Question

If the value of \[a=2i+3j+4k\], \[b=i+j-k\] and \[c=i-j+k\] then compute \[a\times \left( b\times c \right)\] and verify that it is perpendicular to a.

Answer 1

Hint: We will use the fact that dot product of two vectors p and q where \[p=xi+yj+zk\] and \[q={{x}^{'}}i+{{y}^{'}}j+{{z}^{'}}k\] is given by \[p.q=x{{x}^{'}}+y{{y}^{'}}+z{{z}^{'}}\]; and the cross product is given by, \[p\times q=\left| \begin{matrix}
   i & j & k \\
   x & y & z \\
   {{x}^{'}} & {{y}^{'}} & {{z}^{'}} \\
\end{matrix} \right|\].

Complete step-by-step answer:
To compute the cross product of two vectors we have,
Let \[p=xi+yj+zk\] and \[q={{x}^{'}}i+{{y}^{'}}j+{{z}^{'}}k\]
Then cross – product \[p\times q\] is given as,
\[p\times q=\left| \begin{matrix}
   i & j & k \\
   x & y & z \\
   {{x}^{'}} & {{y}^{'}} & {{z}^{'}} \\
\end{matrix} \right|\]
We have, \[a=2i+3j+4k\], \[b=i+j-k\] and \[c=i-j+k\].
First we will compute \[b\times c\] using formula stated above,
We have,
\[b\times c=\left| \begin{matrix}
   i & j & k \\
   1 & 1 & -1 \\
   1 & -1 & 1 \\
\end{matrix} \right|\]
Opening the determinant along \[{{1}^{st}}\] row we get,
\[\begin{align}
  & b\times c=i\left( 1-1 \right)+\left( -j \right)\left( +1+1 \right)+k\left( -1-1 \right) \\
 & \Rightarrow b\times c=-2j-2k \\
\end{align}\]
Let, \[b\times c=d=-2j-2k\].
Then we want to compute \[a\times d\] now.
Using formula stated above we have,
\[a\times d=\left| \begin{matrix}
   i & j & k \\
   2 & 3 & 4 \\
   0 & -2 & -2 \\
\end{matrix} \right|\]
Opening the determinant along \[{{1}^{st}}\] row we get,
\[\begin{align}
  & a\times d=i\left( -6+8 \right)-j\left( -4 \right)+k\left( -4 \right) \\
 & \Rightarrow a\times d=2i+4j-4k \\
\end{align}\]
So, the value of \[a\times \left( b\times c \right)=2i+4j-4k\], which is the required solution.
Finally we have to verify that obtained value of \[a\times \left( b\times c \right)\] is perpendicular to a.
Two vectors are said to be perpendicular if there dot product is zero.
Let \[a\times \left( b\times c \right)=e\]
Then, \[e=2i+4j-4k\]
And \[a=2i+3j+4k\]
Now dot product of two vectors when,
\[p=xi+yj+zk\] and \[q={{x}^{'}}i+{{y}^{'}}j+{{z}^{'}}k\] is given by,
 \[p.q=x{{x}^{'}}+y{{y}^{'}}+z{{z}^{'}}\]
Then using this we have,
\[\begin{align}
  & a.e=2\left( 2 \right)+\left( 3 \right)\left( 4 \right)+\left( 4 \right)\left( -4 \right) \\
 & a.e=4+12-16 \\
 & a.e=16-16 \\
 & a.e=0 \\
\end{align}\]
So, the dot product of \[a\times \left( b\times c \right)\] and a is zero. Hence they both are perpendicular. Hence verified.

Note: Another way to solve this question can be using the formula to vector triple product of cross product of three vectors which is given as, \[\overset{\to }{\mathop{a}}\,\times \left( \overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\, \right)=\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\, \right)\overset{\to }{\mathop{b}}\,-\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\, \right)\overset{\to }{\mathop{c}}\,\], where a, b, c are three vectors. The answer anyway would be the same.