Ncert Solutions For Class 12 Maths Ex 10.4

Ncert Solutions For Class 12 Maths Chapter 10 Ex 10.4

Q 1: Find \(\left | \vec{a }\times \vec{b} \right |\) , if \(\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}\) and \(\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}\)

Solution 1:

We have,

\(\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}\) and \(\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}\)

\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 7 & 7 \\ 3 & – 2 & 2 \end{vmatrix}\)

\(\vec{a} \times \vec{b}\) = \(\hat{i} \left ( – 14 + 14 \right ) – \hat{j} \left ( 2 – 21 \right ) + \hat{k} \left ( – 2 + 21 \right ) = 19 \hat{j} + 19 \hat{k}\)

Therefore,

\(\left |\vec{a} \times \vec{b} \right | = \sqrt{\left ( 19 \right ) ^{ 2} + \left ( 19 \right ) ^{2}} = \sqrt{ 2 \times \left ( 19 \right ) ^{ 2 }} = 19 \sqrt{2}\)

 

 

Q 2: Find a unit vector perpendicular to each of the vector \(\vec{a} + \vec{b} and \vec{a} – \vec{b}\), where \(\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}\)

Solution 2:

We have,

\(\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}\)

Therefore,

\(\vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j} , \vec{a} – \vec{b} = 2 \hat{i} + 4 \hat{k}\) \(\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}\) \(\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \hat{i} \left ( 16 \right ) – \hat{j} \left ( 16 \right ) + \hat{k} \left ( – 8 \right ) = 16 \hat{i} – 16 \hat{j} – 8 \hat{k}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{16 ^{2} + \left ( – 16 \right )^{2} + \left ( – 8 \right ) ^{ 2 }}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{ 2 ^{2} \times 8 ^{ 2} + 2 ^{2} \times 8 ^{2 } + 8 ^{2}}\) \(\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = 8 \sqrt{ 2^{2} + 2 ^{2} + 1 } = 8 \sqrt{9} = 8 \times 3 = 24\)

Therefore, the unit vector perpendicular to each of the vectors \(\vec{a} + \vec{b} and \vec{a} – \vec{b}\) is given by the relation,

= \(\pm \frac{ \left ( \vec{a} + \vec{b} \right ) \times \left ( \vec{a} – \vec{b} \right )}{\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | } = \pm \frac{ 16 \hat{i} – 16 \hat{j} – 8 \hat{k}}{ 24 }\)

= \(\pm \frac{ 2 \hat{i} – 2 \hat{j} – \hat{k} }{ 3 } = \pm \frac{ 2 }{ 3 } \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{ 3} \hat{k}\)

Q 3: If a unit vector \(\vec{a}\) makes angle \(\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}\) and an acute angle θ with \(\vec{k}\) , then find θ and hence, the compounds of \(\vec{a}\)

Solution 3:

Let unit vector \(\vec{a}\) have ( a 1 , a 2 , a 3 ) componenets

\(\vec{a} = a _{1} \hat{i} + a _{2} \hat{ j} + a _{ 3 } \hat{ k }\)

Since, \(\vec{ a }\) is a unit vector , \(\left |\vec{a} \right | = 1\)

Also, it is given that \(\vec{a}\) makes angles \(\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}\) and an acute angle θ with \(\vec{k}\)

Then, we have :

\(\cos \frac{ \pi }{ 3} = \frac{ a _{1}}{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\) \(\frac{ 1 }{ 2 } = a _{1} . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]\)

\(\cos \frac{ \pi }{ 4 } = \frac{ a _{2 } }{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\) \(\frac{ 1 }{ \sqrt{ 2 }} = a _{ 2 } . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]\)

Also , \(\cos \theta = \frac{ a _{ 3 }}{ \left |\vec{a} \right | }\)

\(\boldsymbol{\Rightarrow }\) \(a _{ 3 } = \cos \theta\)

\(\left |a \right | = 1\)

\(\boldsymbol{\Rightarrow }\) \(\sqrt{ {a_{ 1 }}^{ 2 } + {a_{ 2 }}^{ 2 } + {a_{ 3 }}^{ 2 } } = 1\)

\(\boldsymbol{\Rightarrow }\) \(\frac{ 1 }{ 2 }^{ 2 } + \frac{ 1 }{ \sqrt{ 2 }}^{ 2 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\) \(\frac{ 1 }{ 4 } + \frac{ 1 }{ 2 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\) \(\frac{ 3 }{ 4 } + \cos ^{ 2 } \theta = 1\)

