if a vector +b vector =c vector and a square +b square =c square then angle bet a vector and b vector is how much?
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Answer:
Explanation:
May 30, 2017 · Author has 60 answers and 132.9K answer views
a⃗ +b⃗ =c⃗ ............1
|a⃗ |+|b⃗ |=|c⃗ |.........2
(a⃗ +b⃗ ).(a⃗ +b⃗ )=c⃗ .c⃗ from equation 1
|a⃗ |2+|b⃗ |2+2(a⃗ .b⃗ )=|c⃗ |2
|a⃗ |2+|b⃗ |2+2.|a⃗ ||b⃗ |=|c⃗ |2 from equation 2
equating last two equations
2(a⃗ .b⃗ )=2.|a⃗ ||b⃗ |
⟹|a⃗ ||b⃗ |.cosθ=|a⃗ ||b⃗ |
⟹cosθ=1
⟹θ=0
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Punyasloka Sahoo, former Summer Intern at Indian Institute of Technology, Bhubaneswar (2019)
Updated May 16, 2019 · Author has 366 answers and 513.6K answer views
Simple, the angle between the 2 vectors is 0 degree.
First a+b= c
=>(a^2+b^2+2abcos(theta))^(1/2)=| c|
where theta is the angle between a and b.
Again a+b=c
=> (a+b)^2=c^2=| c |^2=a^2+b^2+2abcos(theta)
=>a^2+b^2+2ab=a^2+b^2+2abcos(theta)
((a+b)^2=a^2+b^2+2ab)
=>2abcos(theta)=2ab
=> cos(theta)=1
=>theta=0 degree
11.8K viewsView 19 Upvoters · Answer requested by Dipesh Saili
Kumar Saurav, Math geek.
Answered June 11, 2017 · Author has 486 answers and 1.1M answer views
The magnitude of the resultant of two vectors is given by:[Math Processing Error]
\mC=\mA2+\mB2+2\mA\mBcosθ−−−−−−−−−−−−−−−−−−−−−−−−−−√
In this case,
\mC=\mA+\mB
So,
\mA+\mB=\mA2+\mB2+2\mA\mBcosθ−−−−−−−−−−−−−−−−−−−−−−−−−−√
Squaring,
That is θ=arccos1=0°
There is another case, though.
I simply cancelled 2\mA\mB , assuming it is non-zero.
If the magnitude of either is 0 , then it is a null vector and
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