Find the area of the triangle ABC if adjacent sides of a triangle  are \(\overrightarrow {AB} = \hat i + 2\hat j + 3\hat k\;and\;\overrightarrow {BC} = 3\hat i + 7\hat j + \hat k\)

Concept:

• If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)

• If \(\vec a\;and\;\vec b\) are the adjacent sides of a triangle, then area of triangle is given by: \(\frac{1}{2}\;\left| {\vec a \times \;\vec b} \right|\)

Calculation:

Given: \(\overrightarrow {AB} = \hat i + 2\hat j + 3\hat k\;and\;\overrightarrow {BC} = 3\hat i + 7\hat j + \hat k\) are the adjacent sides of a triangle ABC.

As we know that, If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)

⇒ \(\vec {AB} \times \;\vec {BC} = \hat i(2 - 21) - \hat j(1 - 9) + \hat k(7 - 6)\)

⇒ \(\vec {AB} \times \;\vec {BC} = -19\hat i + 8\hat j+ \hat k\)

⇒ \(|\vec{AB} \times \vec{BC}| = \sqrt{426}\)

So, the area of the required triangle ABC is \(\frac{1}{2}|\vec{AB} \times \vec{BC}| = \frac{\sqrt{426}}{2}\)

Hence, the correct option is 1.