Ex 10.4 Class 12 Maths Ncert Solutions

Given that \( \overrightarrow{a}=\hat{\imath}-7 \hat{\jmath}+7 \hat{k} \) and \( \overrightarrow{b}=3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k} \)
\(\overrightarrow{a} \times \overrightarrow{b}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\hat{\imath}[(-7 \times 2)-(-2 \times 7)]-\hat{\jmath}[(1 \times 2)-(7 \times 3)] +\hat{\mathrm{k}}[(-2 \times 1)-(-7 \times 3)] \)
\(\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(0) \hat{\imath}-(-19) \hat{\jmath}+(19) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=0 \hat{\imath}+19 \hat{\jmath}+19 \hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{0^{2}+19^{2}+19^{2}}=19 \sqrt{2}\)

Given that \( \overrightarrow{a}=3 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \) and
\(\overrightarrow{b}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}\)
Let \( \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \)
\(\Rightarrow \overrightarrow{\mathrm{p}}=(3 \hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}})+(\hat{\imath}+2 \hat{\jmath}-2 \hat{\mathrm{k}})\)
\(\Rightarrow \overrightarrow{\mathrm{p}}=(4 \hat{\imath}+4 \hat{\jmath}+0 \hat{\mathrm{k}})\)
Let \( \overrightarrow{q}=\overrightarrow{a}-\overrightarrow{b} \)
\(\Rightarrow \overrightarrow{\mathrm{q}}=(3 \hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}-2 \hat{\mathrm{k}})\)
\(\Rightarrow \overrightarrow{\mathrm{q}}=(2 \hat{\imath}+0 \hat{\jmath}+4 \hat{\mathrm{k}})\)
Now, we want to find a vector which is perpendicular to both \( \overrightarrow{p} \) and \( \overrightarrow{q} \). It is given by \( \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \).
Let \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \)
\(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}
\hat{i} & \hat{\jmath} & \hat{\mathrm{k}} \\
4 & 4 & 0 \\
3 & 0 & 2
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}[(4 \times 4)-(0 \times 0)]-\hat{\jmath}[(4 \times 4)-(2 \times 0)]+\hat{\mathrm{k}}[(4 \times 0)-(4 \times 2)]\)
\(\Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=(16) \hat{\mathrm{i}}-(16) \hat{\mathrm{j}}+((4) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=16 \hat{\mathrm{i}}-16 \hat{\jmath}-8 \hat{\mathrm{k}}\)
Therefore, \( \overrightarrow{r}=16 \hat{\imath}-16 \hat{\jmath}-8 \hat{k} \)
Now we want to find the unit vector. We know that unit vector means that the magnitude of the vector is 1 (unit).
It is defined as \( \overrightarrow{\mathrm{r}}=\frac{\overrightarrow{r}}{|\overrightarrow{r}|} \)
So, we find the magnitude of \( \overrightarrow{a} \) first.
\(\Rightarrow|\overrightarrow{\mathrm{r}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}=\sqrt{576}=24\)
\(\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}= \pm \frac{(16 \hat{\imath}-16 \hat{\jmath}-8 \widehat{\mathrm{k}})}{24}= \pm\left(\frac{2}{3} \hat{\imath}-\frac{2}{3} \hat{\jmath}-\frac{1}{3} \hat{\mathrm{k}}\right)\)

Let \( \overrightarrow{\mathrm{a}}=x \hat{\imath}+y \hat{\jmath}+z \hat{\mathrm{k}} \)
Given that it is a unit vector,
So, \( \Rightarrow|\overrightarrow{\mathrm{a}}|=\sqrt{x^{2}+y^{2}+z^{2}} \ldots\) (1)
Let \( \alpha, \beta, \theta \) be the angle \( \overrightarrow{\mathrm{a}} \) with \( \hat{\mathrm{i}}, \hat{\jmath}, \hat{\mathrm{k}} \) respectively.
Then,
\(\cos \alpha=\frac{\overrightarrow{a}, \hat{i}}{|\overrightarrow{a}||\hat{1}|}=\frac{x}{1}\)
\(\Rightarrow \alpha=\frac{\pi}{3}(\text { given })\)
\(\Rightarrow x=\cos \alpha=\cos \frac{\pi}{3}=\frac{1}{2}\)
\(\cos \beta=\frac{\overrightarrow{a}, \hat{j}}{|\overrightarrow{a}||\hat{\jmath}|}=\frac{x}{1}\)
\(\Rightarrow \beta=\frac{\pi}{4}(\text { given })\)
\(\Rightarrow x=\cos \beta=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
\(\cos \theta=\frac{\overrightarrow{a}, \mathrm{k}}{|\overrightarrow{a}||\widehat{k}|}=\frac{z}{1}\)
\(\Rightarrow z=\cos \theta\)
Putting value of \( x, y, z \) in equation (1)
\(\sqrt{x^{2}+y^{2}+z^{2}}=1\)
\(\Rightarrow x^{2}+y^{2}+z^{2}=1 \quad \text { (Squaring both sides) }\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+(\cos \theta)^{2}=1\)
\(\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \theta=1\)
\(\Rightarrow \cos ^{2} \theta=1-\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow \cos \theta= \pm \frac{1}{2}\)
As \( \theta \) should be acute, \( \theta=\cos ^{-1} \frac{1}{2}=\frac{\pi}{3} \)
\(\therefore z=\cos \theta=\cos \frac{\pi}{3}=\frac{1}{2}\)
Components of \( \overrightarrow{a} \) are the coefficients of \( \hat{\imath}, \hat{\jmath}, \hat{k} \), which are, \( x=\frac{1}{2} ; y=\frac{1}{\sqrt{2}} ; z=\frac{1}{2} \) with \( \theta=\frac{\pi}{3} \)

