Numerical Problem No. 2.1 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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 =

\to r ₁

$\vec {r₁}$ = -2

ˆ i

$\hat {i}$ - 3

ˆ j

$\hat {j}$

Position vector of point Q = `\vec {r₂}` = 3

ˆ i

$\hat {i}$ + 9

ˆ j

$\hat {j}$

Thus

\to r

$\vec {r}$ =

\to r ₂

$\vec {r₂}$ - `\vec {r₁}`

By substituting the values of

\to r ₂

$\vec {r₂}$ and

\to r ₁

$\vec {r₁}$

\to r

$\vec {r}$ = 3

ˆ i

$\hat {i}$ + 9

ˆ j

$\hat {j}$ - (-2

ˆ i

$\hat {i}$ - 3

ˆ j

$\hat {j}$ )

\to r

$\vec {r}$ = 3

ˆ i

$\hat {i}$ + 9

ˆ j

$\hat {j}$ + 2

ˆ i

$\hat {i}$ + 3

ˆ j

$\hat {j}$ )

\to r

$\vec {r}$ = 5

ˆ i

$\hat {i}$ + 12

ˆ j

$\hat {j}$

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

The magnitude of the position vector

\to r

$\vec {r}$ will be its length which is the distance between the two given points. So, by using the formula

\to r

$\vec {r}$ | =

\sqrt{x^{2} + y^{2}}

$\sqrt {x^2 + y^2}$

by putting values

r =

\sqrt{5^{2} + 12^{2}}

$\sqrt {5^2 + 12^2}$

r =

\sqrt{25 + 144}

$\sqrt {25 + 144}$

r =

\sqrt{169}

$\sqrt {169}$

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
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 1.10 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.2

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