chapter 08: calculus with analytics geometryย ยท mathematics calculus with analytic geometry by sm....
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 1
In each of ๐๐ซ๐จ๐๐ฅ๐๐ฆ ๐ โ ๐, find parametric equations, direction ratios, direction cosines and measures of the direction angles of the straight line through P and Q:
Q#1: ๐ท ๐, โ๐, ๐ , ๐ธ(๐, โ๐๐, ๐)
Solution:
Given points are ๐ 1, โ2,0 ๐๐๐ ๐(5, โ10,1)
Let ๐ (๐ฅ, ๐ฆ, ๐ง) is another point on the straight line then equation of straight line will be written as
๐ฅ โ1
5โ1=
๐ฆ + 2
โ10+2=
๐งโ0
1โ0โน
๐ฅ โ1
4=
๐ฆ + 2
โ8=
๐ง
1 = ๐ก
Parametric equations of given straight line can be written as
๐ฅโ1
4= ๐ก โน ๐ฅ = 1 + 4๐ก
๐ฆ + 2
โ8= ๐ก โน ๐ฆ = โ2 โ 8๐ก
๐ง
1 = ๐ก โน ๐ง = 0 + 1๐ก
Direction ratios of straight line PQ are 4, โ8,1.
Let ๐ is the direction vector , ๐ = 4๐ โ 8๐ + ๐ โน ๐ = 42 + โ8 2 + 12 = 81 = 9
Direction cosines of line through P & Q are
cos ๐ผ =4
9 , cos ๐ฝ =
โ8
9 , cos ๐พ =
1
9
Directional angles of line PQ are
๐ผ = cosโ14
9 , ๐ฝ = cosโ1
โ8
9 , ๐พ = cosโ1
1
9
โน ๐ผ = 63๐37โฒ , ๐ฝ = 152๐44โฒ , ๐พ = 83๐37โฒ
Q#2: ๐ ๐, ๐, โ๐ , ๐(๐, ๐, ๐)
Now do yourself as above
Q#3: ๐ ๐, โ๐, ๐ , ๐(๐, โ๐, ๐)
Do yourself as above
Exercise #8.2
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 2
Q#4: ๐ ๐, ๐, ๐ , ๐(๐, โ๐, ๐๐)
Solution:
Given points are ๐ 3,5,7 & ๐(6, โ8,10)
Let ๐ (๐ฅ, ๐ฆ, ๐ง) is another point on the straight line then equation of straight line will be written as
๐ฅ โ 3
6 โ 3=
๐ฆ โ 5
โ8 โ 5=
๐ง โ 7
10 โ 7
โน ๐ฅ โ 3
3=
๐ฆ โ 5
โ13=
๐ง โ 7
3 = ๐ก
Parametric equations of given straight line can be written as
๐ฅ โ 3
3= ๐ก โน ๐ฅ = 3 + 3๐ก
๐ฆ โ 5
โ13= ๐ก โน ๐ฆ = 5 โ 13๐ก
๐ง โ 7
3= ๐ก โน ๐ง = 7 + 3๐ก
Direction ratios of straight line PQ are 3, โ13,3.
Let ๐ is the direction vector
๐ = 3๐ โ 13๐ + 3๐ โน ๐ = 32 + โ13 2 + 32 = 187
Direction cosines of line through P & Q are
cos ๐ผ =3
187 , cos ๐ฝ =
โ13
187 , cos ๐พ =
3
187
Directional angles of line PQ are
๐ผ = cosโ13
187 , ๐ฝ = cosโ1
โ13
187 , ๐พ = cosโ1
3
187
โน ๐ผ = 77๐19โฒ , ๐ฝ = 159๐19โฒ , ๐พ = 77๐19โฒ
Q#5: Find the direction cosines the coordinate axis.
Solution:
We want to find the direction cosine of x-axis , y-axis and z- axis.
