The sides of a parallelogram are $2^i - 4^j + 5^k$ and $^i - 2^j - 3^k$ . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.

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Solution

Given,
$2^i - 4^j + 5^k$
$^i - 2^j - 3^k$

Consider
$\to A = 2^i - 4^j + 5^k$
and $\to B =^i - 2^j - 3^k$

Then,
$\to P = \to A + \to B$
$= (2^i - 4^j + 5^k) + (^i - 2^j - 3^k)$
$= 3^i - 6^j + 2^k$
Therefore, $\to P = 3^i - 6^j + 2^k$

So, unit vector along the diagonal $= \frac{\to P}{∣ ∣ \to P ∣ ∣}$
$\begin{matrix} = \frac{3^i - 6^j + 2^k}{\sqrt{3^{2} + {(- 6)}^{2} + 2^{2}}} = \frac{3^i - 6^j + 2^k}{7} \end{matrix}$

Now

$\begin{matrix} \to A \times \to B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 4 & 5 1 & - 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (12 + 10) -^j (- 6 - 5) +^k (- 4 + 4) = 22^i + 11^j \end{matrix}$

Area of parallelogram $= ∣ ∣ \to A \times \to B ∣ ∣$
$\begin{matrix} = \sqrt{22^{2} + 11^{2}} = \sqrt{605} = 11 \sqrt{5} s q u a r e u n i t s \end{matrix}$

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The sides of a parallelogram are 2^i−4^j+5^k and ^i−2^j−3^k . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.

The sides of a parallelogram are $2^i - 4^j + 5^k$ and $^i - 2^j - 3^k$ . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.