Question

Three vertices - \( A , B \) and \( D \) of a parallelogram \( ABCD \) are given by, \( A (0,-3,3), B (- \) \( 5, m-3,0) \) and \( D (1,-3,4) \). The area of the parallelogram \( A B C D \) is 6 sq units. Using the vector method, find the value(s) of \( m \). Show your steps.

Answer 1

\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}\)

 = (-5\(\hat i\)+ (m - 3)\(\hat j\) + 0\(\hat k\)) - (0\(\hat i\) - 3\(\hat j\) + 3\(\hat k\))

 = -5\(\hat i\) + m\(\hat j\) - 3\(\hat k\)

\(\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}\)

 = \((\hat i-3\hat j+4\hat k)-(0\hat i-3\hat j+3\hat j)\)

 = \(\hat i+\hat k\) 

∴ \(\overrightarrow{AB}\times\overrightarrow{AD}\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\-5&m&-3\\1&0&1\end{vmatrix}\)

 = \(m\hat i+2\hat j-m\hat k\)

∴ \(|\overrightarrow{AB}\times\overrightarrow{AD}|\) = \(|m\hat i+2\hat j-m\hat j|\) = \(\sqrt{m^2+4+m^2}=\sqrt{2m^2+4}\)

∴ Area of parallelogram = \(|\overrightarrow{AB}\times\overrightarrow{AD}|\) = \(\sqrt{2m^2+4}\) sq. units

∴ \(\sqrt{2m^2+4}=6\)

⇒ 2m² + 4 = 36

⇒ 2m² = 32 ⇒ m² = 16

⇒ m = \(\pm4\)

∴ values of m are -4 and 4.