(I) Let x-axis makes angles 0๐ , 90๐ , 90๐ with x-axis ,y-axis and z- axis
So direction cosines of x-axis are
cos 0๐ = 1 , cos 90๐ = 0 , cos 90๐ = 0
(II) Let y-axis makes angles 90๐ , 0๐ , 90๐ with x-axis ,y-axis and z- axis
So direction cosines of y-axis are
cos 90๐ = 0 , cos 0๐ = 1 , cos 90๐ = 0
(III) Let z-axis makes angles 90๐ , 90๐ , 0๐ with x-axis ,y-axis and z- axis
So direction cosines of z-axis are
cos 90๐ = 0, cos 90๐ = 0 , cos 0๐ = 1
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 3
Q#6: Prove that if measures of the direction angles of a straight line are ๐, ๐ and ๐, then ๐ฌ๐ข๐ง๐ ๐ + ๐ฌ๐ข๐ง๐ ๐ + ๐ฌ๐ข๐ง๐ ๐ = ๐. Solution: Let ๐ผ, ๐ฝ and ๐พ be the directional angles of a straight line then direction cosines of line are cos ๐ผ , cos ๐ฝ & cos ๐พ.
As we know that
cos2 ๐ผ + cos2 ๐ฝ + cos2 ๐พ = 1
1 โ sin2 ๐ผ + 1 โ sin2 ๐ฝ + 1 โ sin2 ๐พ = 1
3 โ sin2 ๐ผ + sin2 ๐ฝ + sin2 ๐พ = 1
3 โ 1 = sin2 ๐ผ + sin2 ๐ฝ + sin2 ๐พ
โน sin2 ๐ผ + sin2 ๐ฝ + sin2 ๐พ = 2 Hence proved.
Q#7: If measures of two of the direction angles of a straight line are ๐๐๐จ and ๐๐๐จ, find measure of the third direction angle.
Solution: Let ๐ผ, ๐ฝ and ๐พ be the directional angles of a straight line ๐ผ = 45๐ , ๐ฝ = 60๐ , ๐พ =?
As we know that
cos2 ๐ผ + cos2 ๐ฝ + cos2 ๐พ = 1
cos2 45๐ + cos2 60๐ + cos2 ๐พ = 1
1
2
2+
1
2
2+ cos2 ๐พ = 1 โน
1
2+
1
4+ cos2 ๐พ = 1
cos2 ๐พ = 1 โ3
4 โน cos2 ๐พ =
1
4 โน cos ๐พ =
1
4 โน cos ๐พ =
1
2 โน ๐พ = cosโ1(
1
2 )
โน ๐พ = 60๐ is required direction angle.
Q#8: The direction cosines ๐, ๐, ๐ of two straight lines are given by the equations ๐ + ๐ + ๐ = ๐ and ๐๐ + ๐๐ โ ๐๐ = ๐. Find measure of the angle between them. Solution: We have to find the angle between two straight lines.
Let ๐1 , ๐1 , ๐1 be the direction cosines of line L and ๐2 , ๐2 , ๐2 be the direction cosines of line M.
Let ๐ be the angle between line L and M then it is described as
cos ๐ = ๐1๐2 + ๐1๐2 + ๐1๐2 -------- (A)
๐ + ๐ + ๐ = 0 -------------------------------- (1)
๐2 + ๐2 โ ๐2 = 0 ------------------------------- (2)
1 โน ๐ = โ(๐ + ๐) --------------------- (3)
Putting in eq.(2)
๐2 + ๐2 โ โ(๐ + ๐) 2 = 0
๐2 + ๐2 โ ๐2 + ๐2 + 2๐๐ = 0
๐2 + ๐2 โ ๐2 โ ๐2 โ 2๐๐ = 0
โ2๐๐ = 0 โน ๐๐ = 0
Either ๐ = 0 or ๐ = 0
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 4
Put ๐ = 0 in eq. (3) โน ๐ = โ๐
โน๐
1=
๐
โ1
๐
0=
๐
1=
๐
โ1 โน
๐2 + ๐2 + ๐2
02 + 12 + (โ1)2=
1
2
So direction cosines of line L are
๐1 = 0 , ๐1 =โ1
2 , ๐1 =
1
2
Put ๐ = 0 in eq. (3) โน ๐ = โ๐
โน๐
1=
๐
โ1
๐
โ1=
๐
1=
๐
0 โน
๐2 + ๐2 + ๐2
โ1 2 + 12 + 02=
1
2
So direction cosines of line M are
๐2 =โ1
2 , ๐2 = 0 , ๐2 =
1
2
Using in equation (A)
cos ๐ = 0 โ1
2 +
โ1
2 0 +
1
2
1
2 โน cos ๐ = 0 + 0 +
1
2 โน cos ๐ =
1
2 โน ๐ = cosโ1
1
2 โน ๐ =
๐
3
Q#9: The direction cosines ๐, ๐, ๐ of two straight lines are given by the equations ๐ + ๐ + ๐ = ๐ and ๐๐๐ + ๐๐๐ โ ๐๐ = ๐. Find measure of the angle between them. Solution: We have to find the angle between two straight lines.
