Question

Compute the values of following expression
[(3i - 2j - 2k) × (i – k) × (i + j + k) × (i - 2j + 3k)]
(a) i-16j+9k
(b) i+7j-5k
(c) i-7j+5k
(d) i+16j-9k

Answer 1

Hint: Proceed the solution of this question, by assuming vectors as $\overrightarrow {\text{a}} $,$\overrightarrow {\text{b}} $,$\overrightarrow {\text{c}} $,$\overrightarrow {\text{d}} $ and first find cross product of first and second vector then cross product of third and fourth vector individually and later on further do their cross product.

Complete step-by-step answer:

In the given question, there are four vectors and it is asked their cross product.

So assuming given vector as

$\overrightarrow {\text{a}} $ = (3i - 2j - 2k)

$\overrightarrow {\text{b}} $ = (i – k)

$\overrightarrow {\text{c}} $ = (i + j + k)

$\overrightarrow {\text{d}} $ = (i - 2j + 3k)


Hence $\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} {\text{ = }}$ (3i - 2j - 2k) × (i – k)

As we know that cross product can be find using determinant, represented as

  \[ \Rightarrow \left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  3&{ - 2}&{ - 2} \\

  1&0&{ - 1}

\end{array}} \right|\]
which can be further solved similar to normal determinant.
We know how to solve a determinant, suppose-


If it is given a general determinant \[\overrightarrow {\text{m}} \times \overrightarrow {\text{n}} = \left( {{{\text{a}}_1}{\text{i}} + {{\text{a}}_2}{\text{j}} + {{\text{a}}_3}{\text{k}}} \right)j \times \left( {{{\text{b}}_1}{\text{i}} + {{\text{b}}_2}{\text{j}} + {{\text{b}}_3}{\text{k}}} \right) = \left| {{\text{ }}\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&k \\

  {{{\text{a}}_1}}&{{{\text{a}}_2}}&{{{\text{a}}_3}} \\

  {{{\text{b}}_1}}&{{{\text{b}}_2}}&{{{\text{b}}_3}}

\end{array}} \right|{\text{ }}\] \[{\text{where }}\overrightarrow {\text{m}} = \left( {{{\text{a}}_1}{\text{i}} + {{\text{a}}_2}{\text{j}} + {{\text{a}}_3}{\text{k}}} \right){\text{ & }}\overrightarrow {\text{n}} = \left( {{{\text{b}}_1}{\text{i}} + {{\text{b}}_2}{\text{j}} + {{\text{b}}_3}{\text{k}}} \right)\]

Hence we can solve it by expanding through the first row.


Therefore, $\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} {\text{ = }}$

\[\left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  3&{ - 2}&{ - 2} \\

  1&0&{ - 1}

\end{array}} \right|\] = i (2)-j (-3+2) +k (2) =2i+j+2k ……..(1)



  $ \Rightarrow \overrightarrow {\text{c}} \times \overrightarrow {\text{d}} {\text{ = }}$ (i + j + k) × (i - 2j + 3k)

As we know that cross product can be find using determinant, so

On representing in determinant form


\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  1&1&1 \\

  1&{ - 2}&3

\end{array}} \right|\]
which can be further solved similar to normal determinant.

Therefore, $\overrightarrow {\text{c}} \times \overrightarrow {\text{d}} {\text{ = }}$\[\left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  1&1&1 \\

  1&{ - 2}&3

\end{array}} \right|\]= i (5)-j (2) +k (-3) = 5i-2j-3k …….(2)

Using equation (1) & (2)


 $ \Rightarrow \left( {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right) \times \left( {\overrightarrow {\text{c}} \times \overrightarrow {\text{d}} } \right){\text{ = }}$ (i + j + k) × (i - 2j + 3k)

On representing in determinant form

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  2&1&2 \\

  5&{ - 2}&{ - 3}

\end{array}} \right|\]
which can be further solved similar to normal determinant.

$ \Rightarrow \left( {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right) \times \left( {\overrightarrow {\text{c}} \times \overrightarrow {\text{d}} } \right){\text{ = }}$ \[\left| {\begin{array}{*{20}{c}}

  {\text{i}}&{\text{j}}&{\text{k}} \\

  2&1&2 \\

  5&{ - 2}&{ - 3}

\end{array}} \right|\]= i (1)-j (-16) +k (-9) = i+16j-9k


Hence option D is correct.

Note: In this particular question we have to solve the determinant to get the answer that we can do either by expanding it through first row or we can learn this result which states as

\[\overrightarrow {\text{m}} \times \overrightarrow {\text{n}} = \left( {{{\text{a}}_1}{\text{i}} + {{\text{a}}_2}{\text{j}} + {{\text{a}}_3}{\text{k}}} \right)j \times \left( {{{\text{b}}_1}{\text{i}} + {{\text{b}}_2}{\text{j}} + {{\text{b}}_3}{\text{k}}} \right) = ({{\text{a}}_2}{{\text{b}}_3} - {{\text{a}}_3}{{\text{b}}_1}){\text{i + }}({{\text{a}}_3}{{\text{b}}_1} - {{\text{a}}_1}{{\text{b}}_3}){\text{j + }}({{\text{a}}_1}{{\text{b}}_2} - {{\text{a}}_2}{{\text{b}}_1}){\text{k }}\]

Along with that we should notice that switching the order of the vectors in the cross product simply changed all the signs in the result. Note as well that this means that the two cross products will point in exactly opposite directions since they only differ by a sign. A cross product of two vectors will always give a vector quantity.