Question

let A = 2m i^ 6m j^ - 3m k^ and B = 4m i^ 2m j^ - 1m k^ . Then A.B = 1) 8m i^ +12mj^ -3m k^ 2) 12m i^ - 14m j^ -20mk^ 3)23 m² 4)17m²

I WILL MARK HIM/HER AS BRAINLIEST!!!!!!!!

Answer 1

Answer -

Given -

$\longrightarrow$$\rm\vec{A} = 2m \hat{\imath} + 6m \hat{\jmath} - 3m\hat{k}$

$\longrightarrow$$\rm\vec{B} = 4m \hat{\imath} + 2m \hat{\jmath} - m\hat{k}$

━━━━━━━━━━━━━

To find -

Scalar product of $\rm\vec{A}$ and $\rm\vec{B}$

$= \rm\vec{A} .\rm\vec{B}$

━━━━━━━━━━━━━

Formula used -

For two vectors

$\longrightarrow$$A = \rm\vec{A} = a_x\hat{\imath} + a_y\hat{\jmath} - a_z\hat{k}$

$\longrightarrow$$B = \rm\vec{B} = b_x\hat{\imath} + b_y\hat{\jmath} - b_z\hat{k}$

$\implies$$\boxed{\rm\vec{A} .\rm\vec{B} = \rm{a_x b_x + a_y b_y + a_z b_z }}$

━━━━━━━━━━━━━

Solution -

$\rm\vec{A} = 2m \hat{\imath} + 6m \hat{\jmath} - 3m\hat{k}$

$\rm\vec{B} = 4m \hat{\imath} + 2m \hat{\jmath} - m\hat{k}$

━━━━━━━━━━━━━

$\implies$$\rm\vec{A} .\rm\vec{B} = \tt(2m \times 4m) +( 6m \times 2m)+ ((- 3)m \times( - 1m))$

$\implies$$\rm\vec{A} .\vec{B} = (8 + 12 + 3)m^2 = 23m^2$

ADDITIONAL INFORMATION -

$\rm{\vec{A} .\vec{B} }= | \vec{A}||\vec{B}| \cos \theta$

where -

$\longrightarrow$$| \vec{A}|$$\implies$ magnitude of $\vec{A}$

$\longrightarrow$$| \vec{B}|$$\implies$ magnitude of $\vec{B}$

$\longrightarrow$$\theta$$\implies$ angle between $\vec{A}$ and $\vec{B}$

━━━━━━━━━━━━━

If $\vec{A}$ and $\vec{B}$ are perpendicular to each other then

$\rm{\vec{A} .\vec{B} }=0$

because cos90° = 0

━━━━━━━━━━━━━

Thanks