Search any question & find its solution
Question:
Answered & Verified by Expert
Area of a rectangle having vertices
$$
\mathbf{A}\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right), \quad \mathrm{B}\left(\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right)
$$
$C\left(\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)$ and $\mathrm{D}\left(-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)$ is
(a) $\frac{1}{2}$ sq. unit
(b) $1 \mathrm{sq}$. unit
(c) 2 sq. units
(d) 4 sq. units
$$
\mathbf{A}\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right), \quad \mathrm{B}\left(\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right)
$$
$C\left(\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)$ and $\mathrm{D}\left(-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)$ is
(a) $\frac{1}{2}$ sq. unit
(b) $1 \mathrm{sq}$. unit
(c) 2 sq. units
(d) 4 sq. units
Solution:
1792 Upvotes
Verified Answer
$\overrightarrow{O A}=-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{O B}=\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}$
$\overrightarrow{O C}=\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k} \overrightarrow{O D}=-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$
$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=2 \hat{i}$
$\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=2 \hat{i}-\hat{j}$
$$
\overline{A B} \times \overline{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 0 \\
2 & -1 & 0
\end{array}\right|=\hat{k}(-2)=-2 \hat{k}
$$
$|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-2)^2}=2$ sq. units
Hence area of rectangle $=2$ sq. units.
Thus option (c) is correct.
$\overrightarrow{O C}=\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k} \overrightarrow{O D}=-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$
$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=2 \hat{i}$
$\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=2 \hat{i}-\hat{j}$
$$
\overline{A B} \times \overline{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 0 \\
2 & -1 & 0
\end{array}\right|=\hat{k}(-2)=-2 \hat{k}
$$
$|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-2)^2}=2$ sq. units
Hence area of rectangle $=2$ sq. units.
Thus option (c) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.