Let ๐1 , ๐1 , ๐1 be the direction cosines of line L and ๐2 , ๐2 , ๐2 be the direction cosines of line M.
Let ๐ be the angle between line L and M then it is described as cos ๐ = ๐1๐2 + ๐1๐2 + ๐1๐2 -------- (A)
๐ + ๐ + ๐ = 0 -------------------------------- (1)
2๐๐ + 2๐๐ โ ๐๐ = 0 -------------------------------- (2)
1 โน ๐ = โ ๐ + ๐ ------------------------------ (3)
Putting in eq. (2) 2๐๐ + 2๐ โ ๐ + ๐ โ ๐ โ ๐ + ๐ = 0
2๐๐ โ 2๐ ๐ + ๐ + ๐ ๐ + ๐ = 0
2๐๐ โ 2๐2 โ 2๐๐ + ๐๐ + ๐2 = 0
โ2๐2 + ๐๐ + ๐2 = 0
2๐2 โ ๐๐ โ ๐2 = 0
2๐2 โ 2๐๐ + ๐๐ โ ๐2 = 0
2๐ ๐ โ ๐ + ๐ ๐ โ ๐ = 0 โน ๐ โ ๐ 2๐ + ๐ = 0
๐ โ ๐ = 0
๐๐ ๐ = ๐ โน๐
1=
๐
1
Putting in eq. (3)
๐ = โ2๐ ๐
โ2=
๐
1
๐๐ ๐
1=
๐
1=
๐
โ2 โน
๐2 + ๐2 + ๐2
12 + 12 + โ2 2=
1
6
So direction cosines of line L are
๐1 =1
6 , ๐1 =
1
6 , ๐1 =
โ2
6
2๐ + ๐ = 0
๐๐ ๐ = โ2๐ โน๐
1=
๐
โ2
Putting in eq. (3)
๐ = ๐ ๐
1=
๐
1
๐๐ ๐
1=
๐
โ2=
๐
1 โน
๐2 + ๐2 + ๐2
12 + 12 + (โ2)2=
1
6
So direction cosines of line M are
๐2 =1
6 , ๐2 =
โ2
6 , ๐2 =
1
6
Using in equation (A)
cos ๐ = 1
6
1
6 +
1
6
โ2
6 +
โ2
6
1
6 =
1
6โ
2
6โ
2
6 =
1โ2โ2
6= โ
3
6 โน cos ๐ = โ
1
2
โน ๐ = cosโ1 โ1
2 โน ๐ = 60๐
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 5
Find equations of the straight line L and M in symmetric forms. Determine whether the pairs of lines intersect.
Find the point of intersection if it exists.
Q#10: ๐ โถ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ ๐, ๐, ๐ , ๐ โ๐, ๐, โ๐ ๐ โถ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ ๐, ๐, โ๐ , ๐ ๐, ๐, ๐ Solution:
The equation of the straight line L through ๐ด 2,1,3 & ๐ต โ1,2, โ4
๐ฅโ2
2+1=
๐ฆโ1
1โ2=
๐งโ3
3+4 โน
๐ฅโ2
3=
๐ฆโ1
โ1=
๐งโ3
7
The equation of the straight line M through ๐ 5,1, โ2 & ๐ 0,4,3
๐ฅโ5
5โ0=
๐ฆโ1
1โ4=
๐ง+2
โ2โ3 โน
๐ฅโ5
5=
๐ฆโ1
โ3=
๐ง+2
โ5
Which are the required equations in symmetric form of L & M.
