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Question: If \[\overrightarrow r = x\widehat i + y\widehat j + z\widehat k,\,find\,(\overrightarrow r \times i...

If r→=xi^+yj^+zk^, find (r→×i)⋅(r→×j)+xy\overrightarrow r = x\widehat i + y\widehat j + z\widehat k,\,find\,(\overrightarrow r \times i) \cdot (\overrightarrow r \times j) + xy .

Explanation

Solution

It is clear that you are given this problem based on vector quantity. What do you understand by a vector? Vector quantity is a quantity that has a magnitude and direction. In this sum, unit vectors are also there. A unit vector is a vector of which the magnitude is one. It is also called a direction vector. i^,j^,k^\widehat i,\widehat j,\widehat k are on the direction x-axis and y-axis and z-axis respectively. You should know about the cross product and dot product of vectors to do this sum.

Step wise solution:
Given data: The r→\overrightarrow r is given by,
r→=xi^+yj^+zk^\overrightarrow r = x\widehat i + y\widehat j + z\widehat k
And we need to find the value of (r→×i)⋅(r→×j)+xy(\overrightarrow r \times i) \cdot (\overrightarrow r \times j) + xy .
We will find out the value of (r→×i^)(\overrightarrow r \times \widehat i)
(r→×i^)=(xi^+yj^+zk^)×i^ (r→×i^)=(xi^)×(i^)+(yj^)×(i^)+(zk^)×(i^) (\overrightarrow r \times \widehat i) = (x\widehat i + y\widehat j + z\widehat k) \times \widehat i\\\ (\overrightarrow r \times \widehat i) = (x\widehat i) \times (\widehat i) + (y\widehat j) \times (\widehat i) + (z\widehat k) \times (\widehat i)
We know the cross product of unit vectors are given as
k^×i^=j^ j^×i^=k^ i^×i^=0 \widehat k \times \widehat i = \widehat j\\\ \widehat j \times \widehat i = \widehat k\\\ \widehat i \times \widehat i = 0
Now,
(r→×i^)=x(i^×i^)+y(j^×i^)+z(k^×i^) ⇒(r→×i^)=0+y(−k^)+z(j^) ⇒(r→×i^)=−yk^+z(j^) (\overrightarrow r \times \widehat i) = x(\widehat i \times \widehat i) + y(\widehat j \times \widehat i) + z(\widehat k \times \widehat i)\\\ \Rightarrow (\overrightarrow r \times \widehat i) = 0 + y( - \widehat k) + z(\widehat j)\\\ \Rightarrow (\overrightarrow r \times \widehat i) = - y\widehat k + z(\widehat j)
To calculate the value of (r→×j^)(\overrightarrow r \times \widehat j)
(r→×j^)=(xi^+yj^+zk^)×j^ ⇒(r→×j^)=x(j^×i^)+y(j^×j^)+z(k^×j^) (\overrightarrow r \times \widehat j) = (x\widehat i + y\widehat j + z\widehat k) \times \widehat j\\\ \Rightarrow (\overrightarrow r \times \widehat j) = x(\widehat j \times \widehat i) + y(\widehat j \times \widehat j) + z(\widehat k \times \widehat j)
We know, the cross-product formula of unit vectors is,
k^×j^=−i^ j^×i^=k^ j^×j^=0 \widehat k \times \widehat j = - \widehat i\\\ \widehat j \times \widehat i = \widehat k\\\ \widehat j \times \widehat j = 0
Now,
(r→×j^)=(xk^+0+z(−i^)) (r→×j^)=xk^−zi^ (\overrightarrow r \times \widehat j) = (x\widehat k + 0 + z( - \widehat i))\\\ (\overrightarrow r \times \widehat j) = x\widehat k - z\widehat i
We will calculate the dot product of (r→×i^)  and  (r→×j^)(\overrightarrow r \times \widehat i)\,\,and\,\,(\overrightarrow r \times \widehat j) .
(r→×i^) ⋅ (r→×j^)=(−yk^+z(j^))⋅(xk^−zi^) ⇒(r→×i^) ⋅ (r→×j^)=−yk^⋅(xk^−zk^)+zj^(xk^−zi^) ⇒(r→×i^) ⋅ (r→×j^)=−xy(k^⋅k^)+yz(k^⋅k^)+xz(j^⋅k^)−z2(j^⋅i^) (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) = ( - y\widehat k + z(\widehat j)) \cdot (x\widehat k - z\widehat i)\\\ \Rightarrow (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) = - y\widehat k \cdot (x\widehat k - z\widehat k) + z\widehat j(x\widehat k - z\widehat i)\\\ \Rightarrow (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) = - xy(\widehat k \cdot \widehat k) + yz(\widehat k \cdot \widehat k) + xz(\widehat j \cdot \widehat k) - {z^2}\left( {\widehat j \cdot \widehat i} \right)
We knew, the rules of the dot product of two unit vectors are
i^⋅i^=j^⋅j^=k^⋅k^=1  i^⋅j^=j^⋅k^=k^⋅i^=0 \widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1\,\\\ \widehat i \cdot \widehat j = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0
and
⇒(r→×i^) ⋅ (r→×j^)=−xy(1)+yz(1)+0+0 ⇒(r→×i^) ⋅ (r→×j^)=−xy+yz \Rightarrow (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) = - xy(1) + yz(1) + 0 + 0\\\ \Rightarrow (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) = - xy + yz
We will find the required value of (r→×i^) ⋅ (r→×j^)+xy(\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) + xy
Now,
(r→×i^) ⋅ (r→×j^)+xy=−xy+yz+xy ⇒(r→×i^) ⋅ (r→×j^)+xy=yz (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) + xy = - {{xy}} + yz + {{xy}}\\\ \Rightarrow (\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) + xy = yz
Hence, the value of (r→×i^) ⋅ (r→×j^)+xy  is  yz(\overrightarrow r \times \widehat i)\, \cdot \,(\overrightarrow r \times \widehat j) + xy\,\,is\,\,yz

Note: Students always make mistakes in the dot products of unit vectors and cross products of that. Also, you must have the basic knowledge of vectors and unit vectors to do this numerical.