describe parallelogram law magnitude and direction of P, Q vectors lies at theta angle
Answers
Answer:
other. What happens when θ=0
0
and θ=90
0
.
Medium
ANSWER
Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude & direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure. Let θ be the angle between P and Q and R be the resultant vectors. According to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.
So, R=P+Q
From triangle OCB
OB
2
=OC
2
+BC
2
OB
2
=(OA+AC)
2
+BC
2
.........(1)
In ΔABC
cosθ=
AB
AC
AC=ABcosθ
AC=ODcosθ=Qcosθ
as [AB=OD=Q]
Also
cosθ=
AB
BC
BC=ABsinθ
BC=ODsinθ=Qsinθ
[as AB=OD=Q]
Magnitude of Resultant substituting AC & BC in equation (1) we get
R
2
=(P+Qcosθ)
2
+(Qsinθ)
2
∴R=
P
2
+2PQcosθ+Q
2
→ Magnitude
From Δ ABC
tanϕ=
OC
BC
=
OA+AC
BC
tanϕ=
P+Qcosθ
Qsinθ
∴ϕ=tan
−1
(
P+Qcosθ
Qsinθ
) .......(2)
When θ=0
o
ϕ=tan
−1
(
P+cos0
Qsin0
)
ϕ=tan
−1
(
1
0
)
ϕ=tan
−1
0
ϕ=0
When θ=90
o
ϕ=tan
−1
(
P+Qcos90
o
Qsin90
o
)
ϕ=tan
−1
(
P+0
Q
)
ϕ=tan
−1
(
P
Q
).
Explanation:
hope it helps you ✌️✌️✌️✌️✌️
Now
C
D=
A
C sinθ=
Q
sinθ
AD=
A
Ccosθ=
Q
cosθ
O
D=
O
A+
A
D=
P
+
Q
cosθ
Putting these values and representing resultant vector OC by
R
, magnitude of the resultant is given by
R
2
=(
P
+
Q
cosθ)
2
+(
Q
sinθ)
2
=
P
2
+
Q
2
+2
P
Q
cosθ
In △ OCD,
tanα=
OD
CD
=
P
+
Q
cosθ
Q
sinθ
Resultant acts in the direction making an angle α=tan
−1
(
P
+
Q
cosθ
Q
sinθ
) with direction of vector P .
