Suppose, in a rectangular coordinate system, a vector A has its tail at the point P (-2, -3) and its tip at Q (3,9). Determine the distance between these two points.

(Ans: 13 Units) 

Given:

Two Points P⃗ (-2, -3) and Q (3, 9)


To Find:

Distance between the point P and Q = r = ?


Solution: 

Position vector of point P = `\vec {r₁}` = -2`\hat {i}` - 3`\hat {j}` 

Position vector of point Q = `\vec {r₂}` = 3`\hat {i}` + 9`\hat {j}` 

Thus 

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

By substituting the values of  `\vec {r₂}` and `\vec {r₁}`

`\vec {r}` = 3`\hat {i}` + 9`\hat {j}`  - (-2`\hat {i}` - 3`\hat {j}` )

`\vec {r}` = 3`\hat {i}` + 9`\hat {j}`  + 2`\hat {i}` + 3`\hat {j}` )

`\vec {r}` = 5`\hat {i}` + 12`\hat {j}`  

Here x-component = 5 and y-component = 12

The magnitude of the position vector `\vec {r}` will be its length which is the distance between the two given points. So, by using the formula

|`\vec {r}`| = `\sqrt {x^2 + y^2}`

by putting values 

r = `\sqrt {5^2 + 12^2}`

r = `\sqrt {25 + 144}`

r = `\sqrt {169}`

r = 13 units  ------Ans
Thus the distance between the two points P and Q is 13 units.

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Numerical Problem 2.1
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List of 11th Physics Numerical Problems of Chapter - 2
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