If \(\rm {a}=\vec{i}-2 \vec{j}+\vec{{k}}\), \(\rm {b}=\vec{i}+\vec{{k}}, {c}=2 \vec{j}-\vec{{k}}\), then the area (in sq. units) of a parallelogram with diagonals a + b and b + c will be:

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This question was previously asked in

AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper

Answer 1

Given:

\(\rm {a}=⃗{i}-2 ⃗{j}+⃗{{k}}\) , \(\rm {b}=⃗{i}+⃗{{k}}, {c}=2 ⃗{j}-⃗{{k}}\)

Concept:

The area of parallelogram is

\(\rm =\frac{1}{2}|{d_1}\times{d_2}|\)

Calculation:

We have,

\(\rm {a}=⃗{i}-2 ⃗{j}+⃗{{k}}\), \(\rm {b}=⃗{i}+⃗{{k}}, {c}=2 ⃗{j}-⃗{{k}}\)

Then

\(\rm {a+b}=2⃗{i}-2 ⃗{j}+2⃗{{k}}\)

\(\rm {b+c}=⃗{i}+2 ⃗{j}\)

The area of parallelogram

\(\rm =\frac{1}{2}|(a+b)\times(b+c)|\)

Now

\(\rm(a+b)\times(b+c)=\begin{bmatrix}⃗{i} & ⃗{j} & ⃗{k} \\ 2 & -2 & 2 \\ 1 & 2 & 0\end{bmatrix} \ \)

\(\rm =-4\vec{i}+2\vec{j}+6\vec{k}\)

Then the area is

\(\rm =\frac{1}{2}|(a+b)\times(b+c)|\)

\(\rm =\frac{1}{2}√{(-4)^2+(2)^2+(6)^2}\)

\(\rm =\frac{1}{2}√{56}\)

= √ 14 sq unit

Hence the option (4) is correct.

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