Correct Answer - Option 1 :
\(\dfrac{\sqrt{165}}{2}\)
Concept:
Let A, B , and C be the vertices of the
\(\triangle ABC\) , then the area of that triangle =
\(\rm \dfrac 1 2 \times |\vec {AB}\times \vec {AC}|\)
Calculations:
Given, triangle whose vertices are at A = (3, -1, 2) = \(\rm 3\vec i -\vec j + 2\vec k\)
B = (1, -1, -3) = \(\rm \vec i -\vec j - 3\vec k\)
and C = (4, -3, 1) = \(\rm 4\vec i -3\vec j + \vec k\)
Let A, B , and C be the vertices of the \(\triangle ABC\) , then the area of that triangle = \(\rm \dfrac 1 2 \times |\vec {AB}\times \vec {AC}|\)
\(\rm \vec {AB} = \rm (\vec i -\vec j - 3\vec k) - (3\vec i -\vec j + 2\vec k)\)
⇒\(\rm \vec {AB}= -2\vec i + 0\vec k -5 \vec k\)
\(\rm \vec {AC} = \rm (4\vec i -3\vec j + \vec k) - (3\vec i -\vec j + 2\vec k)\)
⇒ \(\rm \vec {AC} = \vec i -2\vec j -\vec k\)
Now, \(\rm \vec {AB}\times\vec {AC}= \)\(\begin{vmatrix} \vec i&\vec j & \vec k \\ -2& 0 & -5 \\ 1 & -2 & -1 \end{vmatrix}\)
⇒\(\rm \vec {AB} \times \vec {AC} = \vec i (-10)-\vec j(2+5)+\vec k(4-0)\)
⇒\(\rm \vec {AB} \times \vec {AC} = -10\vec i -7\vec j+4\vec k\)
⇒ \(\rm |\vec {AB} \times \vec {AC}| = \sqrt {(-10)^2+(-7)^2+(4)^2 }\)
⇒\(\rm |\vec {AB} \times \vec {AC}| = \sqrt {165}\)
The area of that triangle = \(\dfrac 1 2 \times |\vec {AB}\times \vec {AC}|\)
⇒ The area of that triangle = \(\dfrac 12 \times \sqrt {165}\)
Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is \(\dfrac {\sqrt {165}}{2}\)
