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Answers
Answer:
3.
Correct option is
B
6π
We have a×(a×c)+b=0
⇒(a.c)a−(a.a)c=−b
⇒c−(a.c)a=b(∵a.a=∣a∣2=1)
⇒∣c−(a.c)a∣2=∣b∣2
⇒∣c∣2+∣a.c∣2∣a∣2−2(a.c)(a.c)=∣b∣2
If θ is the angle between a and c, we get
4+(2cosθ)2−2(2cosθ)2=1
⇒4cos2θ=3⇒cosθ=23 (∵θ is acute )
⇒θ=6π
4.
No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.
5.
No, three non-coplanar vectors cannot ad up to given zero resultant because for non-coplanar vectors the resultant of the two vectors will not lie in the plane of third vector , and so the resultant cannot cancel the third vector to given null vector .
6.
Here A= 30N , B=60N , C=90^°
Let D be the angle between 2 forces. Then tan C =BsinD/A+B cosD
or tan 90^°=60sinD/30+60cosD
or 30+60cosD =0
or cosD=-30/60 =-1/2 =cos120^°
or D=120^°
Effective pulling force is
R= root A^2+B^2+2ABcosD
=root 900+3600+2×30×60 (-1/2)
=30root3N
7.
The resultant of two vectors P and Q is R. If the vector Q is reversed, then the resultant becomes S, then choose the correct option.
A
R2+S2=2(P2−Q2)
B
R2+S2=2(P2+Q2)
C
R2+S2=(P2−Q2)
D
R2−S2=2(P2+Q2)
8.
Let the line x + y = b bisect the chord at P (a, b-a)
Let the line x + y = b bisect the chord at P (a, b-a)∴ equation of chord whose mid-point is P (a,b-a)
Let the line x + y = b bisect the chord at P (a, b-a)∴ equation of chord whose mid-point is P (a,b-a)2a2xa−2b2y(b−a)=2a2a2−2b2(b−a)2
Let the line x + y = b bisect the chord at P (a, b-a)∴ equation of chord whose mid-point is P (a,b-a)2a2xa−2b2y(b−a)=2a2a2−2b2(b−a)2Since it passes through (a,b)
Let the line x + y = b bisect the chord at P (a, b-a)∴ equation of chord whose mid-point is P (a,b-a)2a2xa−2b2y(b−a)=2a2a2−2b2(b−a)2Since it passes through (a,b)∴2aa−2b(b−a)=2a2a2−2b2(b−a)2a2(a21−b21)+a(b1−a1)=0a0,a−a1+b11
9.
∣P^∣=∣Q^∣=1, because they are unit vectors.
∣P^∣=∣Q^∣=1, because they are unit vectors.LHS=∣P^−Q^∣
∣P^∣=∣Q^∣=1, because they are unit vectors.LHS=∣P^−Q^∣=(P^−Q^).(P^−Q^)=(P2+Q2−2P^.Q^)
∣P^∣=∣Q^∣=1, because they are unit vectors.LHS=∣P^−Q^∣=(P^−Q^).(P^−Q^)=(P2+Q2−2P^.Q^)=12+12−2PQcosθ=2−2×1×1cosθ
∣P^∣=∣Q^∣=1, because they are unit vectors.LHS=∣P^−Q^∣=(P^−Q^).(P^−Q^)=(P2+Q2−2P^.Q^)=12+12−2PQcosθ=2−2×1×1cosθ=2(1−cosθ)
∣P^∣=∣Q^∣=1, because they are unit vectors.LHS=∣P^−Q^∣=(P^−Q^).(P^−Q^)=(P2+Q2−2P^.Q^)=12+12−2PQcosθ=2−2×1×1cosθ=2(1−cosθ)=2[2sin2(θ/2)]=2sin(θ/2)=RHS
10.
From polygon law, three vectors having summation zero should form, a closed polygon(triangle). Since, the two vectors are having same magnitude and the third vector is 2 times that of either of two having equal magnitude, i.e., the triangle should be right angled triangle.
Angle between A and B, α=90o
Angle between B and C, β=135o
Angle between A and C, γ=135o.
11.
Here,
vc=Velocity of the cyclist
vr=Velocity of falling rain
In order to protect herself from the rain, the women must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the women
v=vr+(−vc)
v=30+(−10)=20 m/s
tanθ=vrvc=3010
θ=tan−1(31)
θ≈18°
Hence the women must hold the umbrella towards the south, at angle of nearly 18° with the vertical
12.
We draw the graphs of the lines x + 2y = 8 and 2x + y = 8 The inequality (1) and (2) represent the region below the two lines including the point on the respective lines
Since x≥0,y≥0 every point in the shaded region in the first quadrant represent a solution of the given system of inequalities
13.
Let the angle between the swimmer and the perpendicular to the flow of river is θ with the perpendicular to the flow of river and time to cross the river is t
(i)
The direction is given as,
sinθ=105
θ=30∘
(ii)
The resultant velocity is given as,
V=Vsgcos30∘
=10cos30∘
=8.66km/h
(iii)
Displacement in vertical direction is given as,
(Vsgcosθ)t=800×10−3
(10cos30∘)t=800×10−3
t=0.092s
t=5.5min
14.
vx=0.4 m/s
vy=0 m/s
ay=0.15 m/s2,t=2 sec
(vy)final=vy+ayt=0+(0.15)×2=0.3m/s
So, v=vxi^+vyj^=0.4i^+0.3j^