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,

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{aligned} & 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{aligned}$$ Let, $$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{aligned} & a\times d=i\left( -6+8 \right)-j\left( -4 \right)+k\left( -4 \right) \\\ & \Rightarrow a\times d=2i+4j-4k \\\ \end{aligned}$$ So, the value of $$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{aligned} & 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{aligned}$$ So, the dot product of $$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.