Dot product of a vector with itself Calculate V1⋅V1. Express your answer in terms of V1. V1⋅V1=

The dot product of wo vectors V→1 and V→1 is, V→1⋅V→1=|V→1||V→1|cos⁡0∘ =V12 The dot product of single vector with itself is the square of magnitude of the vector.

Dot product of a vector with itself : V→1⋅V→1=|V→1||V→1|cos⁡(θ) But θ=0. (Angle between same vector.) ∴V→1⋅V→1=|V→1||V→1|cos⁡(0) ⇒V→1⋅V→1=|V→1|2=V12

The dot product of wo vectors V→1 and V→1 is, V→1⋅V→1=|V→1||V→1|cos⁡0∘ =V12 The dot product of single vector with itself is the square of magnitude of the vector.

Dot product of a vector with itself : V→1⋅V→1=|V→1||V→1|cos⁡(θ) But θ=0. (Angle between same vector.) ∴V→1⋅V→1=|V→1||V→1|cos⁡(0) ⇒V→1⋅V→1=|V→1|2=V12

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Painevg Answered2021-12-17

Dot product of a vector with itself Calculate

V_{1} \cdot V_{1}

. Express your answer in terms of

V_{1}

V_{1} \cdot V_{1} =

Answer & Explanation

Lynne Trussell

Expert

2021-12-18Added 32 answers

The dot product of wo vectors

{\vec{V}}_{1}

and

{\vec{V}}_{1}

is,

{\vec{V}}_{1} \cdot {\vec{V}}_{1} = | {\vec{V}}_{1} | | {\vec{V}}_{1} | {\cos 0}^{\circ}

= V_{1}^{2}

The dot product of single vector with itself is the square of magnitude of the vector.

Mollie Nash

Expert

2021-12-19Added 33 answers

More explanantion, please!

nick1337

Expert

2021-12-28Added 573 answers

Dot product of a vector with itself :
${\vec{V}}_{1} \cdot {\vec{V}}_{1} = | {\vec{V}}_{1} | | {\vec{V}}_{1} | \cos (θ)$
But $θ = 0$ . (Angle between same vector.)
$∴ {\vec{V}}_{1} \cdot {\vec{V}}_{1} = | {\vec{V}}_{1} | | {\vec{V}}_{1} | \cos (0)$
$\Rightarrow {\vec{V}}_{1} \cdot {\vec{V}}_{1} = | {\vec{V}}_{1} |^{2} = V_{1}^{2}$

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