Mathematics: Post your doubts here!

[Rutzaba]

and after so sooooooooooooooooooooooooooooo many years without practice... i still did it with out looking at the ms yaayyyy feeeling proud f my teachers ! May Allah bless them

 

[Rutzaba]

though i type slow so plz endure me patiently

 

[Thought blocker]

Hi can someone help me with question 10 part 2..

Solve it by tomorrow, sleepy right now.
Well here is the paper you wanted.

 

[Rutzaba]

first as we know seeing plain Q any point that lies on q that is the x y z of any point on the plane would also satisfy the eq of the plain
say like if it is a (x,y,z)= a(2,-3,2)
and the plane of eq is
x+by+cz=d

then it holds true that
2-3b+2z=d eq 1

same holds for b as it also lies on plane we have another eq (by substituting the x y z coords of point b into eq of plane,..)
5-2b+c=d
solve them simultaneously u will get c=b+3 this is one thing we will use.............

other thing formula of angle between two planes...
(n1,n2,n3).(n4,n5,n6) = root of squares of first bracket . root of squares of second bracket . cos angle between them
where the first bracket n1 n2 n3 represents the normal vector of plane 1
secind bracket normal of plane two

dont worry normal vector of a plane is just the coefficient of x y and z in the eq of plane
so for plane p they would be x+y=5 so 1,1,0
other plane x+by+cz=d 1,b,c

(1,b,c).(1,1,0)= root of (1^2 +b^2 +c^2) into root 2 into cos 60
where root two is 1 sqr plus one sqr plus 0 sqr

cos 60 mentioned in question
1+b = root of (1^2 +b^2 +c^2) into root 2 whole divided by two
replace c with b+3

2+2b=root of4b^2 +12b+20
(2+2b)^2 =4b^2 +12b+20
simplify this u will get b=-4
hope u can get the other values bu putting them in the equation c=b+3

 

[Thought blocker]

first as we know seeing plain Q any point that lies on q that is the x y z of any point on the plane would also satisfy the eq of the plain
say like if it is a (x,y,z)= a(2,-3,2)
and the plane of eq is
x+by+cz=d

then it holds true that
2-3b+2z=d eq 1

emma914 if you don't get, I will send detailed explanation b tomorrow.

 

[Emma23]

emma914 if you don't get, I will send detailed explanation b tomorrow.

No I got it..thank you so much

 

[Emma23]

first as we know seeing plain Q any point that lies on q that is the x y z of any point on the plane would also satisfy the eq of the plain
say like if it is a (x,y,z)= a(2,-3,2)
and the plane of eq is
x+by+cz=d

then it holds true that
2-3b+2z=d eq 1

same holds for b as it also lies on plane we have another eq (by substituting the x y z coords of point b into eq of plane,..)
5-2b+c=d
solve them simultaneously u will get c=b+3 this is one thing we will use.............

other thing formula of angle between two planes...
(n1,n2,n3).(n4,n5,n6) = root of squares of first bracket . root of squares of second bracket . cos angle between them
where the first bracket n1 n2 n3 represents the normal vector of plane 1
secind bracket normal of plane two

dont worry normal vector of a plane is just the coefficient of x y and z in the eq of plane
so for plane p they would be x+y=5 so 1,1,0
other plane x+by+cz=d 1,b,c

(1,b,c).(1,1,0)= root of (1^2 +b^2 +c^2) into root 2 into cos 60
where root two is 1 sqr plus one sqr plus 0 sqr

cos 60 mentioned in question
1+b = root of (1^2 +b^2 +c^2) into root 2 whole divided by two
replace c with b+3

2+2b=root of4b^2 +12b+20
(2+2b)^2 =4b^2 +12b+20
simplify this u will get b=-4
hope u can get the other values bu putting them in the equation c=b+3

Thank you so much for this!! so much

 

[Lily9605]

Hi , I've a doubt in question 10 part 3.
Why do we have to use sine of the angle and not cosine?

 

[Thought blocker]

Hi can someone help me with question 10 part 2..

Rutzaba I know you explained it well. But as its new for me, I wana post it as well.
10)
ii)
Plane q contains A and B and intesects P at a 60 degree angle

x + by + cz = d

(1,b,c) &sdot (3,1,-1) = 0
the line containing the points A and B is in the plane. Which means that its dot product with the normal vector = 0
3 + b - c = 0
c = (3+b)

(1,b,c)⋅(1,1,0) = ||1,b,c||||1,1,0|| cos 60
1+ b = (1^2 + b^2 + c^2)^1/2(2^1/2) * 0.5
2*(1+b)^2 = 1 + b^2 + c^2
2*(1+b)^2 = 1 + b^2 + (3+b)^2
2 + 4b + 2b^2 = 1 + b^2 + 9 + 6b + b^2
-8 = 2b

b = -4
c = -1

d = (1,-4,-1)⋅A
d = (1,-4,-1)⋅(2,-3,2)
d = 12
q: x - 4y - z = 12

 

[shaminou]

Thanks

 

[psychiatrist]

Hi can someone help me with question 10 part 2..

