Question #a648f

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Question #a648f

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Answer 1 · 2017-06-07T04:22:43

Q-1

The projection or component of $\to A$ $vecA$ along $\to B$ $vecB$ is $\frac{\to A \cdot \to B}{{∣ ∣ ∣ \to B ∣ ∣ ∣}^{2}} \to B$ $(vecA*vecB)/absvecB^2vecB$

So the component of $\to a = 3 ˆ i + 4 ˆ j$ $veca=3hati+4hatj$ along the direction of $ˆ i + ˆ j$ $hati+hatj$ will be

$\frac{(3 ˆ i + 4 ˆ j) \cdot (ˆ i + ˆ j)}{{∣ ∣ ˆ i + ˆ j ∣ ∣}^{2}} (ˆ i + ˆ j)$ $((3hati+4hatj)*(hati+hatj))/abs(hati+hatj)^2(hati+hatj)$

$= \frac{3 + 4}{{(\sqrt{2})}^{2}} (ˆ i + ˆ j)$ $=(3+4)/(sqrt2)^2(hati+hatj)$

$= \frac{7}{2} (ˆ i + ˆ j)$ $=7/2(hati+hatj)$

And the component of $\to a = 3 ˆ i + 4 ˆ j$ $veca=3hati+4hatj$ along the direction of $ˆ i - ˆ j$ $hati-hatj$ will be

$\frac{(3 ˆ i + 4 ˆ j) \cdot (ˆ i - ˆ j)}{{∣ ∣ ˆ i - ˆ j ∣ ∣}^{2}} (ˆ i - ˆ j)$ $((3hati+4hatj)*(hati-hatj))/abs(hati-hatj)^2(hati-hatj)$

$= \frac{3 - 4}{{(\sqrt{2})}^{2}} (ˆ i - ˆ j)$ $=(3-4)/(sqrt2)^2(hati-hatj)$

$= - \frac{1}{2} (ˆ i - ˆ j)$ $=-1/2(hati-hatj)$

Q-2

Let $(x, y)$ $(x,y)$ be the cartesian coordinates of the point A represeted by the radius vector $\to r = a t ˆ i - b t^{2} ˆ j$ $vecr=athati-bt^2hatj$

So $x = a t and y = - b t^{2}$ $x=at and y=-bt^2$

Eliminating t from above relation we get

$y = - b {(\frac{x}{a})}^{2}$ $y=-b(x/a)^2$

$\Rightarrow y = - \frac{b}{a^{2}} x^{2}$ $=>y=-b/a^2x^2$

$\Rightarrow x^{2} = - \frac{a^{2}}{b} y$ $=>x^2=-a^2/by$

This is the equation of the trajectory represented by the radius vector $\to r = a t ˆ i - b t^{2} ˆ j$ $vecr=athati-bt^2hatj$

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