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Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 1

Last Updated : 28 Mar, 2021
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Question 1. If \vec{a}= \hat{i}+3\hat{j}-2\hat{k}   and \vec{b}= -\hat{i}+3\hat{k} , find |\vec{a} \times \vec{b}|

Solution:

Given, \vec{a}= \hat{i}+3\hat{j}-2\hat{k}    and \vec{b}= -\hat{i}+3\hat{k}    .

=> \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}

=> \vec{a} \times \vec{b}   \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 3 & -2\\ -1 & 0 & 3 \end{vmatrix}

=> \vec{a} \times \vec{b}   \hat{i}[(3)(3)-(0)(-2)]-\hat{j}[(1)(3)-(-1)(-2)]+\hat{k}[(1)(0)-(-1)(3)]

=> \vec{a} \times \vec{b}   \hat{i}[9-0]-\hat{j}[3-2]+\hat{k}[0-(-3)]

=> \vec{a} \times \vec{b}   9\hat{i}-\hat{j}+3\hat{k}

Now, |\vec{a} \times \vec{b}|    

=> |\vec{a} \times \vec{b}|   \sqrt{(9)^2+(-1)^2+(3)^2}

=> |\vec{a} \times \vec{b}|   \sqrt{81+1+9}

=> |\vec{a} \times \vec{b}|   = √91

Question 2(i). If \vec{a}= 3\hat{i}+4\hat{j}    and \vec{b}= \hat{i}+\hat{j}+\hat{k}   , find the value of |\vec{a} \times \vec{b}|    

Solution:

Given, \vec{a}= 3\hat{i}+4\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}   

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 4 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(4)(1)-(1)(0)]-\hat{j}[(3)(1)-(1)(0)]+\hat{k}[(3)(1)-(1)(4)]

=> \vec{a} \times \vec{b}  \hat{i}[4-0]-\hat{j}[3-0]+\hat{k}[3-4]

=> \vec{a} \times \vec{b}  4\hat{i}-3\hat{j}-\hat{k}

Now, |\vec{a} \times \vec{b}|   

=> |\vec{a} \times \vec{b}|  \sqrt{(4)^2+(-3)^2+(-1)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{16+1+9}

=> |\vec{a} \times \vec{b}|  \sqrt{26}

Question 2(ii). If \vec{a}= 2\hat{i}+\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}  , find the magnitude of |\vec{a} \times \vec{b}|   

Solution:

Given, \vec{a}= 2\hat{i}+\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}

=> \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}

=> \vec{a} \times \vec{b}   \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(1)-(1)(0)]-\hat{j}[(2)(1)-(1)(0)]+\hat{k}[(2)(1)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[1-0]-\hat{j}[2-0]+\hat{k}[2-1]

=> \vec{a} \times \vec{b}  \hat{i}-2\hat{j}+\hat{k}

Now, |\vec{a} \times \vec{b}|   

=> |\vec{a} \times \vec{b}|  \sqrt{(1)^2+(-2)^2+(1)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{1+4+1}

=> |\vec{a} \times \vec{b}|  = √6

Question 3(i). Find a unit vector perpendicular to both the vectors 4\hat{i}-\hat{j}+3\hat{k}   and -2\hat{i}+\hat{j}-2\hat{k}   

Solution:

Given 4\hat{i}-\hat{j}+3\hat{k}    and -2\hat{i}+\hat{j}-2\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}   

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & 3\\ -2 & 1 & -2 \end{vmatrix}

=> \vec{a} \times \vec{b}    \hat{i}[(-1)(-2)-(1)(3)]-\hat{j}[(4)(-2)-(-2)(3)]+\hat{k}[(4)(1)-(-2)(-1)]

=> \vec{a} \times \vec{b}    \hat{i}[2-3]-\hat{j}[-8+6]+\hat{k}[4-2]

