{5}$
• B. 2
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

2

Answer 2

Step 1: Calculate $\vec{a} \times (\hat{i} + \hat{j})$:
$\vec{a} \times (\hat{i} + \hat{j}) =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\\2 & 1 & -1 \\\1 & 1 & 0\end{vmatrix}= -\hat{i} + \hat{k}$
Step 2: Calculate $(\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}$:
$(\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i} = (-\hat{i} + \hat{k}) \times \hat{i} = \hat{k} + \hat{j}$
Step 3: Calculate $((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$:
$((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i} = (\hat{k} + \hat{j}) \times \hat{i} = \hat{j} - \hat{k}$
Thus, $\vec{b} = \hat{j} - \hat{k}$.
Step 4: Find the projection of $\vec{a}$ on $\vec{b}$:
$\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Calculating $\vec{a} \cdot \vec{b}$ and $|\vec{b}|$:
$\vec{a} \cdot \vec{b} = (2)(0) + (1)(1) + (-1)(-1) = 1 + 1 = 2$
$|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
$\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Therefore, the square of the projection is:
$(\sqrt{2})^2 = 2$