The area of the parallelogram whose diagonals are ^i−3^j+2^k,−^i+2

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Question

The area of the parallelogram whose diagonals are

^i - 3^j + 2^k, -^i + 2^j

is

Options:

A . 4√29sq.units

B . 12√21sq.units

C . 10√3sq.units

D . 12√270sq.units

Answer: Option B
:
B
Vector area =

\frac{1}{2} (a \times b) = \frac{1}{2} ∣ ∣∣∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 3 & 2 1 & 2 & 0 \end{matrix} ∣ ∣∣∣ ∣ = \frac{1}{2} [(0 - 4) -^j (0 + 2) +^k (2 - 3)] = \frac{1}{2} (- 4^i - 2^j -^k)

Area =

\frac{1}{2} \sqrt{16 + 4 + 1} = \frac{1}{2} \sqrt{21} s q .

units

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