\(\boldsymbol{\Rightarrow }\) \(\cos ^{ 2 } \theta = 1 – \frac{ 3 }{ 4 } = \frac{ 1 }{ 4 }\)

\(\boldsymbol{\Rightarrow }\) \(\cos \theta = \frac{ 1 }{ 2 } \\ \theta = \cos^{-1} \left ( \frac{ 1 }{ 2 } \right ) \\ \theta = \frac{ \pi }{ 3 }\)

Therefore,

\(a_{ 3 } = \cos \frac{ \pi }{ 3 } = \frac{1}{ 2 }\)

Therefore, \(\theta = \frac{ \pi }{ 3} and the components of \vec{a} are \left ( \frac{ 1 }{ 2} , \frac{ 1 }{ \sqrt{ 2}} , \frac{ 1 }{2} \right )\)

Q 4: Show that:

\(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )\)

Solution:

To prove:

\(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )\)

= \(\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = \left ( \vec{a} -\vec{ b} \right ) \times \vec{a} + \left ( \vec{a} – \vec{b} \right ) \times \vec{b}\) . . . . . . . . . . . . [ By distributivity of vector product over addition ]

= \(\vec{a} \times \vec{a} – \vec{b} \times \vec{a} + \vec{a} \times \vec{b} – \vec{b} \times \vec{b}\) . . . . . . . . . . . . . . . [ again, by distributivity of vector product over addition ]

= \(\vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} – \vec{0}\)

= \(2 \vec{a} \times \vec{b}\)

Q 5: Find \(\lambda and \mu if \left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}\)

Solution:

\(\left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}\)

\(\boldsymbol{\Rightarrow }\) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda \mu & \end{vmatrix}\) = \(0 \hat{i} + 0 \hat{j} + 0 \hat{k}\)

\(\boldsymbol{\Rightarrow }\) \(\hat{i} \left ( 6 \mu – 27 \mu \right ) – \hat{j} \left ( 2 \mu – 27 \right ) + \hat{k} \left ( 2 \lambda – 6 \right ) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}\)

On comparing the corresponding components, we have :

\(6 \mu – 27 \lambda = 0 \\ 2 \mu – 27 = 0 \\ 2 \lambda – 6 = 0\)

Now,

\(2 \lambda – 6 = 0 \Rightarrow \lambda = 3\)

\(\boldsymbol{\Rightarrow }\) \(2 \mu – 27 = 0 \\ \Rightarrow \mu = \frac{ 27 }{ 2 }\)

Therefore, \(\lambda = 3 and \mu = \frac{ 27 }{ 2 }\)

Q 6: Given that:

\(\vec{a}.\vec{b} = 0 and \vec{a} \times \vec{b} = \vec{0}\)

What can you conclude about the vectors \(\vec{a} and \vec{b}\) ?

Solution:

\(\vec{a}.\vec{b} = 0\)

Then ,

  1. i) either \(\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \perp \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0\)
  2. ii) Either \(\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \parallel \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0\)

But, \(\vec{a} and \vec{b}\) cannot be perpendicular and parallel simultaneously.

Therefore, \(\left |\vec{a} \right | = 0 or \left |\vec{b} \right | = 0\)

Q 7: Let the vectors \(\vec{a }, \vec{b} ,\vec{ c} given as a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}\)

Then show that = \(\vec{a} \times \left ( \vec{ b } + \vec{ c } \right ) = \vec{a} \times \vec{b} + \vec{a }\times \vec{ c }\)

Solution:

We have,

\(a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}\) \(\left ( \vec{b} + \vec{c} \right ) = \left ( b_{ 1 } + c _{1} \right ) \hat{i} + \left ( b_{ 2 } + c_{ 2 } \right ) \hat{j} + \left ( b_{ 3 } + c_{ 3 } \right ) \hat{k}\)

Now, \(\vec{a} \times \left ( \vec{b} + \vec{c} \right )\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{ 1 } & a_{ 2} & a_{ 3 } \\ b_{1} + c_{1} & b_{2} + c_{ 2 } & b_{ 3 } + c_{ 3 } \end{vmatrix}\)

