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.

 

Related Links
NCERT Solutions Class 7 Maths NCERT Solutions Class 10 Maths
NCERT Solutions Class 11 Maths NCERT Solutions Class 12 Maths
NCERT Solutions Class 11 Physics NCERT Solutions Class 11 Chemistry
NCERT Solutions Class 11 Biology NCERT Solutions Class 6 Maths
Ncert Books NCERT Solutions for Class 5 Maths
NCERT Solutions for Class 4 Maths NCERT Solutions Class 8 Maths