Question

if r1 vector =4m along n-e and r2 vector =2m due north,find. |r1 vector +r2 vector| amd |r1 vector-r2 vector |

Answer 1

given, $\vec{r_1}=4$ along N- E

so, we can write it in vector form, $\vec{r_1}=4cos45^{\circ}\hat{i}+4sin45^{\circ}\hat{j}$

or, $\vec{r_1}=2\sqrt{2}\hat{i}+2\sqrt{2}\hat{j}$

similarly, $\vec{r_2}=2$ due North.

so, we can write it in vector form, $\vec{r_2}=2\hat{j}$

now,$\vec{r_1}+\vec{r_2}=2\sqrt{2}\hat{i}+2\sqrt{2}\hat{j}+2\hat{j}$

= $2\sqrt{2}\hat{i}+(2\sqrt{2}+2)\hat{j}$

magnitude of $\vec{r_1}+\vec{r_2}$ = √{(2√2)² + (2√2+2)²}

= √{8 + 8 + 4 + 8√2}

= √{20 + 8√2}

$\vec{r_1}-\vec{r_2}=2\sqrt{2}\hat{i}+(2\sqrt{2}-2)\hat{j}$

magnitude of $\vec{r_1}-\vec{r_2}$= √{(2√2)² + (2√2-2)²}

= √{8 + 8 + 4 - 8√2}

= √{20 - 8√2}

Answer 2

Explanation:

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