Question

Find the area of the parallelogram whose diagonals are represented by the vectors : d₁ = 3i + j - 2k and d₂ = i + 3j + 4k.

Answer 1

It is clear by parallelogram law of vector addition that diagonals of parallelogram are given by \(\vec d_1=\vec a+\vec b\) and \(\vec d_2=\vec a-\vec b\).

Therefore, 

(By addition and subtraction of above two equations)

Now, we know that the area of the parallelogram whose adjacent are  \(\vec a\) and  \(\vec b\) is given by \(|\vec a\times\vec b|.\) 

Therefore, the area of the parallelogram is

It means the area of the parallelogram whose diagonals are  \(\vec d_1\) and  \(\vec d_2\) is given by \(\cfrac12|\vec d_1\times\vec d_2|.\) 

Now, we have \(\vec d_1=3\hat i+\hat j-2\hat k\) and  \(\vec d_2=\hat i-3\hat j+4\hat k.\) 

Therefore, \(\vec d_1\times\vec d_2\) 

(By expanding along row R₁)

Now,

Therefore, the area of the parallelogram is 

Hence, the area of the parallelogram whose diagonals are  \(\vec d_1=3\hat i+\hat j-2\hat k\) and  \(\vec d_2=\hat i-3\hat j+4\hat k\) is 5\(\sqrt3\).