Vectors 4 Question 11
11. Consider the cube in the first octant with sides $O P$, $O Q$ and $O R$ of length 1, along the $X$-axis, $Y$-axis and $Z$-axis, respectively, where $O(0,0,0)$ is the origin. Let $S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT. If $\mathbf{p}=\mathbf{S P}, \mathbf{q}=\mathbf{S Q}, \mathbf{r}=\mathbf{S R}$ and $\mathbf{t}=\mathbf{S T}$, then the value of $|(\mathbf{p} \times \mathbf{q}) \times(\mathbf{r} \times \mathbf{t})|$ is ….. .
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Answer:
Correct Answer: 11. (0.5)
Solution:
- (0.5) Here, $P(1,0,0), Q(0,1,0), R(0,0,1), T=(1,1,1)$ and $S=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.
Now,
$$ \vec{p}=\overrightarrow{S P}=\overrightarrow{O P}-\overrightarrow{O S} $$
$$ =\left(\frac{1}{2} \hat{i}-\frac{1}{2} \hat{j}-\frac{1}{2} \hat{k}\right)=\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) $$
$\vec{q}=\overrightarrow{S Q}=\frac{1}{2}(-\hat{i}+\hat{j}-\hat{k})$
$$ \vec{r}=\overrightarrow{S R}=\frac{1}{2}(-\hat{i}-\hat{j}+\hat{k}) $$
and $\quad \vec{t}=\overrightarrow{S T}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$
$\vec{p} \times \vec{q}=\frac{1}{4}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ -1 & 1 & -1\end{array}\right|=\frac{1}{4}(2 \hat{i}+2 \hat{j})$
and $\quad \vec{r} \times \vec{t}=\frac{1}{4}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ -1 & -1 & 1 \ 1 & 1 & 1\end{array}\right|=\frac{1}{4}(-2 \hat{i}+2 \hat{j})$
Now, $(\vec{p} \times \vec{q}) \times(\vec{r} \times \vec{t})=\frac{1}{16}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 0 \ -2 & 2 & 0\end{array}\right|$
$$ =\frac{1}{16}(8 \hat{k})=\frac{1}{2} \hat{k} $$
$\therefore|(p \times q) \times(\vec{r} \times \vec{t})|=\left|\frac{1}{2} \hat{k}\right|=\frac{1}{2}=0.5$
