Q219 / 270 · NDA 2024 (Phase 1) · Mathematics
For any vector $\vec{r}$, what is $(\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})+(\vec{r} \cdot \hat{j})(\vec{r} \times \hat{j})+(\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})$ equal to?
- Correct answerA$\overrightarrow{0}$
- B$\vec{r}$
- C$2 \vec{r}$
- D$3 \vec{r}$
Solution
$(\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})+(\vec{r} \cdot \hat{j})(\vec{r} \times \hat{j})+(\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})$ ...(i)
We have,
$\begin{aligned} \vec{r} & =a \hat{i}+b \hat{j}+c \hat{k} \\ (\vec{r} \hat{i}) & =(a \hat{i}+b \hat{j}+c \hat{k}) \cdot \hat{i}=a \\ (\vec{r} \cdot \hat{j}) & =b \\ (\vec{r} \cdot \hat{k}) & =c\end{aligned}$
and $\qquad \vec{r} \times \hat{i}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a & b & c \\
1 & 0 & 0
\end{array}\right|$
$=-\hat{j}(-c)-b \hat{k}=c \hat{j}-b \hat{k}$
Similarly, $\quad \vec{r} \times \hat{j}=-c \hat{i}+a \hat{k}$
and$\vec{r} \hat{k}=b \hat{i}-a \hat{j}$
Putting these values in (i) we get
$\begin{aligned}
& a(\hat{j}-b \hat{k})+b(-c \hat{i}+a \hat{k})+c(b \hat{i}-a \hat{j}) \\
&=a c \hat{j}-a b \hat{k}-b c \hat{i}+a b \hat{k}+b c \hat{i}-a c \hat{j}=0
\end{aligned}$
Hence, the correct answer is option: 'A'.