Question

If \[a=3i-2j+2k\],\[b=6i+4j-2k\] and \[c=3i-2j-4k\],then $a.(b\times c)$is [Karnataka CET 2001]
A. $122$
B. $-144$
C. $120$
D. $-120$

Answer 1

Hint: Let $\vec{a},\vec{b}$and$\vec{c}$are three vectors. Then the scalar product of $\vec{a}$ and $(\vec{b}\times \vec{c})$ that is $a.(b\times c)$called as the triple product of $\vec{a},\vec{b}$and $\vec{c}$.
If \[a={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k,b={{b}_{1}}i+{{b}_{2}}j+{{b}_{3}}k\] and \[c={{c}_{1}}i+{{c}_{2}}j+{{c}_{3}}k\], then
$\vec{b}\times \vec{c}=\left| \begin{matrix}
   i & j & k \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|=({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})i-({{b}_{1}}{{c}_{3}}-{{b}_{3}}{{c}_{1}})j+({{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}})k$
And
\[\vec{a}.(\vec{b}\times \vec{c})={{a}_{1}}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})+{{a}_{2}}({{b}_{3}}{{c}_{1}}-{{b}_{1}}{{c}_{3}})+{{a}_{3}}({{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}})\]
$=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$



Formula Used:\[\vec{a}.(\vec{b}\times \vec{c})={{a}_{1}}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})+{{a}_{2}}({{b}_{3}}{{c}_{1}}-{{b}_{1}}{{c}_{3}})+{{a}_{3}}({{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}})\]
$=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$



Complete step by step solution:Given that \[a=3i-2j+2k\],\[b=6i+4j-2k\] and \[c=3i-2j-4k\]
So, we have the values of the coefficients here,
$\begin{align}
  & {{a}_{1}}=3,{{a}_{2}}=-2,{{a}_{1}}=2 \\
 & {{b}_{1}}=6,{{b}_{2}}=4,{{b}_{3}}=-2 \\
 & {{c}_{1}}=3,{{c}_{2}}=-2,{{c}_{3}}=-4 \\
\end{align}$

We are Putting these values in the formula,
\[\vec{a}.(\vec{b}\times \vec{c})=\left| \begin{matrix}
   3 & -2 & 2 \\
   6 & 4 & -2 \\
   3 & -2 & -4 \\
\end{matrix} \right|\]
By solving this we will have
\[\vec{a}.(\vec{b}\times \vec{c})=3(-16-4)+2(-24+6)+2(-12-12)\]
\[\vec{a}.(\vec{b}\times \vec{c})=-144\]
Hence, we can see that we are getting the value not equal zero. So these vectors are not coplanar too.



Option ‘B’ is correct

Note: This question is only formula-based type question. So, while solving such question try to note the correct values of each coefficient so that while putting their value in the formula, you will get the correct answers. Sometimes students do mistakes while solving with minus in the determinants. So, take care of that too.