The area of the parallelogram whose diagonals are ^ i − 3 ^ j + 2 ^ k , − ^ i + 2 ^ j is Options: A . &nbsp 4 √ 29 s q .units B . &nbsp 1 2 √ 21 s q .units C . &nbsp 10 √ 3 s q .units D . &nbsp 1 2 √ 270 s q .units

Answer: Option B : B Vector area = 1 2 ( a × b ) = 1 2 ∣ ∣ ∣ ∣ ∣ ^ i ^ j ^ k 1 − 3 2 1 2 0 ∣ ∣ ∣ ∣ ∣ = 1 2 [ ( 0 − 4 ) − ^ j ( 0 + 2 ) + ^ k ( 2 − 3 ) ] = 1 2 ( − 4 ^ i − 2 ^ j − ^ k ) Area = 1 2 √ 16 + 4 + 1 = 1 2 √ 21 s q . units

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

Discussion Forum : Vector Algebra

Question -

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

Options:

A .

4 \sqrt{29} s q

.units

B .

\frac{1}{2} \sqrt{21} s q

.units

C .

10 \sqrt{3} s q

.units

D .

\frac{1}{2} \sqrt{270} s q

.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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