Now we write the equations of L & M in parametric forms.
๐ฟ๐๐ก ๐ฅ โ 2
3=
๐ฆ โ 1
โ1=
๐ง โ 3
7 = ๐ก ๐ ๐๐ฆ
๐๐๐ก ๐ฅ โ 5
5=
๐ฆ โ 1
โ3=
๐ง + 2
โ5 = ๐ ๐ ๐๐ฆ
Now parametric equations of lines L & M are
๐ฅ โ 2
3= ๐ก
๐ฆ โ 1
โ1 = ๐ก โน ๐ฟ โถ
๐ฅ = 2 + 3๐ก๐ฆ = 1 โ 1๐ก๐ง = 3 + 7๐ก
๐ง โ 3
7 = ๐ก
๐ฅ โ 5
5= ๐
๐ฆ โ 1
โ3 = ๐ โน ๐ โถ
๐ฅ = 5 + 5๐ ๐ฆ = 1 โ 3๐
๐ง = โ2 โ 5๐
๐ง + 2
โ5 = ๐
Let the lines L & M intersect at P ๐ฅ, ๐ฆ, ๐ง so this point will lie on both lines L & M.
Comparing above equations
โน 2 + 3๐ก = 5 + 5๐ โน 3๐ก โ 5๐ = 3 ------------ (1)
1 โ ๐ก = 1 โ 3๐ โน ๐ก โ 3๐ = 0 ------------ (2)
3 + 7๐ก = โ2 โ 5๐ โน 7๐ก + 5๐ = โ5 ----------- (3)
Now multiply eq. (2) by 3 and subtracting
4๐ = 3 โน ๐ =๐
๐
Put in eq. (2)
๐ก โ 3 3
4 = 0 โน ๐ =
๐
๐
Now putting these values in equation (3) we have
7 9
4 + 5
3
4 = โ5 โน
63
4+
15
4= โ5 โน
78
4โ โ5
We see that these values of t & s do not satisfy equation (3)
Hence, given straight lines L & M do not intersect. So point of intersection doesnโt exist.
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 6
Q#11: ๐ โถ ๐ซ = ๐๐ข + ๐๐ฃ โ ๐ค + ๐ญ ๐๐ข โ ๐๐ฃ โ ๐๐ค
๐ โถ ๐ซ = ๐๐ข + ๐๐ฃ + ๐๐ค + ๐ฌ ๐๐๐ข โ ๐๐ฃ + ๐๐ค
Solution:
For given straight line L
๐ = (3 + 6๐ก)๐ + 2 โ 4๐ก ๐ + (โ1 โ 3๐ก)๐
For given straight line M
๐ = (5 + 14๐ )๐ + 4 โ 6๐ ๐ + (โ7 + 2๐ )๐
Now parametric equations for line L are
๐ฅ = 3 + 6๐ก ๐ฆ = 2 โ 4๐ก ---------(A)
๐ง = โ1 โ 3๐ก
Parametric equations for given line M are
๐ฅ = 5 + 14๐ ๐ฆ = 4 โ 6๐ -------- (B)
๐ง = 7 + 2๐
Now equations of straight lines L & M are in symmetric form
๐ฅ โ 3
6=
๐ฆ โ 2
โ4=
๐ง + 1
โ3 โ โ โ โ ๐ฟ
๐ฅ โ 5
14=
๐ฆ โ 4
โ6=
๐ง โ 7
2 โ โ โ โ ๐
Let the lines L & M intersect at ๐ฅ, ๐ฆ, ๐ง so this point will lie on both lines L & M.