=> \vec{a} \times \vec{b}  -\hat{i}+2\hat{j}+2\hat{k}

Unit vector is given by 

=> \hat{p}   = \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> |\vec{a} \times \vec{b}|  \sqrt{(-1)^2+(2)^2+(2)^2}

=> |\vec{a} \times \vec{b}|  = 3

=> Unit vector is,

=> \hat{p}  \dfrac{1}{3}(-\hat{i}+2\hat{j}+2\hat{k})

Question 3(ii). Find a unit vector perpendicular to the plane containing the vectors \vec{a}=2\hat{i}+\hat{j}+\hat{k}   and \vec{b}=\hat{i}+2\hat{j}+\hat{k}   .

Solution:

 Given, \vec{a}=2\hat{i}+\hat{j}+\hat{k}   and \vec{b}=\hat{i}+2\hat{j}+\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 1\\ 1 & 2 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(1)-(2)(1)]-\hat{j}[(2)(1)-(1)(1)]+\hat{k}[(2)(2)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[1-2]-\hat{j}[2-1]+\hat{k}[4-1]

=> \vec{a} \times \vec{b}  -\hat{i}-\hat{j}+3\hat{k}

Unit vector is given by 

=> \hat{p}  \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> |\vec{a} \times \vec{b}|  \sqrt{(-1)^2+(-1)^2+(3)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{11}

=> Unit vector is,

=> \hat{p}  \dfrac{1}{\sqrt{11}}(-\hat{i}-\hat{j}+3\hat{k})

Question 4. Find the magnitude of vector \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})   

Solution:

Given \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})   

=> \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})      

=> \vec{a}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 4 & 3\\ 1 & 1 & -1 \end{vmatrix}

=> \vec{a}  \hat{i}[(4)(-1)-(1)(3)]-\hat{j}[(0)(-1)-(1)(3)]+\hat{k}[(0)(1)-(1)(4)]

=> \vec{a}  \hat{i}[-4-3]-\hat{j}[0-3]+\hat{k}[0-4]

=> \vec{a}  -7\hat{i}+3\hat{j}-4\hat{k}

Unit vector is,

=> |\vec{a}|  \sqrt{(-7)^2+(3)^2+(-4)^2}

=> |\vec{a}|  \sqrt{49+9+16}

=> |\vec{a}|  = √74

Question 5. If \vec{a}=4\hat{i}+3\hat{j}+\hat{k}    and \vec{b}=\hat{i}-2\hat{k}   , then find |2\hat{b}\times\vec{a}|   

Solution:

Given, \vec{a}=4\hat{i}+3\hat{j}+\hat{k}    and \vec{b}=\hat{i}-2\hat{k}   

=> \hat{b}  \dfrac{\vec{b}}{|\vec{b}|}

=> \hat{b}  \dfrac{(\hat{i}-2\hat{k})}{\sqrt{1^2+(-2)^2}}

=> \hat{b}  \dfrac{(\hat{i}-2\hat{k})}{\sqrt{5}}

=> 2\hat{b}  \dfrac{2}{\sqrt{5}}{(\hat{i}-2\hat{k})}

=> 2\hat{b}\times\vec{a}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \dfrac{2}{\sqrt{5}} & 0 &\dfrac{-4}{\sqrt{5}}\\ 4 & 3 & 1 \end{vmatrix}

=> 2\hat{b}\times\vec{a}   \hat{i}[(0)(1)-(3)(-\dfrac{4}{\sqrt{5}})]-\hat{j}[(\dfrac{2}{\sqrt{5}})(1)-(4)(\dfrac{-4}{\sqrt{5}})]+\hat{k}[(\dfrac{2}{\sqrt{5}})(3)-(4)(0)]

=> 2\hat{b}\times\vec{a}  \hat{i}[0+\dfrac{12}{\sqrt{5}}]-\hat{j}[\dfrac{2}{\sqrt{5}}+\dfrac{16}{\sqrt{5}}]+\hat{k}[\dfrac{6}{\sqrt{5}}-0]