This is how The Godfather taught me.
Given:
.. plane p has equation x+y = 5
.. plane q has equation x+by+cz = d
.. plane p intersect plane q at an angle of 60°
.. plane q contains the points A=(2, -3, 2) and B=(5, -2, 1)

Find
.. the values of b, c, and d, hence the equation of plane q

Solution
We know that the equation of plane q will be satisfied at points A and B, so we can write equations for this:
.. 2 - 3b + 2c = d … q equation for point A
.. 5 - 2b + c = d … q equation for point B
Equating these, we get
.. 2 - 3b + 2c = 5 - 2b + c
.. c = b + 3 … add 3b-c-2 to both sides

The dot product of the normal vectors of the two planes will equal the product of the cosine of the angle between them and the magnitudes of the two vectors.
.. (i + j)•(i + bj + ck) = |i + j| * |i + bj + ck| * cos(60°)
.. 1 + b = (√2)(√(1+b^2+c^2))(1/2) … evaluate
.. (1 + b)^2 = (1/2)(1 + b^2 + (b+3)^2) … square both sides and use c=b+3
.. 0 = b + 4 … subtract the left side and simplify
.. -4 = b … subtract 4 to get the value of b.
.. c = b + 3 = -4 + 3 = -1 … use the equation above to find c.

Now, we know b = -4 and c = -1. We can compute d from the equation of plane q at point B
.. 5 - 2(-4) + (-1) = d = 12

(b, c, d) = (-4, -1, 12)
The equation of plane q is
.. x - 4y - z = 12

 

[Thought blocker]

Thanks

Anytime.

 

[Emma23]

This is how The Godfather taught me.
Given:
.. plane p has equation x+y = 5
.. plane q has equation x+by+cz = d
.. plane p intersect plane q at an angle of 60°
.. plane q contains the points A=(2, -3, 2) and B=(5, -2, 1)

Find
.. the values of b, c, and d, hence the equation of plane q

Solution
We know that the equation of plane q will be satisfied at points A and B, so we can write equations for this:
.. 2 - 3b + 2c = d … q equation for point A
.. 5 - 2b + c = d … q equation for point B
Equating these, we get
.. 2 - 3b + 2c = 5 - 2b + c
.. c = b + 3 … add 3b-c-2 to both sides

The dot product of the normal vectors of the two planes will equal the product of the cosine of the angle between them and the magnitudes of the two vectors.
.. (i + j)•(i + bj + ck) = |i + j| * |i + bj + ck| * cos(60°)
.. 1 + b = (√2)(√(1+b^2+c^2))(1/2) … evaluate
.. (1 + b)^2 = (1/2)(1 + b^2 + (b+3)^2) … square both sides and use c=b+3
.. 0 = b + 4 … subtract the left side and simplify
.. -4 = b … subtract 4 to get the value of b.
.. c = b + 3 = -4 + 3 = -1 … use the equation above to find c.

Now, we know b = -4 and c = -1. We can compute d from the equation of plane q at point B
.. 5 - 2(-4) + (-1) = d = 12

(b, c, d) = (-4, -1, 12)
The equation of plane q is
.. x - 4y - z = 12

Thank you!

 

[Lily9605]

could someone please explain question 6 part 1.

 

[Thought blocker]

could someone please explain question 6 part 1.

 

[Lily9605]

Thank you soo much !

 

[Lily9605]

I also have a doubt in question 10. I know how to seperate the variables but the mark scheme says to use
Use 2 cos^2 (2x) = 1+ cos 2x or equivalent but where does the 2cos^2(2x) come from?

 

[Rutzaba]

Thank you so much for this!! so much

welkum swthrt

 

[Thought blocker]

I also have a doubt in question 10. I know how to seperate the variables but the mark scheme says to use
Use 2 cos^2 (2x) = 1+ cos 2x or equivalent but where does the 2cos^2(2x) come from?

Rutzaba I have to reached to this level. Can you help her out?

 

[Rutzaba]

I also have a doubt in question 10. I know how to seperate the variables but the mark scheme says to use
Use 2 cos^2 (2x) = 1+ cos 2x or equivalent but where does the 2cos^2(2x) come from?

from 4 cos sqr x
U kno cos sqr x= (1+ cos2x)/2
4(1+ cos2x)/2
2(1+ cos2x) this is wat they are saying