Comparing equations (A) & (B)
โน 3 + 6๐ก = 5 + 14๐ โน 6๐ก โ 14๐ = 2 โน 3๐ก โ 7๐ = 1 ------------- (1)
2 โ 4๐ก = 4 โ 6๐ โน 4๐ก โ 6๐ = โ2 ------------ (2)
โ1 โ 3๐ก = 7 + 2๐ โน 3๐ก + 2๐ = โ8 ------------- (3)
Subtracting eq. (1) & (3)
โ9๐ = 9 โน ๐ = โ๐
Putting value of s in equation (1)
3๐ก โ 7 โ1 = 1 โน 3๐ก = 1 โ 7 โน 3๐ก = โ6 โน ๐ = โ๐
Now putting values of s & t in equation (2)
4 โ2 โ 6 โ1 = โ2 โน โ8 + 6 = โ2 โน โ2 = โ2
We see that these values of s & t satisfy equation (2)
Hence given straight lines L & M intersect.
For point of intersection we put ๐ = โ1 in equation (B)
Hence point of intersection of given straight line is ๐ฅ, ๐ฆ, ๐ง = โ9,10,5 .
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 7
Q#12: ๐ โถ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ ๐, โ๐, ๐ and parallel to๐ = ๐, ๐, ๐ ๐ โถ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ โ๐, ๐, ๐ and parallel to๐ = ๐, ๐, ๐ Solution:
Symmetric form of given straight line L through the point ๐ด 2, โ1,0 and parallel to ๐ = 4,3, โ2
๐ฅโ2
4=
๐ฆ+1
3=
๐งโ0
2
๐๐๐ก ๐ฅโ2
4=
๐ฆ+1
3=
๐ง
โ2 = ๐ก
Symmetric form of given straight line M through
the point ๐ โ1,3,5 and parallel to ๐ = 1,7,3 .
๐ฅ + 1
1=
๐ฆ โ 3
7=
๐ง โ 5
3
๐๐๐ก ๐ฅ + 1
1=
๐ฆ โ 3
7=
๐ง โ 5
3 = ๐
Now parametric equations of lines L & M are
๐๐๐ ๐๐๐๐ ๐ฟ ๐ฅ โ 2
4= ๐ก โน ๐ฅ = 2 + 4๐ก
๐ฆ+1
3= ๐ก โน ๐ฆ = โ1 + 3๐ก
๐ง
โ2 = ๐ก โน ๐ง = 0 โ 2๐ก
๐๐๐ ๐๐๐๐ ๐ ๐ฅ + 1
1= ๐ โน ๐ฅ = โ1 + 1๐
๐ฆโ3
7= ๐ โน ๐ฆ = 3 + 7๐
๐งโ5
3= ๐ โน ๐ง = 5 + 3๐
NOW DO YOURSELF AS ABOVE
Find the distance of the given point P from the given line L.
Q#13: ๐ = ๐, โ๐, ๐ , ๐ โถ ๐ฑ = ๐ + ๐ญ๐ฒ = ๐ โ ๐๐ญ
๐ณ = โ๐ + ๐๐ญ
Solution: Given point and given line are
๐ = 3, โ2,1 , ๐ฟ โถ ๐ฅ = 1 + ๐ก๐ฆ = 3 โ 2๐ก
๐ง = โ2 + 2๐ก
A point on the line L is ๐ด = (1,3, โ2)
Now ๐ด๐ = ๐ 3, โ2,1 โ ๐ด 1,3, โ2 = 3 โ 1, โ2 โ 3,1 + 2
๐ด๐ = 2๐ โ 5๐ + 3๐
Now direction vector of the given line L is ๐ = 1๐ โ 2๐ + 2๐
Let d be the required distance of a point from a line L then by using formula
๐ = ๐ด๐ ร๐
๐ โ โ โ โ ๐ด
โด ๐ด๐ ร ๐ = ๐ ๐ ๐
2 โ5 31 โ2 2
= โ10 + 6 ๐ โ 4 โ 3 ๐ + โ4 + 5 ๐ โน ๐ด๐ ร ๐ = โ4๐ โ ๐ + ๐
๐ด๐ ร ๐ = โ4 2 + โ1 2 + 12 = 16 + 1 + 1 = 18 & ๐ = 12 + โ2 2 + 22 = 1 + 4 + 4 = 9 = 3
Putting in equation (A) ๐ = 18
3 ๐๐ ๐๐๐๐ข๐๐๐๐ ๐๐๐ ๐ก๐๐๐๐
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 8
๐#๐๐: ๐ = ๐, โ๐, ๐ , ๐ โถ ๐ฑ โ ๐
๐=
๐ฒ + ๐
โ๐=
๐ณ + ๐
๐
Solution: A point on the line L is ๐ด = (1, โ3, โ1)
๐ด๐ = ๐ 0, โ2,1 โ ๐ด 1, โ3, โ1 = 0 โ 1, โ2 + 3,1 + 1
๐ด๐ = โ๐ + ๐ + 2๐
Now direction vector of the given line L is ๐ = 4๐ โ 2๐ + 5๐
DO YOURSELF AS ABOVE
Q#15: If the edges of a rectangular parallelepiped are ๐, ๐, ๐; show that the angles between the four diagonals are given by
๐๐ซ๐๐๐จ๐ฌ ยฑ๐๐ ยฑ ๐๐ ยฑ ๐๐
๐๐ + ๐๐ + ๐๐
Solution:
Consider a parallelepiped as shown in the figure.