=> 2\hat{b}\times\vec{a}  \dfrac{12}{\sqrt{5}}\hat{i}-\dfrac{18}{\sqrt{5}}\hat{j}+\dfrac{6}{\sqrt{5}}\hat{k}

Now, |2\hat{b}\times\vec{a}|   

=> |2\hat{b}\times\vec{a}|  \sqrt{(\dfrac{12}{\sqrt{5}})^2+(\dfrac{-18}{\sqrt{5}})^2+(\dfrac{6}{\sqrt{5}})^2}

=> |2\hat{b}\times\vec{a}|  \sqrt{\dfrac{144}{5} + \dfrac{324}{5}+\dfrac{36}{5}}

=> |2\hat{b}\times\vec{a}|  \sqrt{\dfrac{504}{5}}

Question 6. If \vec{a}=3\hat{i}-\hat{j}-2\hat{k}   and \vec{b}=2\hat{i}+3\hat{j}+\hat{k}  , find (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})   

Solution:

Given, \vec{a}=3\hat{i}-\hat{j}-2\hat{k}    and \vec{b}=2\hat{i}+3\hat{j}+\hat{k}   

=> 2\vec{a}  2(3\hat{i}-\hat{j}-2\hat{k})

=> 2\vec{a}  6\hat{i}-2\hat{j}-4\hat{k}

=> 2\vec{b}  2(2\hat{i}+3\hat{j}+\hat{k})

=> 2\vec{b}  4\hat{i}+6\hat{j}+2\hat{k}

=> \vec{a}+2\vec{b}  (3\hat{i}-\hat{j}-2\hat{k})+(4\hat{i}+6\hat{j}+2\hat{k})

=> \vec{a}+2\vec{b}  (3+4)\hat{i}+(-1+6)\hat{j}+(-2+2)\hat{k}

=> \vec{a}+2\vec{b}  7\hat{i}+5\hat{j}

=> 2\vec{a}-\vec{b}  (6\hat{i}-2\hat{j}-4\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})

=> 2\vec{a}-\vec{b}  (6-2)\hat{i}+(-2-3)\hat{j}+(-4-1)\hat{k}

=> 2\vec{a}-\vec{b}  4\hat{i}-5\hat{j}-5\hat{k}

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 7 & 5 & 0\\ 4 & -5 & -5 \end{vmatrix}

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \hat{i}[(5)(-5)-(-5)(0)]-\hat{j}[(7)(-5)-(4)(0)]+\hat{k}[(7)(-5)-(4)(5)]

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \hat{i}[-25-0]-\hat{j}[-35-0]+\hat{k}[-35-20]

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  -25\hat{i}+35\hat{j}-55\hat{k}

Question 7(i). Find a vector of magnitude 49, which is perpendicular to both the vectors 2\hat{i}+3\hat{j}+6\hat{k}    and 3\hat{i}-6\hat{j}+2\hat{k}   

Solution:

Given, \vec{a} =2\hat{i}+3\hat{j}+6\hat{k}    and \vec{b}=3\hat{i}-6\hat{j}+2\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}

=> \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}

=>  \vec{a} \times \vec{b}   = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 3 & 6\\ 3 & -6 & 2 \end{vmatrix}

=>  \vec{a} \times \vec{b}   = \hat{i}[(3)(2)-(-6)(6)]-\hat{j}[(2)(2)-(3)(6)]+\hat{k}[(2)(-6)-(3)(3)]

=>  \vec{a} \times \vec{b}   = \hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9]

=>  \vec{a} \times \vec{b}  42\hat{i}+14\hat{j}-21\hat{k}

Magnitude of vector is given by,

=> |\vec{a} \times \vec{b}|  \sqrt{(42)^2+(14)^2+(-21)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{1764+196+441}

