A vector D makes a point of 20°, and D makes a point of 110° with the X-hub. The extents of these vectors are 3 m and 4 m individually. See the result.
The point between D and E from the x-pivot is 20° and 110° individually.
Their extents are 3 units and 4 units individually.
Subsequently the point among D and E is = 110 – 20 = 90°
Now, X2 = D2 + E2 + 2DEcosѲ
= 32 + 42 + 2.3.4 Cos(90)
Or then again, X = 5
Let ϕ be the point among X and D,
Then, at that point, tan ϕ = Esinѳ/(D+CosѲ) =4/3 , or ϕ = 53°.
The resultant makes a point of (53+20)° = 73° with the x pivot.
Allow An and B to be the two vectors of size 10 units each. On the off chance that they are leaned to the X-pivot at points 30 and 60 separately, view as the resultant.
D and E are leaned at points of 30 degrees and 60 degrees regarding the x pivot
Point between them = (60 – 30) = 90 degrees
Considering that |D| = |E| = 10 units, we get
X2 = D2 + E2 + 2DEcosѲ
= 102 + 102 + 2.10.10 Cos(30)
X = 20cos15°
Or then again, X = 19.3 units
Furthermore, tan ϕ = Bsinѳ/(D+CosѲ) , or ϕ = 15°.
Subsequently, this resultant makes a point of (15+30) = 45 degrees with the x pivot
Add vectors A,B and C each having extent of 100 unit and leaned to the X-pivot at
points 45°, 135° and 315° separately.
Vectors A, B and C are situated at 45°, 135° and 315° separately.
Let A = Ax i + Ay j +Az k, B = Bx i + By j +Bz k, and C = Cx i + Cy j +Cz k, and we can write that,
Ax = Cx =100cos(45°)=100/√2 , by considering their components
Now Ay = 100sin(45°)= 100/√2,
By = 100sin(1350)= 100/√2
Similarly, Cy = -100/√2
Net x part = 100/√2+100/√2-100/√2=100/√2
Net y part = 100/√2+100/√2-100/√2=100/√2
R2 =x2 +y2 =1002
R=100 and tan ϕ = (100/√2)/( 100/√2)=1, and ϕ = 45°
Let a=4i+3j and b=3i+4j . (a) Find the sizes of (a)a , (b) b, (c) a+b and (d) .a-b
a = 4i+3j, b = 3i+4j
a+b = 7i+7j and a-b = i-j
|a+b|= 7√2 and |a-b| = = √2
Allude to the figure. (a) Find the size, (b) x and y parts and (c) the point with the X-pivot of the resultant of OA, BC and DE.
x part of OA = 2cos30° = √3
x part of BC = 1.5 cos 120° = – 0.75
x part of DE = 1 cos 270° = 0
y part of OA = 2 sin 30° = 1
BC = 1.5, we know that sin 120° = 1.3
DE = 1, we know that sin 270° = – 1
Rx = x part of resultant = 3 – 0.75 + 0 = 0.98 m
Ry = resultant y part = 1 + 1.3 – 1 = 1.3 m
So, R = (R2x +R2y )1/2 = 1.6 m
It makes and point Θ with positive x-hub then, at that point, tan Θ = Ry/Rx = 1.32 ϴ = tan-1 (1.32)
Two vectors have extents of 3 units and 4 units individually. What should be the point between them
assuming the greatness of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.
|a| = 3 and |b| = 4
Allow Θ to be the point between them.
By, applying the relation X2 = D2 + E2 + 2DEcosѲ,
a)X = 1,
1 = 9+16+24Cos ϴ
Or, ϴ = 180°
For, X = 5, we have
25 = 9+16+24Cos ϴ
Or, cos ϴ = 0;
ϴ = 90°
c)For X =7,
49= 9+16+24Cos ϴ,
Or cos ϴ = 1,
And ϴ = 0°
A government operative report about a presumed vehicle peruses as follows. “The vehicle moved 2.00 km towards the east, made a contrary left turn, ran for 500 m, made a contrary right turn, ran for 4.00 km and ended.” Track down the dislodging of the vehicle.
Abdominal muscle = 2i + 0.5j + 4i = 6i + 0.5j
As the vehicle went ahead, took a left and afterwards a right.
So, CD = (62 +0.52 )1/2 = 6.02km
So, ϕ = tan-1 (DE\CE) = tan-1 (0.5/6) = tan-1 (1/12)
A mosquito net more than a 7 ft * 4 ft bed is 3 ft high. The net has an opening at one corner of the bed through which a mosquito enters the bed. It flies and reclines the slantingly across from the upper corner of the net.
(a)Track down the extent of the uprooting of the mosquito.
(b) Taking the opening as the beginning, the length of the bed as the X-pivot, its width as the Y-hub, and in an upward direction up as the Z-hub compose the parts of the removal vector.
The dislodging vector is given by r= 7i+4j +3k
Greatness of uprooting of the mosquitoes= (72 +42 +32 )1/2 = 74ft
The parts of relocation along with x pivot, y hub and z hub are 7 units, 4 units and 3 units individually.
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