Question

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

Answer 1

Answer:

Explanation:

magnitude=root36+4=root40

Answer 2

The Magnitude of this vector will be :

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

Explanation :

Given :

End points of given vector are :

• (4 , -4 , 0)
• (-2 , -2 , 0)

To Find :

Magnitude of the vector .

Solution :

Now,

$\textbf{Position vector of 1st point :}\\ \\ \\\implies \sf 4 \: \hat i - 4 \: \hat j + 0 \: \hat k$

$\textbf{Position vector of 2nd point :}\\ \\ \\\implies \sf - 2 \: \hat i - 2 \: \hat j + 0 \: \hat k$

Let's consider the vector as 'r',

That will be difference in position vector by the 2 points.

So,

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

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

Hence,

The magnitude of this vector :

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