=> |\vec{a} \times \vec{b}|  \sqrt{2401}

=> |\vec{a} \times \vec{b}|  49

=> Vector is, 42\hat{i}+14\hat{j}-21\hat{k}

Question 7(ii). Find the vector whose length is 3 and which is perpendicular to the vector 3\hat{i}+\hat{j}-4\hat{k}   and 6\hat{i}+5\hat{j}-2\hat{k}   

Solution:

Given, 3\hat{i}+\hat{j}-4\hat{k}    and 6\hat{i}+5\hat{j}-2\hat{k}   

A vector perpendicular to 2 vectors is given by   \vec{a} \times \vec{b}

=>  \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}

=>  \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 1 & -4\\ 6 & 5 & -2 \end{vmatrix}

=>  \vec{a} \times \vec{b}  \hat{i}[(1)(-2)-(5)(-4)]-\hat{j}[(3)(-2)-(6)(-4)]+\hat{k}[(3)(5)-(6)(1)]

=>  \vec{a} \times \vec{b}  \hat{i}[-2+20]-\hat{j}[-6+24]+\hat{k}[15-6]

=>  \vec{a} \times \vec{b}  18\hat{i}-18\hat{j}+9\hat{k}

Magnitude of vector is given by,

=> |\vec{a} \times \vec{b}|  \sqrt{(18)^2+(-18)^2+(9)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{324+324+81}

=> |\vec{a} \times \vec{b}|  \sqrt{729}

=> |\vec{a} \times \vec{b}|  = 27

=> Unit vector is,

=> \hat{p}  \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> \hat{p}  \dfrac{1}{27}(18\hat{i}-18\hat{j}+9\hat{k})

Required vector is, 

=> 3\times\dfrac{1}{27}(18\hat{i}-18\hat{j}+9\hat{k})=2\hat{i}-2\hat{j}+\hat{k}      

Question 8(i). Find the parallelogram determined by the vectors: 2\hat{i}    and 3\hat{j}   

Solution:

Given that, \vec{a}=2\hat{i}  and \vec{b} =3\hat{j}   

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 0 & 0\\ 0 & 3 & 0 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(0)(0)-(3)(0)]-\hat{j}[(2)(0)-(0)(0)]+\hat{k}[(2)(3)-(0)(0)]

=> \vec{a} \times \vec{b}  \hat{i}[0-0]-\hat{j}[0-0]+\hat{k}[6-0]

=> \vec{a} \times \vec{b}  6\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(0)^2+(0)^2+(6)^2}

=> |\vec{a} \times \vec{b}|  6

=> Area = 6 square units.

Question 8(ii). Find the parallelogram determined by the vectors: 2\hat{i}+\hat{j}+3\hat{k}    and \hat{i}-\hat{j}  .

Solution:

Given that, \vec{a}=2\hat{i}+\hat{j}+3\hat{k}   and \vec{b}=\hat{i}-\hat{j}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(0)-(-1)(3)]-\hat{j}[(2)(0)-(1)(3)]+\hat{k}[(2)(-1)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[0+3]-\hat{j}[0-3]+\hat{k}[-2-1]

=> \vec{a} \times \vec{b}  3\hat{i}+3\hat{j}-3\hat{k}

Thus, the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(3)^2+(3)^2+(-3)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{9+9+9}

=> Area = 3\sqrt{3}

Question 8(iii). Find the area of the parallelogram determined by the vectors: 3\hat{i}+\hat{j}-2\hat{k}    and \hat{i}-3\hat{j}+4\hat{k}   

Solution:

Given that, \vec{a} = 3\hat{i}+\hat{j}-2\hat{k}   and \vec{b} = \hat{i}-3\hat{j}+4\hat{k}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 1 & -2\\ 1 & -3 & 4 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(4)-(-3)(-2)]-\hat{j}[(3)(4)-(1)(-2)]+\hat{k}[(3)(-3)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[4-6]-\hat{j}[12+2]+\hat{k}[-9-1]