Here the lengths of the edges OA, OB & OC are a, b, c respectively.
The coordinates of the vertices of parallelepiped are
๐ = 0,0,0 , ๐โฒ = ๐, ๐, ๐ , ๐ด = ๐, 0,0 , ๐ดโฒ = 0, ๐, ๐
๐ต = 0, ๐, 0 , ๐ตโฒ = ๐, 0, ๐ , ๐ถ = 0,0, ๐ , ๐ถโฒ = ๐, ๐, 0
The four diagonals of given parallelepiped are
๐๐โฒ , ๐ด๐ดโฒ , ๐ต๐ตโฒ , ๐ถ๐ถ โฒ
Now
๐๐โฒ = ๐๐ + ๐๐ + ๐๐
๐ด๐ดโฒ = โ๐๐ + ๐๐ + ๐๐
๐ต๐ตโฒ = ๐๐ โ ๐๐ + ๐๐
๐ถ๐ถ โฒ = ๐๐ + ๐๐ โ ๐๐
So direction cosines of
๐๐โฒ โน ๐1 =๐
๐2 + ๐2 + ๐2 , ๐1 =
๐
๐2 + ๐2 + ๐2 , ๐1 =
๐
๐2 + ๐2 + ๐2
๐ด๐ดโฒ โน ๐2 =โ๐
๐2 + ๐2 + ๐2 , ๐2 =
๐
๐2 + ๐2 + ๐2 , ๐2 =
๐
๐2 + ๐2 + ๐2
๐ต๐ตโฒ โน ๐3 =๐
๐2 + ๐2 + ๐2 , ๐3 =
โ๐
๐2 + ๐2 + ๐2 , ๐3 =
๐
๐2 + ๐2 + ๐2
๐ถ๐ถ โฒ โน ๐4 =๐
๐2 + ๐2 + ๐2 , ๐4 =
๐
๐2 + ๐2 + ๐2 , ๐4 =
โ๐
๐2 + ๐2 + ๐2
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 9
Let ๐ผ be the angle between ๐๐โฒ ๐๐๐ ๐ด๐ดโฒ
cos ๐ผ = ๐1๐2 + ๐1๐2 + ๐1๐2
= ๐
๐2+๐2+๐2
โ๐
๐2+๐2+๐2 +
๐
๐2+๐2+๐2
๐
๐2+๐2+๐2 +
๐
๐2+๐2+๐2
๐
๐2+๐2+๐2
cos ๐ผ =โ๐2+๐2+๐2
๐2+๐2+๐2 โ โ โ โ โ โ โ (1)
Let ๐ฝ be the angle between ๐ด๐ดโฒ ๐๐๐ ๐ต๐ตโฒ
cos ๐ฝ = ๐2๐3 + ๐2๐3 + ๐2๐3
๐๐๐ ๐ฝ =โ๐2 โ ๐2 + ๐2
๐2 + ๐2 + ๐2 โ โ โ โ โ (2)
Let ๐พ be the angle between ๐ต๐ตโฒ ๐๐๐ ๐ถ๐ถ โฒ
cos ๐พ = ๐3๐4 + ๐3๐4 + ๐3๐4
cos ๐พ =๐2 โ ๐2 โ ๐2
๐2 + ๐2 + ๐2 โ โ โ โ โ โ(3)
Let ๐ be the angle between ๐๐โฒ ๐๐๐ ๐ถ๐ถ โฒ
cos ๐ = ๐1๐4 + ๐1๐4 + ๐1๐4
cos ๐ =๐2 + ๐2 โ ๐2
๐2 + ๐2 + ๐2 โ โ โ โ โ (4)
From (1),(2),(3) & (4) we see that the angles between four diagonals are
cos ๐๐๐๐๐ =ยฑ๐2ยฑ๐2ยฑ๐2
๐2+๐2+๐2 โน ๐๐๐๐๐ = cosโ1
ยฑ๐2ยฑ๐2ยฑ๐2
๐2+๐2+๐2 ๐๐๐๐ ๐๐๐๐ฃ๐๐.
