Question

what is the dot product of two vectors of magnitude 3and 5 if angle between them is 60degree
​

Answer 1

Given :

• Magnitude of two vectors are 3 and 5
• Angle between the two vectors = 60°

To find :

• The Dot product of the two vectors

Solution :

Let the two vectors be $\sf{\overrightarrow{A}}$ , $\sf{\overrightarrow{B}}$ and angle between them as θ

The angle bewtween the two vectors is given by ,

$\star \: {\boxed{ \sf{ \purple{ \cos( \theta) = \frac{ \overrightarrow{a}. \overrightarrow{b}}{ | \overrightarrow{a}| | \overrightarrow{b}| } }}}}$

Where ,

• θ is the angle between them
• $\sf{{ | \overrightarrow{a}| | \overrightarrow{b}| }}$ is the product of their magnitudes
• $\sf{\overrightarrow{a}. \overrightarrow{b}}$ is the dot product of the two vectors

We have ,

• θ = 60°
• $\sf{{ | \overrightarrow{a}| }}$ = 3
• ${| \overrightarrow{b}| }$ = 5

By substituting the values ,

$: \implies \sf\cos(60) = \dfrac{\overrightarrow{a}. \overrightarrow{b}}{(3)(5)} \\ \\ : \implies \sf \: \cos(60) = \frac{\overrightarrow{a}. \overrightarrow{b}}{15} \\ \\ : \implies \sf \: \frac{1}{2} = \frac{\overrightarrow{a}. \overrightarrow{b}}{15} \\ \\ : \implies \sf \: 15 = 2(\overrightarrow{a}. \overrightarrow{b}) \\ \\ : \implies \sf \: \frac{15}{2} = \overrightarrow{a}. \overrightarrow{b} \\ \\ : \implies \sf \: 7.5 = \overrightarrow{a}. \overrightarrow{b} \\ \\ : \implies { \underline{\boxed { \sf {\: \overrightarrow{a}. \overrightarrow{b} = 7.5}}}}$

Hence , The dot product of the two given vectors is 7.5

Answer 2

Answer:

Given :

Magnitude of two vectors are 3 and 5

Angle between the two vectors = 60°

To find :

The Dot product of the two vectors

Solution :

Let the two vectors be \sf{\overrightarrow{A}}

A

, \sf{\overrightarrow{B}}

B

and angle between them as θ

The angle bewtween the two vectors is given by ,

\star \: {\boxed{ \sf{ \purple{ \cos( \theta) = \frac{ \overrightarrow{a}. \overrightarrow{b}}{ | \overrightarrow{a}| | \overrightarrow{b}| } }}}}⋆

cos(θ)=

∣

a

∣∣

b

∣

a

.

b

Where ,

θ is the angle between them

\sf{{ | \overrightarrow{a}| | \overrightarrow{b}| }}∣

a

∣∣

b

∣ is the product of their magnitudes

\sf{\overrightarrow{a}. \overrightarrow{b}}

a

.

b

is the dot product of the two vectors

We have ,

θ = 60°

\sf{{ | \overrightarrow{a}| }}∣

a

∣ = 3

{| \overrightarrow{b}| }∣

b

∣ = 5

By substituting the values ,

\begin{gathered} : \implies \sf\cos(60) = \dfrac{\overrightarrow{a}. \overrightarrow{b}}{(3)(5)} \\ \\ : \implies \sf \: \cos(60) = \frac{\overrightarrow{a}. \overrightarrow{b}}{15} \\ \\ : \implies \sf \: \frac{1}{2} = \frac{\overrightarrow{a}. \overrightarrow{b}}{15} \\ \\ : \implies \sf \: 15 = 2(\overrightarrow{a}. \overrightarrow{b}) \\ \\ : \implies \sf \: \frac{15}{2} = \overrightarrow{a}. \overrightarrow{b} \\ \\ : \implies \sf \: 7.5 = \overrightarrow{a}. \overrightarrow{b} \\ \\ : \implies { \underline{\boxed { \sf {\: \overrightarrow{a}. \overrightarrow{b} = 7.5}}}}\end{gathered}

:⟹cos(60)=

(3)(5)

a

.

b

:⟹cos(60)=

15

a

.

b

:⟹

2

1

=

15

a

.

b

:⟹15=2(

a

.

b

)

:⟹

2

15

=

a

.

b

:⟹7.5=

a

.

b

:⟹

a

.

b

=7.5

Hence , The dot product of the two given vectors is 7.5