=> \vec{a} \times \vec{b}  -2\hat{i}-14\hat{j}-10\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|    \sqrt{(-2)^2+(-14)^2+(-10)^2}

=> |\vec{a} \times \vec{b}|    \sqrt{4+196+100}

=> Area = 10\sqrt{3}

Question 8(iv). Find the area of the parallelogram determined by the vectors: \hat{i}-3\hat{j}+\hat{k}   and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given that, \vec{a} = \hat{i}-3\hat{j}+\hat{k}   and \vec{b} = \hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -3 & 1\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(-3)(1)-(1)(1)]-\hat{j}[(1)(1)-(1)(1)]+\hat{k}[(1)(1)-(1)(-3)]

=> \vec{a} \times \vec{b}  \hat{i}[-3-1]-\hat{j}[1-1]+\hat{k}[1+3]

=> \vec{a} \times \vec{b}  -4\hat{i}+4\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(-4)^2+(0)^2+(4)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{16+16}

=> Area = 4\sqrt{2}

Question 9(i). Find the area of the parallelogram whose diagonals are: 4\hat{i}-\hat{j}-3\hat{k}   and -2\hat{i}+\hat{j}-2\hat{k}   

Solution:

Given, \vec{a}=4\hat{i}-\hat{j}-3\hat{k}   and \vec{b}=-2\hat{i}+\hat{j}-2\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & -3\\ -2 & 1 & -2 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(-1)(-2)-(1)(-3)]-\hat{j}[(4)(-2)-(-2)(-3)]+\hat{k}[(4)(1)-(-2)(-1)]

=> \vec{a} \times \vec{b}  \hat{i}[2+3]-\hat{j}[-8-6]+\hat{k}[4-2]

=> \vec{a} \times \vec{b}  5\hat{i}+14\hat{j}+2\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(5)^2+(14)^2+(2)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{225}

=> Area = 15/2 = 7.5 square units

Question 9(ii). Find the area of the parallelogram whose diagonals are: 2\hat{i}+\hat{k}    and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given, \vec{a}=2\hat{i}+\hat{k}   and \vec{b}=\hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 0 & 1\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(0)(1)-(1)(1)]-\hat{j}[(2)(1)-(1)(1)]+\hat{k}[(2)(1)-(1)(0)]

=> \vec{a} \times \vec{b}  \hat{i}[0-1]-\hat{j}[2-1]+\hat{k}[2-0]

=> \vec{a} \times \vec{b}  -\hat{i}-\hat{j}+2\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(-1)^2+(-1)^2+(2)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{6}

=> Area = \dfrac{\sqrt{6}}{2}

Question 9(iii). Find the area of the parallelogram whose diagonals are: 3\hat{i}+4\hat{j}   and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given, \vec{a}=3\hat{i}+4\hat{j}   and \vec{b}=\hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 4 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(4)(1)-(1)(0)]-\hat{j}[(3)(1)-(1)(0)]+\hat{k}[(3)(1)-(1)(4)]

=> \vec{a} \times \vec{b}  \hat{i}[4-0]-\hat{j}[3-0]+\hat{k}[3-4]

=> \vec{a} \times \vec{b}  4\hat{i}-3\hat{j}-\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(4)^2+(-3)^2+(-1)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{26}

=> Area = \dfrac{\sqrt{26}}{2}

Question 9(iv). Find the area of the parallelogram whose diagonals are: 2\hat{i}+3\hat{j}+6\hat{k}    and 3\hat{i}-6\hat{j}+2\hat{k}   

Solution:

Given, \vec{a}=2\hat{i}+3\hat{j}+6\hat{k}  and \vec{b}=3\hat{i}-6\hat{j}+2\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \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}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 3 & 6\\ 3 & -6 & 2 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(3)(2)-(-6)(6)]-\hat{j}[(2)(2)-(3)(6)]+\hat{k}[(2)(-6)-(3)(3)]

=> \vec{a} \times \vec{b}  \hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9]

