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Mathematics: Post your doubts here!

N

Nushad Nahue

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M2, s04, Q4 part 1.
According to the figure I can tell that the force in the rod is 0.7m from the hinge and has to act upward for equilibrium.
But mark scheme says ,

Distance of the rod from the hinge is (0.7) 2.4/2.5 or 0.7cos16.26° (=0.672). How did come to be? Please help!

Capture.PNG
 
D

DudeWithoutName

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Holmes said:
View attachment 63626
View attachment 63627
View attachment 63628

Help me out anyone........
Zaki ali asghar
Click to expand...
Hey! First question, to find the angle between the two planes we can find the angle between the two normals. That is equal to the angle between the two planes.
The first normal is i+j+3k and the second is 2i-2j+k. Using the dot product between these two we have: (1)(2)+(1)(-2)+(3)(1)=sqrt(1^2+1^2+3^2)*sqrt(2^2+2^2+1^2)cos(theta)
theta=72.5

Second question, before i proceed, do you know how use the vector product to find common perpendiculars or do you use another method?
 
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DudeWithoutName

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Holmes said:
View attachment 63629
View attachment 63630
View attachment 63631
Can anyone tell me how to shade the region?
and how to represent Re z >2
Help me Maths nerds.

Zaki ali asghar
Click to expand...
Modulus of (z-u) being less than 2 is basically a circle with center u of radius 2 and the wanted region is whats inside the circle because its less than.
Re z>2 means real part must be bigger than 2 so the x value of the complex number on the argand diagram must be greater than 2 so draw the line x=2 and the final shaded region should be

The region inside the circle that is to the right of line x=2
 
Holmes

Holmes

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DudeWithoutName said:
Hey! First question, to find the angle between the two planes we can find the angle between the two normals. That is equal to the angle between the two planes.
The first normal is i+j+3k and the second is 2i-2j+k. Using the dot product between these two we have: (1)(2)+(1)(-2)+(3)(1)=sqrt(1^2+1^2+3^2)*sqrt(2^2+2^2+1^2)cos(theta)
theta=72.5

Second question, before i proceed, do you know how use the vector product to find common perpendiculars or do you use another method?
Click to expand...
Common perpendiculars??? finding the intersection line of these two planes(other method)
 
Holmes

Holmes

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438
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DudeWithoutName said:
Modulus of (z-u) being less than 2 is basically a circle with center u of radius 2 and the wanted region is whats inside the circle because its less than.
Re z>2 means real part must be bigger than 2 so the x value of the complex number on the argand diagram must be greater than 2 so draw the line x=2 and the final shaded region should be

The region inside the circle that is to the right of line x=2
Click to expand...
Thanks for replying thanks a lot.
 
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