Question

The angle between vectors a=3i+4j+5k and b=6i+8j+10k

Answer 1

HELLO DEAR,

let the angle between two vector is $\Theta$

GIVEN:-

$\bold{\vec{A} = 3i + 4j + 5k}$

and,

$\bold{\vec{B} = 6i + 8j + 10k}$

now,

so, $\bold{\vec{A}.\vec{B}}$ = (3i + 4j + 5k).(6i + 8j + 10k)

$\bold{\Rightarrow \vec{A}.\vec{B}}$ = (3.6 + 4.8 + 5.10)

$\bold{\Rightarrow \vec{A}.\vec{B}}$ = (18 + 32 + 50)

$\bold{\Rightarrow \vec{A}.\vec{B}}$ = 100

and,

magnitude of $\bold{|\vec{A}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}}$

magnitude of $\bold{|\vec{B}| = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{200}}$

we know:-

$\sf{cos\Theta = \frac{\vec{A}.\vec{B}}{A.B}}$

$\bold{\Rightarrow cos\Theta = \frac{100}{\sqrt{50} \times \sqrt{200}}}$

$\bold{\Rightarrow cos\Theta = \frac{100}{\sqrt{10000}}}$

$\bold{\Rightarrow cos\Theta = \frac{100}{100}}$

$\bold{\Rightarrow cos\Theta = 1}$

$\bold{\Rightarrow cos\Theta = cos0\degree}$

$\bold{\Rightarrow \Theta = 0\degree}$

I HOPE ITS HELP YOU DEAR,

THANKS