Question

two vectors one pointing towards the north and other pointing southwest have the same magnitude of 8 units, find the difference of two vectors

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Answer 1

Given:

Two vectors one pointing towards the north and other pointing southwest have the same magnitude of 8 units.

To find:

Difference of two vectors.

Calculation:

Let vector A have magnitude of 8 units directed towards North.

Let vector B have magnitude of 1 units directed towards South-west.

• Hence angle between the vectors will be 135°.

So, we can say :

$| \vec{A} - \vec{B}| = \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos( {180}^{ \circ} - \theta ) }$

$= > | \vec{A} - \vec{B}| = \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos( {180}^{ \circ} - {135}^{ \circ} ) }$

$= > | \vec{A} - \vec{B}| = \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos( {45}^{ \circ} )}$

$= > | \vec{A} - \vec{B}| = \sqrt{ {(8)}^{2} + {(8)}^{2} + 2(8)(8) \cos( {45}^{ \circ} )}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 2{(8)}^{2} + 2 {(8)}^{2} \cos( {45}^{ \circ} )}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 128 + 128\cos( {45}^{ \circ} )}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 128 \{1 + \cos( {45}^{ \circ} ) \}}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 128 \{1 + \dfrac{1}{ \sqrt{2} } \}}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 128 \{1 + 0.707 \}}$

$= > | \vec{A} - \vec{B}| = \sqrt{ 128 \{1.707 \}}$

$= > | \vec{A} - \vec{B}| = \sqrt{90.52}$

$= > | \vec{A} - \vec{B}| = 9.51 \: units$

So, final answer is:

$\boxed{ \red{ \bold{ | \vec{A} - \vec{B}| = 9.51 \: units}}}$