Vector or Cross Product
The vector or cross product is another fundamental operation between two vectors in three-dimensional space. Unlike the scalar or dot product, the result of the vector product is another vector.
Class 12 Vector or Cross Solution RD Sharma Solution
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} and6\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.
Summary
This set introduces the concept of cross product and its basic properties.
Key points include:
- Definition of cross product
- Calculation of cross product using component form
- Magnitude and direction of cross product
- Basic properties like anticommutativity and distributivity
- Geometric applications of cross product
Practice Questions:
1). If a = 2i - j + 3k and b = i + 2j - k, find a × b.
2). Prove that i × j = k, j × k = i, and k × i = j.
3). If a = 3i + 2j - k and b = i - 2j + 3k, find |a × b|.
4). Show that a × (b + c) = a × b + a × c for any vectors a, b, and c.
5). Find a unit vector perpendicular to both a = i + j + k and b = i - j + k.
6). Prove that (a × b) · a = 0 for any vectors a and b.
7). If a = 2i + 3j - k and b = i + 2j + 4k, find the angle between a and b using cross product.
8). Show that |a × b| = |a||b| sin θ, where θ is the angle between a and b.
9). Find the area of the parallelogram formed by the vectors a = 3i - 2j + k and b = i + j + 2k.
10). Prove that a × (b × c) = (a · c)b - (a · b)c for any vectors a, b, and c.
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