What is the area of the parallelogram having diagonals 3î + ĵ - 2k̂ and î - 3ĵ + 4k̂?

Concept:

Considering a parallelogram ABCD, AC and BD are the diagonals that bisect each other at O.

We know that, the diagonal of the parallelogram bisects the parallelogram into two triangles of equal area.

Area of the parallelogram = 2 × area of ∆BCD.

In ∆BCD,

Base = BD and Height = CE = OC × sin θ = ½ × AC × sin θ

Area of triangle BCD = ½ × base × height = 1/2 × |\(\overrightarrow {BD} \)| × |\(\frac{{\overrightarrow {AC} }}{2}\) sin θ|

So, area of the parallelogram ABCD = |\(\overrightarrow {BD} \)| × |\(\frac{{\overrightarrow {AC} }}{2}\) sin θ | = 1/2 × |\(\overrightarrow {BD} \times \overrightarrow {AC} \)|

Calculation:

Given:

Let us assume the diagonals AC and BD as,

\(\overrightarrow {AC} \) = 3î + ĵ - 2k̂

\(\overrightarrow {BD}\) = î - 3ĵ + 4k̂

To find: Area of the parallelogram?

Area of the parallelogram = ½ × |\(\overrightarrow {BD} \times \overrightarrow {AC} \)|

= ½ × \(\left| {\begin{array}{*{20}{c}} i&j&k\\ 3&1&-2\\ 1&-3&4 \end{array}} \right|\)

= ½ × |î {4 – 6} ĵ – {12 – (-2)} + k̂ {-9 – 1}|

= ½ × |-2î - 14ĵ – 10 k̂|

= ½ × \(\sqrt {{2^2} + {{14}^2} + {{10}^2}} \)

= ½ × √(4 + 196 + 100)

= ½ × √(300)

= ½ × 10√3

= 5√3