We solve for the left-hand side,
L.H.S. \( = \)
\((\overrightarrow{a}-\overrightarrow{b}) \times(\overrightarrow{a}+\overrightarrow{b})=(\overrightarrow{a} \times \overrightarrow{a})+(\overrightarrow{a} \times \overrightarrow{b})-(\overrightarrow{b} \times \overrightarrow{a})-(\overrightarrow{b} \times \overrightarrow{b})\)
We know \( \overrightarrow{a} \times \overrightarrow{a}=0 \) and \( \overrightarrow{b} \times \overrightarrow{b}=0 \)
Also, \( \overrightarrow{b} \times \overrightarrow{a}=-(\overrightarrow{a} \times \overrightarrow{b}) \)
Putting these in our equation, we get
\(\Rightarrow(\overrightarrow{a}-\overrightarrow{b}) \times(\overrightarrow{a}+\overrightarrow{b})=0+(\overrightarrow{a} \times \overrightarrow{b})+(\overrightarrow{a} \times \overrightarrow{b})=0\)
\(\Rightarrow(\overrightarrow{a}-\overrightarrow{b}) \times(\overrightarrow{a}+\overrightarrow{b})=2(\overrightarrow{a} \times \overrightarrow{b})\)
which is equal to R.H.S.
Hence proved.

Let
\( \overrightarrow{a}=2 \hat{\imath}+6 \hat{\jmath}+27 \hat{k} \) and \( \overrightarrow{b}=\hat{\imath}+\lambda \hat{\jmath}+\mu \hat{k} \)
Given that \( \overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{0}=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k} \)
\(\overrightarrow{a} \times \overrightarrow{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{\jmath} & \hat{k} \\
2 & 6 & 27 \\
3 & \lambda & \mu
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\hat{\imath}[(6 \times \mu)-(27 \times \lambda)]-\hat{\jmath}[(2 \times \mu)-(27 \times 1)]\)\(+\hat{\mathrm{k}}[(2 \times \lambda)-(1 \times 6)]\)
\(\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(6 \mu-27 \lambda) \hat{\imath}-(2 \mu-27) \hat{\jmath}+(2 \lambda-6) \hat{\mathrm{k}}\)
\(\Rightarrow 0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}=(6 \mu-27 \lambda) \hat{\imath}-(2 \mu-27) \hat{\jmath}+(2 \lambda-6) \hat{\mathrm{k}}\)
By comparing the y-coefficients,
\(\Rightarrow(2 \mu-27)=0\)
\(\Rightarrow \mu=\frac{27}{2}\)
By comparing the z-coefficients,
\(\Rightarrow(2 \lambda-6)=0\)
\(\Rightarrow \lambda=\frac{6}{2}=3\)
These values should also satisfy the equation we will get from comparing the x -coefficients,
\(\Rightarrow 6 \mu-27 \lambda=0\)
\(\Rightarrow\left(6 \times \frac{27}{2}\right)-27 \times 3=0\)
\(\Rightarrow 0=0\)
Therefore, \( \mu=\frac{27}{2} \) and \( \lambda=3 \).

