Question

If $\hat{i}$, $\hat{j}$, $\hat{k}$ are unit orthonormal vectors and $\bar{a}$ is a vector, if $\bar{a} \times \bar{r} = \hat{j}$, then $\bar{a} \cdot \bar{r}$ :

• A. 0
• B. $\in [-1, 1]$
• C. $\in (0, 1)$
• D. $\in R$

Answer 1

Answer: (D)

(D)

Answer 2

Given the equation $\bar{a} \times \bar{r} = \hat{j}$. This implies that $\hat{j}$ is perpendicular to both $\bar{a}$ and $\bar{r}$. Therefore, $\bar{a} \cdot \hat{j} = 0$ and $\bar{r} \cdot \hat{j} = 0$.

Let $\bar{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\bar{r} = r_1\hat{i} + r_2\hat{j} + r_3\hat{k}$. From $\bar{a} \cdot \hat{j} = 0$, we get $a_2 = 0$. So, $\bar{a} = a_1\hat{i} + a_3\hat{k}$. From $\bar{r} \cdot \hat{j} = 0$, we get $r_2 = 0$. So, $\bar{r} = r_1\hat{i} + r_3\hat{k}$.

Now, let's compute the cross product $\bar{a} \times \bar{r}$: $\bar{a} \times \bar{r} = (a_1\hat{i} + a_3\hat{k}) \times (r_1\hat{i} + r_3\hat{k})$ $\bar{a} \times \bar{r} = a_1r_1(\hat{i} \times \hat{i}) + a_1r_3(\hat{i} \times \hat{k}) + a_3r_1(\hat{k} \times \hat{i}) + a_3r_3(\hat{k} \times \hat{k})$ Using $\hat{i} \times \hat{i} = \hat{k} \times \hat{k} = 0$, $\hat{i} \times \hat{k} = -\hat{j}$, and $\hat{k} \times \hat{i} = \hat{j}$: $\bar{a} \times \bar{r} = a_1r_3(-\hat{j}) + a_3r_1(\hat{j}) = (a_3r_1 - a_1r_3)\hat{j}$ We are given $\bar{a} \times \bar{r} = \hat{j}$. Thus, $(a_3r_1 - a_1r_3)\hat{j} = 1\hat{j}$ This implies $a_3r_1 - a_1r_3 = 1$.

Now, let's compute the dot product $\bar{a} \cdot \bar{r}$: $\bar{a} \cdot \bar{r} = (a_1\hat{i} + a_3\hat{k}) \cdot (r_1\hat{i} + r_3\hat{k})$ $\bar{a} \cdot \bar{r} = a_1r_1(\hat{i} \cdot \hat{i}) + a_1r_3(\hat{i} \cdot \hat{k}) + a_3r_1(\hat{k} \cdot \hat{i}) + a_3r_3(\hat{k} \cdot \hat{k})$ Using $\hat{i} \cdot \hat{i} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$: $\bar{a} \cdot \bar{r} = a_1r_1(1) + a_1r_3(0) + a_3r_1(0) + a_3r_3(1) = a_1r_1 + a_3r_3$ We have the condition $a_3r_1 - a_1r_3 = 1$. We need to determine the possible values of $a_1r_1 + a_3r_3$.

Let's test with specific vectors. Choose $\bar{a} = \hat{i}$. Then $a_1=1, a_3=0$. The condition $a_3r_1 - a_1r_3 = 1$ becomes $0 \cdot r_1 - 1 \cdot r_3 = 1$, which means $r_3 = -1$. The vector $\bar{r}$ must be of the form $\bar{r} = r_1\hat{i} - \hat{k}$ (since $r_2=0$). Then, $\bar{a} \cdot \bar{r} = \hat{i} \cdot (r_1\hat{i} - \hat{k}) = r_1(\hat{i} \cdot \hat{i}) - (\hat{i} \cdot \hat{k}) = r_1(1) - 0 = r_1$. Since $r_1$ can be any real number, $\bar{a} \cdot \bar{r}$ can take any real value.