Q#16: A straight line makes angles of measure ๐, ๐, ๐, ๐ with the four diagonals of a cube. Prove that
๐๐จ๐ฌ๐๐ + ๐๐จ๐ฌ๐๐ + ๐๐จ๐ฌ๐๐ + ๐๐จ๐ฌ๐๐ =๐
๐
Solution:
Let โโ๐โโ be the length of each side of a cube
Points of the each corner of the cube are
๐ = 0,0,0 , ๐ = ๐, ๐, ๐ , ๐ด = ๐, 0,0 , ๐ต = 0, ๐, 0 , ๐ถ =(0,0, ๐)
๐ดโฒ = 0, ๐, ๐ , ๐ตโฒ = ๐, 0, ๐ , ๐ถโฒ = ๐, ๐, 0
Now ๐๐ , ๐ด๐ดโฒ , ๐ต๐ตโฒ , ๐ถ๐ถ โฒ are the diagonals of a cube
๐๐ = ๐, ๐, ๐ โ 0,0,0 = ๐๐ + ๐๐ + ๐๐
๐ด๐ดโฒ = 0, ๐, ๐ โ ๐, 0,0 = โ๐๐ + ๐๐ + ๐๐
๐ต๐ตโฒ = ๐, 0, ๐ โ 0, ๐, 0 = ๐๐ โ ๐๐ + ๐๐
๐ถ๐ถ โฒ = ๐, ๐, 0 โ 0,0, ๐ = ๐๐ + ๐๐ โ ๐๐
Length of each diagonal is ๐2 + ๐2 + ๐2 = 3๐2 = 3๐
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 10
Now direction cosines of each diagonals are
๐๐ โน ๐1 =๐
3๐ , ๐1 =
๐
3๐ , ๐1 =
๐
3๐
๐๐ โน ๐1 =1
3 , ๐1 =
1
3 , ๐1 =
1
3
๐ด๐ดโฒ โน ๐2 =โ1
3 , ๐2 =
1
3 , ๐2 =
1
3
๐ต๐ตโฒ โน ๐3 =1
3 , ๐3 =
โ1
3 , ๐3 =
1
3
๐ถ๐ถ โฒ โน ๐4 =1
3 , ๐4 =
1
3 , ๐4 =
โ1
3
Let ๐, ๐, ๐ be the direction cosines of line L which makes angles ๐ผ, ๐ฝ, ๐พ & ๐ฟ with each diagonals
๐ผ is the angle between line L and ๐๐
cos ๐ผ = ๐๐1 + ๐๐1 + ๐๐1 = ๐ 1
3 + ๐
1
3 + ๐
1
3 =
๐ + ๐ + ๐
3 โ โ โ (1)
๐ฝ is the angle between line L and ๐ด๐ดโฒ
cos ๐ฝ = ๐๐2 + ๐๐2 + ๐๐2 = ๐ โ1
3 + ๐
1
3 + ๐
1
3 =
โ๐ + ๐ + ๐
3 โ โ โ (2)
๐พ is the angle between line L and ๐ต๐ตโฒ
cos ๐พ = ๐๐3 + ๐๐3 + ๐๐3 = ๐ 1
3 + ๐
โ1
3 + ๐
1
3 =
๐ โ ๐ + ๐
3 โ โ โ (3)
๐ฟ is the angle between line L and ๐ถ๐ถ โฒ
cos ๐ฟ = ๐๐4 + ๐๐4 + ๐๐4 = ๐ 1
3 + ๐
1
3 + ๐
โ1
3 =
๐ + ๐ โ ๐
3 โ โ โ (4)
Squaring eq. (1),(2),(3) & (4) and adding
