Question

The volume of a parallelopiped with diagonals of three non parallel adjacent faces given by the vectors $\hat{i},\hat{j},\hat{k}$ is

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

Answer 1

The correct option is C $\frac{1}{2}$

Consider the face with edges $\overrightarrow{q}$ and $\overrightarrow{r}$ and diagonal $\overrightarrow{i}$
$\therefore \overrightarrow{q}+\overrightarrow{r}=\hat{i}$
$\overrightarrow{r}+\overrightarrow{p}=\hat{j}$
$\overrightarrow{p}+\overrightarrow{q}=\hat{k}$
Adding $\overrightarrow{q}+\overrightarrow{r}+\overrightarrow{r}+\overrightarrow{p}+\overrightarrow{p}+\overrightarrow{q}=\hat{i}+\hat{j}+\hat{k}$
$2(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
We have $\overrightarrow{p}+\overrightarrow{r}=\hat{j}$
$\Rightarrow \overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{q}+\hat{j}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{q}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})-\hat{j}$
$=\frac{\hat{i}-\hat{j}+\hat{k}}{2}$
we have $\overrightarrow{q}+\overrightarrow{r}=\hat{i}$
$\Rightarrow \overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{p}+\hat{i}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{p}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})-\hat{i}$
$=\frac{1}{2}(-\hat{i}+\hat{j}+\hat{k})$
we have $\overrightarrow{p}+\overrightarrow{q}=\hat{k}$
$\Rightarrow \overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \hat{k}+\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\overrightarrow{r}=\frac{1}{2}(\hat{i}+\hat{j}-\hat{k})$
Volume $=[\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}]$
$=\frac{1}{8}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$
$\Rightarrow \frac{1}{8}[-1\left(0\right)-1(-1-1)+1(1+1)]$
$=\frac{1}{8}[2+2]=\frac{4}{8}=\frac{1}{2}$