=> \vec{a} \times \vec{b}  42\hat{i}+14\hat{j}-21\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(42)^2+(14)^2+(-21)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{2401}

=> Area = \dfrac{49}{2}

=> Area = 24.5

Question 10. If  \vec{a}=2\hat{i}+5\hat{j}-7\hat{k}  \vec{b}=-3\hat{i}+4\hat{j}+\hat{k}    and \vec{c}=\hat{i}-2\hat{j}-3\hat{k}  , compute (\vec{a}\times\vec{b})\times\vec{c}    and \vec{a}\times(\vec{b}\times\vec{c})    and verify these are not equal.

Solution:

Given \vec{a}=2\hat{i}+5\hat{j}-7\hat{k}  \vec{b}=-3\hat{i}+4\hat{j}+\hat{k}   and \vec{c}=\hat{i}-2\hat{j}-3\hat{k}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 5 & -7\\ -3 & -4 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(5)(1)-(4)(-7)]-\hat{j}[(2)(1)-(-3)(-7)]+\hat{k}[(2)(4)-(-3)(5)]

=> \vec{a} \times \vec{b}  \hat{i}[5+28]-\hat{j}[2-21]+\hat{k}[8+15]

=> \vec{a} \times \vec{b}  33\hat{i}+19\hat{j}+23\hat{k}

=> (\vec{a}\times\vec{b})\times\vec{c}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 33 & 19 & 23\\ 1 & -2 & -3 \end{vmatrix}

=> (\vec{a}\times\vec{b})\times\vec{c}   = \hat{i}[(19)(-3)-(-2)(23)]-\hat{j}[(33)(-3)-(1)(23)]+\hat{k}[(33)(-2)-(1)(19)]

=> (\vec{a}\times\vec{b})\times\vec{c}   = \hat{i}[-57+46]-\hat{j}[-99-23]+\hat{k}[-66-19]

=> (\vec{a}\times\vec{b})\times\vec{c}  -11\hat{i}+122\hat{j}-85\hat{k}

=> \vec{b} \times \vec{c}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -3 & 4 & 1\\ 1 & -2 & -3 \end{vmatrix}

=> \vec{b} \times \vec{c}  \hat{i}[(4)(-3)-(-2)(1)]-\hat{j}[(-3)(-3)-(1)(1)]+\hat{k}[(-3)(-2)-(1)(4)]

=> \vec{b} \times \vec{c}  \hat{i}[-12+2]-\hat{j}[9-1]+\hat{k}[6-4]

=> \vec{b} \times \vec{c}  -10\hat{i}-8\hat{j}+2\hat{k}

=> \vec{a}\times(\vec{b}\times\vec{c})  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 5 & -7\\ -10 & -8 & 2 \end{vmatrix}

=> \vec{a}\times(\vec{b}\times\vec{c})  \hat{i}[(5)(2)-(-8)(-7)]-\hat{j}[(2)(2)-(-10)(-7)]+\hat{k}[(2)(-8)-(-10)(5)]

=> \vec{a}\times(\vec{b}\times\vec{c})  \hat{i}[10-56]-\hat{j}[4-70]+\hat{k}[-16+50]

=> \vec{a}\times(\vec{b}\times\vec{c})  -46\hat{i}+66\hat{j}+34\hat{k}

=> (\vec{a}\times\vec{b})\times\vec{c}   is not equal to \vec{a}\times(\vec{b}\times\vec{c})

=> Hence verified.

