Question 4: Motion in 2D & Vectors

Question

If $$\alpha, \beta$$ and $$\gamma$$ are the angles between the vectors $$\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$$, and $$\overrightarrow{\mathrm{R}}$$ and $$\alpha=90^{\circ}$$ as shown in figure. the product of $$(\vec{Q} 	imes \vec{R}) \cdot \vec{Q}$$ is equal to

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

$$
R Q^2 \sin 	heta \cos 	heta
$$

Answer

$$
R Q^2 \cos 	heta
$$

Answer

$$
R Q^2 \sin	heta
$$

Answer

Zero

Motion in 2D & Vectors

If $\alpha, \beta and \gamma are the angles between the vectors \overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}, and \overrightarrow{\mathrm{R}} and \alpha=90^{\circ} as shown in figure. the product of (\vec{Q} \times \vec{R}) \cdot \vec{Q}$ is equal to

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