Given that
\( \overrightarrow{a} \cdot \overrightarrow{b}=0 \)
\( \Rightarrow|\overrightarrow{a}| \cdot|\overrightarrow{b}| \cos \theta=0 \) (where \( \theta \) is the angle between the vectors)
\( \Rightarrow|\overrightarrow{a}|=0 \) or \( |\overrightarrow{b}|=0 \) or \( \cos \theta=0 \)
Also given that,
\(\overrightarrow{a} \times \overrightarrow{b}=0\)
\( \Rightarrow|\overrightarrow{a}| \cdot|\overrightarrow{b}| \sin \theta=0 \) (where \( \theta \) is the angle between the vectors)
\( \Rightarrow|\overrightarrow{a}|=0 \) or \( |\overrightarrow{b}|=0 \) or \( \sin \theta=0 \)
As there is no value of \( \theta \) for which both \( \sin \theta \) and \( \cos \theta \) are zero.
Therefore, the condition for which \( \overrightarrow{a} \cdot \overrightarrow{b}=0 \) and \( \overrightarrow{a} \times \overrightarrow{b}=0 \) is:
\(|\overrightarrow{a}|=0 \text { or }|\overrightarrow{b}|=0 \text {. }\)

Given that
\(\overrightarrow{\mathrm{a}}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{\mathrm{k}},\)
\(\overrightarrow{\mathrm{b}}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{\mathrm{k}} \text { and }\)
\(\overrightarrow{\mathrm{c}}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{\mathrm{k}}\)
Solving for left hand side,
First, we calculate
\( \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\left(b_{1}+c_{1}\right) \hat{\imath}+\left(b_{2}+c_{2}\right) \hat{\jmath}+\left(b_{3}+c_{3}\right) \hat{\mathrm{k}} \)
\(\begin{array}{l}
\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\
a_{1} & a_{2} & a_{3} \\
b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3}
\end{array}\right| \\
\Rightarrow \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|+\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\
a_{1} & a_{2} & a_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right| \\
\Rightarrow \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})
\end{array}\)
Which is equal to R.H.S.
Hence proved.

\( |\overrightarrow{a} \times \overrightarrow{b}|=|\overrightarrow{a}| \cdot|\overrightarrow{b}| \cdot \sin \theta \) (where \( \theta \) is the angle between the vectors)
If \( \overrightarrow{a}=\overrightarrow{0} \)
\(\Rightarrow \overrightarrow{a} \times \overrightarrow{b}=(0) \cdot|\overrightarrow{b}| \cdot \sin \theta=0\)
Similarly, If \( \overrightarrow{b}=\overrightarrow{0} \),
\(\Rightarrow \overrightarrow{a} \times \overrightarrow{b}=|\overrightarrow{a}| \cdot(0) \cdot \sin \theta=0\)
Now, If \( \overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{0} \)
\( |\overrightarrow{a}| \cdot|\overrightarrow{b}| \cdot \sin \theta=0 \) (where \( \theta \) is the angle between the vectors)
\(\Rightarrow|\overrightarrow{a}|=0 \text { or }|\overrightarrow{b}|=0 \text { or } \sin \theta=0\)
\( \sin \theta=0 \), this implies \( \theta=0^{\circ} \)
This implies that the converse is not always true. The vectors may not be zero but the angle between is \( 0^{\circ} \), i.e. the vectors are parallel.

\(\text { So, } \overrightarrow{A}=\hat{\imath}+\hat{\jmath}+2 \hat{k}\)
\(\overrightarrow{B}=2 \hat{\imath}+3 \hat{\jmath}+5 \hat{k}\)
\(\overrightarrow{C}=\hat{\imath}+5 \hat{\jmath}+5 \hat{k}\)
Now we want to find the \( \overrightarrow{\mathrm{AC}} \)
\(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{C}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AC}}=(\hat{\imath}+5 \hat{\jmath}+5 \hat{k})-(\hat{\imath}+\hat{\jmath}+2 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{AC}}=(0 \hat{\imath}+4 \hat{\jmath}+3 \hat{k})\)
We know that Area \( (\Delta \mathrm{ABC})=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \)
So, we find \( |\overrightarrow{A B} \times \overrightarrow{A C}| \) first.
\(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\
1 & 2 & 3 \\
3 & 4 & 3
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}[(3 \times 2)-(4 \times 3)]-\hat{\jmath}[(1 \times 3)-(0 \times 3)]\)
\(+\hat{\mathrm{k}}[(4 \times 1)-(2 \times 0)]\)
\(\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=(-6) \hat{\imath}-(3) \hat{\jmath}+(4) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=-6 \hat{\imath}-3 \hat{\jmath}+4 \hat{\mathrm{k}}\)
\(\therefore \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}=\sqrt{61}\)
\(\therefore \text { Area }(\Delta \mathrm{ABC})=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{\sqrt{61}}{2} \text { sq. units }\)