Question 11. If |\vec{a}|=2   |\vec{b}|=5    and |\vec{a}\times\vec{b}|=8   , find \vec{a}.\vec{b}    

Solution:

We know that,

=> \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}

=> |\vec{a}\times\vec{b}|  =  |\vec{a}||\vec{b}||\sin\theta||\hat{n}|

We know that |\hat{n}|  is 1, as \hat{n}  is a unit vector

=> 8 = 2\times5\times\sin\theta\times1

=> 10\sin\theta  = 8

=> \sin\theta  = \dfrac{4}{5}

Also,

=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta

And \sin^2\theta + \cos^2\theta = 1

=> \cos\theta  = \sqrt{1-\sin^2\theta}

=> \cos\theta  = \sqrt{1-(\dfrac{4}{5})^2}

=> \cos\theta  = \sqrt{\dfrac{9}{16}}

=> \cos\theta  = \dfrac{3}{5}

=> \vec{a}.\vec{b}  = 2\times5\times\dfrac{3}{5}

=> \vec{a}.\vec{b}  = 6

Question 12. Given \vec{a} = \dfrac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})   \vec{b} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})   \vec{c} = \dfrac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})   \hat{i}   \hat{j}   , \hat{k}    being a right-handed orthogonal system of unit vectors in space, show that \vec{a}   \vec{b}    and \vec{c}    is also another system.

Solution:

To show that \vec{a} \vec{b}  and \vec{c}  is a right-handed orthogonal system of unit vectors, we need to prove:

(1) |\vec{a}|=|\vec{b}|=|\vec{c}|=1

=> |\vec{a}| = \dfrac{1}{7}\sqrt{2^2+3^2+6^2}

=> |\vec{a}| = \dfrac{1}{7}\sqrt{4+9+36}

=> |\vec{a}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{a}| = \dfrac{1}{7}\times7

=> |\vec{a}| = 1

=> |\vec{b}| = \dfrac{1}{7}\sqrt{3^2+(-6)^2+2^2}

=> |\vec{b}| = \dfrac{1}{7}\sqrt{9+36+4}

=> |\vec{b}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{b}| = \dfrac{1}{7}\times7

=> |\vec{b}| = 1

=> |\vec{c}| = \dfrac{1}{7}\sqrt{6^2+2^2+(-3)^2}

=> |\vec{c}| = \dfrac{1}{7}\sqrt{36+4+9}

=> |\vec{c}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{c}| = \dfrac{1}{7}\times7

=> |\vec{c}| = 1

(2) \vec{a}\times\vec{b} = \vec{c}

=> \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}

=>\vec{a}\times\vec{b}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & 3 & 6\\3 & -6& 2\end{vmatrix}

=>\vec{a}\times\vec{b} = \dfrac{1}{49}(\hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9])

=>\vec{a}\times\vec{b} = \dfrac{1}{49}(42\hat{i}+14\hat{j}-21\hat{k})

=> \vec{a}\times\vec{b} = \dfrac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})=\vec{c}

(3) \vec{b}\times\vec{c} =\vec{a}

=>\vec{b}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\b_1 & b_2 & b_3\\c_1 & c_ 2& c_3\end{vmatrix}

=>\vec{b}\times\vec{c}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3 & -6 & 2\\6 & 2& -3\end{vmatrix}

=>\vec{b}\times\vec{c} = \dfrac{1}{49}(\hat{i}[18-4]-\hat{j}[-9-12]+\hat{k}[6+36])

=>\vec{b}\times\vec{c} = \dfrac{1}{49}(14\hat{i}+21\hat{j}+42\hat{k})

=>\vec{b}\times\vec{c} = \dfrac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})=\vec{a}

(4) \vec{c}\times\vec{a} = \vec{b}

=>\vec{c}\times\vec{a} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\a_1 & a_ 2& a_3\end{vmatrix}

=> \vec{c}\times\vec{a}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\6 & 2 & -3\\2 & 3& 6\end{vmatrix}

=>\vec{c}\times\vec{a} = \dfrac{1}{49}(\hat{i}[12+9]-\hat{j}[36+6]+\hat{k}[18-4])

=>\vec{c}\times\vec{a} = \dfrac{1}{49}(21\hat{i}-42\hat{j}+14\hat{k})

=>\vec{c}\times\vec{a} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})=\vec{b}

Hence proved.



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