Correct Answer - Option 1 : 15/2 sq units
Concept:
Area of triangle when two vectors are given:
\(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)
Cross product:
\(\begin{array}{l} \rm \vec{a}=x_{1} \hat{1}+y_{1} \hat{j}+z_{1} \hat{k}\\ \rm \vec{b}=x_{2} \hat{1}+y_{2} \hat{j}+z_{2} \hat{k} \\ \rm \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \rm \hat{i} &\rm j &\rm \hat{k} \\ \rm x_{1} &\rm y_{1} & \rm z_{1} \\ \rm x_{2} & \rm y_{2} &\rm z_{2} \rm \end{array}\right| \end{array}\)
Calculation:
Here, Let A = (0,2,2), B = (2,0,-1) and C =(3,4,0)
AB = (2-0, 0-2, -1-2) = (2, -2, -3) and
AC = (3-0, 4-2, 0-2) = (3, 2, -2)
Area of triangle = \(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)
\(\begin{array}{l} =\rm \frac{1}{2}\left|\begin{array}{ccc} \rm \hat i &\rm \hat j & \rm \hat k \\ 2 & -2 & -3 \\ 3 & 2 & -2 \end{array}\right| \\ =\frac{1}{2}|[\rm \hat i(4+6)+ \rm \hat j(-4+9)+ \rm \hat k(4+6)]| \end{array}\)
=1/2(10 i + 5 j + 10 k)
= 1/2 √(100 + 25 + 100)
= 15/2 sq unit
Hence, option (1) is correct.
