find dot product of a vector dot b vector is a is equals to 3 b is equals to 2 and theta is equal to 60 degree
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Answered by
0
(
A
+2
B
).(2
A
−3
B
)
=
A
.2
A
−
A
.3
B
+2
B
.2
A
−2
B
.3
B
=2∣A∣
2
−3
A
.
B
+4
B
A−6∣B∣
2
=2∣A∣
2
+∣A∣∗∣B∣ Cos ϕ−6∣B∣
2
we use the following in the above proof:
\begin{lgathered}\vec{A}.\vec{B}=|A|*|B|* Cos\ \phi,\ where\ \phi=angle\ between\ vectors\ \vec{A}\ and\ \vec{B}\\\vec{A}.\vec{B}=\vec{B}.\vec{A}\\\vec{nA}=n*\vec{A}\end{lgathered}
A
.
B
=∣A∣∗∣B∣∗Cos ϕ, where ϕ=angle between vectors
A
and
B
A
.
B
=
B
.
A
nA
=n∗
A
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Answered by
0
a.b = |a| |b| cosx
a.b = 3.2.cos60 = 3
hope it helps...........
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