\(= \hat{i} \left [ a_{ 2} \left ( b_{ 3 } + c_{3 } \right ) – a_{ 3 } \left ( b_{ 2 } + c_{ 2 } \right ) \right ] – \hat{j} \left [ a_{ 1 } \left ( b_{ 3 } + c_{ 3 } \right ) – a_{ 3 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ] + \hat{k} \left [ a_{ 1 } \left ( b_{ 2 } + c _{ 2 }\right ) – a_{ 2 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ]\)

\(= \hat{i} \left [ a_{ 2 } b_{3} + a_{2} c_{3} – a_{3} b_{2} – a_{3} c_{2} \right ] + \hat{j} \left [ – a_{1} b_{3} – a_{1} c_{3} + a_{3} b_{1} + a_{3} c_{1} \right ] + \hat{k} \left [ a_{1} b_{2} + a_{1} c_{2} – a_{2}b _{1} – a_{2} c_{1} \right ]\) . . . . . . . . . . . . . . . . . ( 1 )

\(\boldsymbol{\Rightarrow }\) \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1 } & a_{2} & a_{3} \\ b_{ 1} & b_{ 2} & b_{3} \end{vmatrix}\)

= \(\hat{i} \left [ a_{ 2 } b_{ 3 } – a_{ 3 } b_{2} \right ] + \hat{j} \left [ a_{3} b_{1} – a_{1} b_{3} \right ] + \hat{k} \left [ a_{1} b_{2} – a_{2} b_{1} \right ]\) . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )

\(\boldsymbol{\Rightarrow }\) \(\vec{a} \times \vec{ c } = \begin{vmatrix} \hat{ I } & \hat{ j } & \hat{ k } \\ a_{1 } & a_{2} & a_{3} \\ c_{ 1} & c_{ 2 } & c_{3 } \end{vmatrix}\)

\(\hat{i} \left [ a_{ 2 } c_{ 3 } – a_{ 3 } c_{2} \right ] + \hat{j} \left [ a_{3} c_{1} – a_{1} c_{3} \right ] + \hat{k} \left [ a_{1} c_{2} – a_{2} c_{1} \right ]\) . . . . . . . . . . . . . . . . . . . . . . . . . . ( 3 )

On adding (2) and (3), we get:

\(\left ( \vec{a} \times \vec{b} \right ) + \left ( \vec{a} \times \vec{ c} \right ) = \hat{i} \left [ a_{ 2 } b_{ 3 } + a_{ 2 } c_{ 3 } + – a _{ 3 } b_{ 2 } – a_{ 3 } c_{ 2 } \right ] + \hat{j} \left [ a_{ 3 } b_{ 1 } + a_{ 3 } c_{ 1 } + – a_{ 1 } b_{ 3 } – a_{ 1 } c_{ 3 } \right ] + \hat{ k } \left [ a_{ 1 } b_{ 2 } + a_{ 1 } c_{ 2 } + – a_{ 2 } b_{ 1 } – a_{ 2 } c_{ 1 } \right ]\) . . . . . . . . . . . . . . . . . . . . . . . . . . ( 4 )

Now, from (1) and (4), we have:

\(\vec{a} \times \left ( \vec{b} + \vec{c} \right ) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\)

Hence, the given result is proved.

 

 

Question 8: Suppose, any of the vector \(\overrightarrow{m} \;or\; \overrightarrow{n} = \overrightarrow{0}\), then \(\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}\).

Is the above given statement true?

Give reason in support to your answer with an example.

Answer 8:

By taking any two non – zero vectors, for the condition \(\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}\).

Suppose, \(\overrightarrow{m} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \\ Then,\)

\(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 4 & 6 \end{vmatrix}\) \(\hat{i} (4 – 4) – \hat{j} (6 – 6) + \hat{k} (4 – 4) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{0} \\ Now,\; we\; find\; that, \\ \left [ \overrightarrow{m} \right ] = \sqrt{(1) ^{2} + (2) ^{2} + (3) ^{2}} = \sqrt{14} \\ So,\; \overrightarrow{m} \neq \overrightarrow{0} \\ \left [ \overrightarrow{n} \right ] = \sqrt{(2) ^{2} + (4) ^{2} + (6) ^{2}} = \sqrt{56} \\ So,\; \overrightarrow{n} \neq \overrightarrow{0}\)

Thus, the above given statement is not true.

Question 9: P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5) are the vertices of a triangle. Obtain the area.

