Question #a648f

Jun 7, 2017

Q-1

The projection or component of $\vec{A}$ along $\vec{B}$ is $\frac{\vec{A} \cdot \vec{B}}{\left\mid \vec{B} \right\mid} ^ 2 \vec{B}$

So the component of $\vec{a} = 3 \hat{i} + 4 \hat{j}$ along the direction of $\hat{i} + \hat{j}$ will be

$\frac{\left(3 \hat{i} + 4 \hat{j}\right) \cdot \left(\hat{i} + \hat{j}\right)}{\left\mid \hat{i} + \hat{j} \right\mid} ^ 2 \left(\hat{i} + \hat{j}\right)$

$= \frac{3 + 4}{\sqrt{2}} ^ 2 \left(\hat{i} + \hat{j}\right)$

$= \frac{7}{2} \left(\hat{i} + \hat{j}\right)$

And the component of $\vec{a} = 3 \hat{i} + 4 \hat{j}$ along the direction of $\hat{i} - \hat{j}$ will be

$\frac{\left(3 \hat{i} + 4 \hat{j}\right) \cdot \left(\hat{i} - \hat{j}\right)}{\left\mid \hat{i} - \hat{j} \right\mid} ^ 2 \left(\hat{i} - \hat{j}\right)$

$= \frac{3 - 4}{\sqrt{2}} ^ 2 \left(\hat{i} - \hat{j}\right)$

$= - \frac{1}{2} \left(\hat{i} - \hat{j}\right)$

Q-2

Let $\left(x , y\right)$ be the cartesian coordinates of the point A represeted by the radius vector $\vec{r} = a t \hat{i} - b {t}^{2} \hat{j}$

So $x = a t \mathmr{and} y = - b {t}^{2}$

Eliminating t from above relation we get

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

$\implies y = - \frac{b}{a} ^ 2 {x}^{2}$

$\implies {x}^{2} = - {a}^{2} / b y$

This is the equation of the trajectory represented by the radius vector $\vec{r} = a t \hat{i} - b {t}^{2} \hat{j}$