Given \( \overrightarrow{a}=\hat{\imath}-\hat{\jmath}+\hat{k} \), and
\(\overrightarrow{b}=2 \hat{\imath}-7 \hat{\jmath}+\hat{k}\)
We know that Area (parallelogram) \( =|\overrightarrow{a} \times \overrightarrow{b}| \)
So, we find \( |\overrightarrow{a} \times \overrightarrow{b}| \)
\(\overrightarrow{a} \times \overrightarrow{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{\jmath} & \hat{k} \\
1 & -1 & 3 \\
2 & -7 & 1
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}[(-1 \times 1)-(-7 \times 3)]-\hat{\jmath}[(1 \times 1)-(2 \times 3)] +\hat{k}[(-7 \times 1)-(-1 \times 2)] \)
\(\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(20) \hat{\imath}-(-5) \hat{\jmath}+(-5) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=20 \hat{\mathrm{i}}-5 \hat{\jmath}+5 \hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{20^{2}+5^{2}+(-5)^{2}}=\sqrt{450}=15 \sqrt{2}\)
\(\therefore \text { Area (parallelogram) }=\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=15 \sqrt{2} \text { sq. units }\)

\(|\overrightarrow{a} \times \overrightarrow{b}|=|\overrightarrow{a}| \cdot|\overrightarrow{b}| \sin \theta\)
Given that:
\(|\overrightarrow{a}|=3\)
\( |\overrightarrow{b}|=\frac{\sqrt{2}}{3} \) and
\( |\overrightarrow{a} \times \overrightarrow{b}|=1 \) (Magnitude of unit vector is 1 )
Putting the values in the equation (1), we get
\( |\overrightarrow{a} \times \overrightarrow{b}|=|\overrightarrow{a}| \cdot|\overrightarrow{b}| \sin \theta \) (where \( \theta \) is the angle between the vectors)
\(\Rightarrow 1=3 \times \frac{\sqrt{2}}{3} \times \sin \theta\)
\(\Rightarrow \frac{1}{\sqrt{2}}=\sin \theta\)
\(\Rightarrow \theta=\sin ^{-1} \frac{1}{\sqrt{2}}=\frac{\pi}{4}\)
Therefore, the correct option is (B).

Let
\(\overrightarrow{a}=-\hat{\imath}+\frac{1}{2} \hat{\jmath}+4 \hat{k},\)
\(\overrightarrow{b}=\hat{\imath}+\frac{1}{2} \hat{\jmath}+4 \hat{k}\)
\(\overrightarrow{c}=\hat{\imath}-\frac{1}{2} \hat{\jmath}+4 \hat{k}\)
\(\overrightarrow{d}=-\hat{\imath}-\frac{1}{2} \hat{\jmath}+4 \hat{k}\)
Now we want to find the \( \overrightarrow{\mathrm{AB}} \)
\(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=\left(\hat{\imath}+\frac{1}{2} \hat{\jmath}+4 \hat{\mathrm{k}}\right)-\left(-\hat{\imath}+\frac{1}{2} \hat{\jmath}+4 \hat{\mathrm{k}}\right)\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}})\)
Now we want to find the \( \overrightarrow{\mathrm{AC}} \)
\(\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{D}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AD}}=\left(-\hat{\imath}-\frac{1}{2} \hat{\jmath}+4 \hat{\mathrm{k}}\right)-\left(-\hat{\imath}+\frac{1}{2} \hat{\jmath}+4 \hat{\mathrm{k}}\right)\)
\(\Rightarrow \overrightarrow{\mathrm{AD}}=(0 \hat{\imath}-1 \hat{\jmath}+0 \hat{\mathrm{k}})\)
We know that Area (rectangle) \( =|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}| \)
So, we find \( |\overrightarrow{A B} \times \overrightarrow{A D}| \),
\(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=\left|\begin{array}{rrr}
\hat{i} & \hat{\jmath} & \hat{\mathrm{k}} \\
2 & 0 & 0 \\
0 & -1 & 0
\end{array}\right|\)
Expanding along first row,
\(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=\hat{\imath}[(0 \times 0)-(-1 \times 0)]-\hat{\jmath}[(2 \times 0)-(0 \times 0)]+\hat{\mathrm{k}}[(-1 \times 2)-(0 \times 0)]\)
\(\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=(0) \hat{\imath}-(0) \hat{\jmath}+(-2) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=0 \hat{\mathrm{\imath}}+0 \hat{\jmath}-2 \hat{\mathrm{k}}\)
\(\therefore \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{0^{2}+0^{2}+(-2)^{2}}=2\)
\(\therefore \text { Area (rectangle) }=|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}|=2 \text { sq. units }\)