Question

A question in vectors

Answer 1

If A and B are two non-zero vectors such that
\[\left| A+B \right|=\frac{ \left| A-B \right| }{ 2 }\]
and
\[\left| A \right|=2\left| B \right|\]
then the angle between A and B is

Answer 2

Let's do it analytically first , sure i would love the graphical appraoch

Answer 3

@ikram002p @ganeshie8 r u there

Answer 4

\[\left| A+B \right|=\frac{ \left| A-B \right| }{ 2 }\]

\[2\left| A+B \right|= \left| A-B \right| \]

Answer 5

square both sides

Answer 6

\[4(A^2 + B^2 + 2A.B)= A^2 + B^2 - 2A.B \]

Answer 7

3a^2+3b^2+10ab=0

Answer 8

\[3(A^2 + B^2 )+ 10A.B= 0 \]

Answer 9

since \(A^2 = |A|^2\),


\[3(|A|^2 + |B|^2 )+ 10A.B= 0 \]

Answer 10

substitute the given value : |A| = 2|B| above ^

Answer 11

Was just doing that wait

Answer 12

\[3((2|B|)^2 + |B|^2 )+ 10A.B= 0 \]

Answer 13

\[15|B|^2+ 10A.B= 0 \]

Answer 14

expand the dot product A.B also

Answer 15

\[15|B|^2 + 10|A| |B|\cos \theta = 0 \]

Answer 16

substitute |A| again ^

Answer 17

mod B^2 is same as B^2 right?

15B^2 =-20B^2Costheta
15/20 = cos theta
cos^-1(3/4)=theta

Answer 18

yep \(\large \vec{B}^2 = \vec{B} \bullet \vec{B} = |\vec{B}| |\vec{B}| \cos (0) = |\vec{B}| |\vec{B} | = |\vec{B}|^2\)

Answer 19

and angle should be

cos^-1(\(\color{red}{-}\)3/4)=theta

right ?

Answer 20

nice