A vector perpendicular to a plane containing a triangle ABC having vertices as A(1,1,0), B(2,1,1) and C (0,3,2), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-2\hat{\imath} - 3\hat{\jmath} + 2\hat{k}\)
Two sides of the triangle lie in the plane:
\(\overrightarrow{AB} = (2-1)\hat{\imath} + (1-1)\hat{\jmath} + (1-0)\hat{k} = \hat{\imath} + \hat{k}\)
\(\overrightarrow{AC} = (0-1)\hat{\imath} + (3-1)\hat{\jmath} + (2-0)\hat{k} = -\hat{\imath} + 2\hat{\jmath} + 2\hat{k}\)
Their cross product is perpendicular to both, and so to the plane:
\( \displaystyle \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & 1 \\ -1 & 2 & 2 \end{vmatrix} \)
\( = (0 - 2)\hat{\imath} - (2 + 1)\hat{\jmath} + (2 - 0)\hat{k} \)
\( = -2\hat{\imath} - 3\hat{\jmath} + 2\hat{k} \)
Check: \((-2)(1) + (-3)(0) + 2(1) = 0\) and \((-2)(-1) + (-3)(2) + 2(2) = 0\).