Question

The magnitude of a given vector with end points (4, – 4, 0) and (– 2, – 2, 0) must be​

Answer 1

Answer:

Given:

End points of given vector are :

• (4 , -4 , 0)
• (-2 , -2 , 0)

To find:

Magnitude of the vector

Concept:

First let's place these points in 3D coordinate geometrical graphs . Then vector can be created by observing each component ( X , Y and X component)

Calculation:

Position vector of 1st point :

$= 4 \: \hat i - 4 \: \hat j + 0 \: \hat k$

Position vector of 2nd point :

$= - 2 \: \hat i - 2 \: \hat j + 0 \: \hat k$

So vector be r , it will be difference in position vector of the 2 points .

$\therefore \: \vec r = \{ - 2 - 4 \}\: \hat i + \{- 2 - ( - 4) \}\: \hat j$

$= > \vec r = - 6 \: \hat i + 2 \: \hat j$

So magnitude of this vector shall be :

$| \vec r| = \sqrt{ {6}^{2} + {2}^{2} } = \sqrt{40} = 6.32 \: units$

Answer 2

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QUESTION -

The magnitude of a given vector with end points (4, – 4, 0) and (– 2, – 2, 0) must be -

SOLUTION -

The end points the given vector are ( 4, -4, 0 ) and ( -2, -2 , 0 )

This can be expressed as -

$\vec{A} = 4 \hat{ I } - 4 \hat { j } + 0 \hat { k }$

$\vec{B} = -2 \hat{ I } - 2 \hat{ j } + 0 \hat{ k }$

Net Vector -

$\vec{ B - A } = -6 \hat { I } - 2 \hat { J }$

Magnitude = √ [ { Coefficient I } ^ 2 + { Coefficient j } ^ 2 + { coefficient k } ^ 2 ]

=>

$Magnitude = \sqrt{ [ { -6 } ^ 2 + { -2 } ^ 2] } = \sqrt{ 40 } = 2 \sqrt{ 10 }$