Question

A question on vectors
@ganeshie8

Answer 1

If A=i+j+k
B= -3i +4j -2k

Find the angle between this vectors

Answer 2

\(\large \vec{A} .\vec{ B} = |\vec{A}| |\vec{B}| \cos (\theta)\)

seen this formula for dot product before ?

Answer 3

Is it applicable for three directions

Answer 4

Well , then how to handle the L.H.S
R.H.S is trivial

Answer 5

thats a good question ! angle and lengths will remain same irrespective of whether you're working them in 2D or 3D

Answer 6

for LHS we use algebra definition of dot product :

\[\large \langle a_1, a_2, a_3\rangle\bullet \langle \color{red}{b_1, b_2, b_3}\rangle = a_1\color{red}{b_1} + a_2\color{red}{b_2} + a_3\color{red}{b_3}\]

Answer 7

multiply component by component
add them

Answer 8

If A=i+j+k = \(\langle 1, 1, 1\rangle \)
B= -3i +4j -2k = \(\langle -3, 4, -2\rangle \)

Answer 9

\[\large \vec{A} \bullet \vec{ B} = |\vec{A}| |\vec{B}| \cos (\theta)\]

Answer 10

\[\large \langle 1,1,1 \rangle \bullet \langle -3,4,-2 \rangle = | \langle 1,1,1 \rangle|~ |\langle -3,4,-2 \rangle | \cos (\theta)\]

Answer 11

evaluate the LHS..

Answer 12

-1

Answer 13

yep !

\[\large \langle 1,1,1 \rangle \bullet \langle -3,4,-2 \rangle = | \langle 1,1,1 \rangle|~ |\langle -3,4,-2 \rangle | \cos (\theta)\]

\[\large -1 = | \langle 1,1,1 \rangle|~ |\langle -3,4,-2 \rangle | \cos (\theta)\]

Answer 14

what about the RHS