Question 6: Motion in 2D & Vectors

Question

The sides of a parallelogram are represented by vectors $$\vec{p}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$$ and $$\vec{q}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$$. Then, the area of the parallelogram is

Accepted Answer

Area of a parallelogram $$=|\vec{a} 	imes \vec{b}|$$ where, a and b are sides of parallelogram.
Given, $$\vec{a} = \vec{p} = 5\widehat i - 4\widehat j + 3\widehat k$$
and $$\vec{b} = \vec{q} = 3\widehat i + 2\widehat j - \widehat k$$
$$	herefore \vec{a} 	imes \vec{b} = \left| {\matrix{
   {\widehat i} & {\widehat j} & {\widehat k}  \cr 
   5 & { - 4} & 3  \cr 
   3 & 2 & { - 1}  \cr

} } ight|$$
$$\Rightarrow \vec{a} 	imes \vec{b} = \widehat i(4 - 6) - \widehat j( - 5 - 9) + \widehat k(10 + 12)$$
$$\Rightarrow \vec{a} 	imes \vec{b} =  - 2\widehat i + 14\widehat j + 22\widehat k$$
Thus, $$|\vec{a} 	imes \vec{b}| = \sqrt {{{(2)}^2} + {{(14)}^2} + {{(22)}^2}}  = \sqrt {684} $$ sq units

Answer

$$\sqrt{684}$$ sq units

Answer

$$\sqrt{72}$$ sq units

Answer

171 sq units

Answer

72 sq units

Motion in 2D & Vectors

The sides of a parallelogram are represented by vectors $\vec{p}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} and \vec{q}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$. Then, the area of the parallelogram is

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