Two particles are located at it `\vec {r₁}` = 3`\hat {i}` + 7`\hat {j}` and `\vec {r₂}` = -2`\hat {i}` + 3`\hat {j}` respectively. Find both the magnitude of the vector `\vec {r₂}` - `\vec {r₁}` and its orientation with respect to the x-axis. [Ans: 6.4, 219°]


Given:

`\vec {r₁}` = 3`\hat {i}` + 7`\hat {j}` 
`\vec {r₂}` = -2`\hat {i}` + 3`\hat {j}`


To Find:

Magnitude of the Position Vector = |`\vec {r}`| = 
|`\vec {r₂}` - `\vec {r₁}`| = ?
Orientation with respect to X-asis = Ó¨ = ?


Solution: 

As we are given 

`\vec {r}` = `\vec {r₂}` - `\vec {r₁}` 

`\vec {r}` = (-2`\hat {i}` + 3`\hat {j}`) - (3`\hat {i}` + 7`\hat {j}`)

`\vec {r}` = -5`\hat {i}` - 4 `\hat {j}`

Here x-component = -5 and the y-component = -4 . So, to find its magnitude |`\vec {r}`| we have

|`\vec {r}`| = `\sqrt {x^2 + y^2}`

by putting values 

r = `\sqrt {(-5)^2 + (-4)^2}`

r = `\sqrt {25 + 16}`

r = `\sqrt {41}`

r = 6.4    --------Ans (1)

Thus the magnitude of the vector `\vec {r₂}` - `\vec {r₁}` 6.4

For direction (orientation w.r.t. X-axis) we have the formula

tan Ó¨ = `\frac {y}{x}`

or

Ó¨ =  tan⁻¹ `\frac {y}{x}`

by putting value of x and y


Ó¨ =  tan⁻¹ `\frac {-4}{-5}`

Ó¨ =  tan⁻¹ (0.8)

Ó¨ =  39⁰

As x-component = -5 and y-component = -4 (both are negative) therefore it lies in the 3rd quadrant. so we will 180⁰ to the angle for the 3rd quadrant. i.e.

Ó¨ =  180⁰  + 39⁰

Ó¨ =  219⁰  --------Ans (2)
 
Thus the Orientation of the vector `\vec {r₂}` - `\vec {r₁}` with respect to x-asis  is 219⁰ 


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Numerical Problem 2.4
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List of 11th Physics Numerical Problems of Chapter - 2
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