Correct Answer - Option 4 :
\(10\hat i + 2\hat j - 16\hat k\)
Concept:
If \(\vec a \ and \ \vec b\) are two vectors then \(\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)
Calculation:
Given: \(\rm \vec{a}= 3i + j +2 k \) and \(\rm \vec{b}= i +3 j + k \) .
⇒ \(\vec u = \rm \vec{a}+\vec{b}=( 3i+j +2k ) +(i+3j+k)= 4i + 4j + 3k\)
⇒ \(\vec v = \rm \vec{a}-\vec{b}=( 3i+j +2k ) -(i+3j+k)= 2i - 2j + k\)
As we need to find find the vector which is perpendicular to both \(\vec u \ and \ \vec v\) i.e we need to find \(\vec u \times \vec v \)
As we know that, if \(\vec a \ and \ \vec b\) are two vectors then \(\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)
\(\Rightarrow \rm \vec{u}\times\vec{v} = \rm \begin{vmatrix} i & j &k \\ 4 &4 &3 \\ 2&-2 &1 \end{vmatrix}\)
⇒ \(\rm \vec{u}\times\vec{v} = \hat i (4 + 6) - \hat j (4 - 6) + \hat k(- 8 - 8)\)
⇒ \(\rm \vec{u}\times\vec{v} = 10\hat i + 2\hat j - 16\hat k\)
Hence, option 4 is correct.