Answer 9:

Given:

Vertices of a triangle P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5)

The contiguous sides \(\overrightarrow{PQ} \;and\; \overrightarrow{QR}\) of the triangle is given as,

\(\overrightarrow{PQ} = (2 – 1) \hat{i} + (3 – 1) \hat{j} + (5 – 2) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{QR} = (1 – 2) \hat{i} + (5 – 3) \hat{j} + (5 – 5) \hat{k} = – \hat{i} + 2 \hat{j} \\ Area\; of\; \Delta \;triangle = \frac{1}{2} \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right |\) \(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ – 1 & 2 & 0 \end{vmatrix} \\ \hat{i} (- 6) – \hat{j} (3) + \hat{k} (2 + 2) = – 6 \hat{i} -3 \hat{j} + 4 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 6) ^{2} + (- 3) ^{2} + 4 ^{2}} = \sqrt{36 + 9 + 16} = \sqrt{61}\)

Thus, \(\frac{\sqrt{61}}{2}\) is the area of triangle ABC.

Question 10: \(\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}\) are the adjacent sides of a vector. Obtain the area of the parallelogram.

Answer 10:

The area of the parallelogram whose contiguous sides are \(\overrightarrow{m} \;and\; \overrightarrow{n} is \left | \overrightarrow{m} \times \overrightarrow{n} \right |\)

Given,

\(\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}\) are the adjacent sides of a vector.

\(\overrightarrow{m} \times \overrightarrow{n} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 1 & 3 \\ 2 & – 7 & 1 \end{vmatrix} \\ \hat{i} (- 1 + 21) – \hat{j} (1 – 6) + (- 7 + 2) \hat{k} = 20 \hat{i} + 5 \hat{j} – 5 \hat{k} \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = \sqrt{20 ^{2} + 5 ^{2} + 5 ^{2}} = \sqrt{400 + 25 + 25} = 15 \sqrt{2}\)

\(15 \sqrt{2}\) is the area of the given parallelogram.

Question 11: Suppose, vectors \(\overrightarrow{m} \;and\; \overrightarrow{n} \; in\; such\; a\; way\; that \left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; \overrightarrow{m} \times \overrightarrow{n}\) is a unit vector, suppose the angle between the two vectors is

\((i)\; \frac{\pi }{6}, \;(ii)\; \frac{\pi }{4},\; (iii)\; \frac{\pi }{3} \;(iv) \frac{\pi }{2}\)

Answer 11:

Given, \(\left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; as\; we\; know\; \overrightarrow{m} \times \overrightarrow{n} = 1 \\ \overrightarrow{m} \times \overrightarrow{n} = \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | \;sin \theta \hat{s}\), where s is a unit vector perpendicular to both \(\overrightarrow{m} \;and\; \overrightarrow{n}\) and \(\Theta\) is the angle between \(\overrightarrow{m} \;and\; \overrightarrow{n}\)

Now, \(\overrightarrow{m} \;and\; \overrightarrow{n} \;is\; a \;unit \; vector \;if\; \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \hat{s} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \right | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sin\; \Theta = 1 \\ sin\; \Theta = \frac{1}{\sqrt{2}} \\ \Theta = \frac{\pi }{4}\)

The correct answer is option (ii)

Question 12: The area of the rectangle with P, Q, R and S as the vertices with positive vectors \(– \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \; respectively\; is \\ (i) \frac{1}{2}, \;\;\;\; (ii) 1 \\ (iii) 2, \;\;\;\; (iv) 4\)

Answer 12:

Given,

The area of the rectangle PQRS with P, Q, R and S as the vertices with positive vectors such as:

\(\overrightarrow{OP} = – \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k},\; \overrightarrow{OQ} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \;\overrightarrow{OR} = \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; \overrightarrow{OS} – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k}\)

The contiguous sides \(\overrightarrow{PQ} \;and\; \overrightarrow{QR}\) of the rectangle is given as,

\(\overrightarrow{PQ} = (1 + 1) \hat{i} + (\frac{1}{2} – \frac{1}{2}) \hat{j} + (4 – 4) \hat{k} = 2 \hat{i} \;and\; \overrightarrow{QR} = (1 – 1) \hat{i} + \left (- \frac{1}{2} – \frac{1}{2} \right ) \hat{j} + (4 – 4) \hat{k} = – \hat{j}\) \(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & – 1 & 0 \end{vmatrix} \\ \hat{k} (- 2) = – 2 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 2) ^{2}} = 2\)

Area of a rectangle is 2 square units.

Option (iii) is the correct answer.