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Find the magnitude and direction of r(1)...

Find the magnitude and direction of `r_(1)-r_(2)` when `|r_(1)|=5` and points North-East while `|r_(2)|=5` but points North-West.

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To find the magnitude and direction of the vector \( \mathbf{r_1} - \mathbf{r_2} \), where \( |\mathbf{r_1}| = 5 \) (pointing North-East) and \( |\mathbf{r_2}| = 5 \) (pointing North-West), we can follow these steps: ### Step 1: Represent the vectors in component form - The vector \( \mathbf{r_1} \) pointing North-East can be represented in terms of its components. Since it makes a 45-degree angle with both the North and East axes: \[ \mathbf{r_1} = 5 \left( \cos(45^\circ), \sin(45^\circ) \right) = 5 \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = \left( \frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2} \right) \] - The vector \( \mathbf{r_2} \) pointing North-West can also be represented similarly: \[ \mathbf{r_2} = 5 \left( \cos(135^\circ), \sin(135^\circ) \right) = 5 \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = \left( -\frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2} \right) \] ### Step 2: Calculate \( \mathbf{r_1} - \mathbf{r_2} \) - To find \( \mathbf{r_1} - \mathbf{r_2} \), we subtract the components of \( \mathbf{r_2} \) from \( \mathbf{r_1} \): \[ \mathbf{r_1} - \mathbf{r_2} = \left( \frac{5\sqrt{2}}{2} - \left(-\frac{5\sqrt{2}}{2}\right), \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2} \right) \] \[ = \left( \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}, 0 \right) = \left( 5\sqrt{2}, 0 \right) \] ### Step 3: Find the magnitude of \( \mathbf{r_1} - \mathbf{r_2} \) - The magnitude of the resultant vector \( \mathbf{r_1} - \mathbf{r_2} \) is simply the length of the vector: \[ |\mathbf{r_1} - \mathbf{r_2}| = \sqrt{(5\sqrt{2})^2 + 0^2} = 5\sqrt{2} \] ### Step 4: Determine the direction of \( \mathbf{r_1} - \mathbf{r_2} \) - The direction of the resultant vector \( \mathbf{r_1} - \mathbf{r_2} \) is along the positive x-axis (East direction) since the y-component is zero. ### Final Answer - The magnitude of \( \mathbf{r_1} - \mathbf{r_2} \) is \( 5\sqrt{2} \) and the direction is East. ---
To find the magnitude and direction of the vector \( \mathbf{r_1} - \mathbf{r_2} \), where \( |\mathbf{r_1}| = 5 \) (pointing North-East) and \( |\mathbf{r_2}| = 5 \) (pointing North-West), we can follow these steps: ### Step 1: Represent the vectors in component form - The vector \( \mathbf{r_1} \) pointing North-East can be represented in terms of its components. Since it makes a 45-degree angle with both the North and East axes: \[ \mathbf{r_1} = 5 \left( \cos(45^\circ), \sin(45^\circ) \right) = 5 \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = \left( \frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2} \right) \] ...
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