๐๐๐ 2๐ผ + ๐๐๐ 2๐ฝ + ๐๐๐ 2๐พ + ๐๐๐ 2๐ฟ = ๐ + ๐ + ๐
3
2
+ โ๐ + ๐ + ๐
3
2
+ ๐ โ ๐ + ๐
3
2
+ ๐ + ๐ โ ๐
3
2
=๐2+๐2+๐2+2๐๐ +2๐๐ +2๐๐
3+
๐2+๐2+๐2โ2๐๐ +2๐๐โ2๐๐
3
+๐2+๐2+๐2โ2๐๐โ2๐๐ +2๐๐
3+
๐2+๐2+๐2+2๐๐โ2๐๐ โ2๐๐
3
๐๐๐ 2๐ผ + ๐๐๐ 2๐ฝ + ๐๐๐ 2๐พ + ๐๐๐ 2๐ฟ =4๐2 + 4๐2 + 4๐2
3=
4 ๐2 + ๐2 + ๐2
3
๐๐๐ 2๐ผ + ๐๐๐ 2๐ฝ + ๐๐๐ 2๐พ + ๐๐๐ 2๐ฟ =4
3 ๐๐๐๐ ๐๐๐๐ฃ๐๐
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 11
Q#17: Find equations of the straight line passing through the point ๐(๐, โ๐, ๐) and parallel to the straight line joining the points ๐(๐, ๐, ๐) and ๐(๐, ๐, ๐). Solution: Consider two lines ๐ฟ1 & ๐ฟ2
Let ๐ฟ1 be the required equation of the straight line passing through the point ๐ 0, โ3,2
The points ๐ด(3,4,7) and ๐ต(2,7,5) on line ๐ฟ2. So direction ratios from ๐ด(3,4,7) to ๐ต(2,7,5) are
๐ด๐ต = 2,7,5 โ 3,4,7 = โ๐ + 3๐ โ 2๐
Then โ1,3, โ2 are direction ratios of line ๐ฟ2.
By given condition both lines are parallel so both lines having same direction ratios.
Now required equation through the point ๐ 0, โ3,2
๐ฅ โ 0
โ1=
๐ฆ + 3
3=
๐ง โ 2
โ2 โน
๐ฅ
1=
๐ฆ + 3
โ3=
๐ง โ 2
2 ๐๐๐๐ข๐๐๐๐ ๐๐๐ข๐๐ก๐๐๐.
Q#18: Find equations of the straight line passing through the point ๐(๐, ๐, โ๐) and perpendicular to each of straight lines
๐ฑ โ ๐
๐=
๐ฒ
๐=
๐ณ + ๐
๐ ๐๐ง๐
๐ฑ
๐=
๐ฒ + ๐
โ๐=
๐ณ + ๐
๐
Solution:
Given equations of lines are
๐ฅ โ 3
2=
๐ฆ
2=
๐ง + 1
2 โ โ โ โ(๐ฟ1)
๐ฅ
3=
๐ฆ + 1
โ1=
๐ง + 2
2 โ โ โ โ ๐ฟ2
Direction ratios of line ๐ฟ1 are 2,2,2 โน ๐ 1 = 2๐ + 2๐ + 2๐
Direction ratios of line ๐ฟ2 are 3, โ1,2 โน ๐ 2 = 3๐ โ ๐ + 2๐
Suppose L be the required line through the point ๐(2,0, โ2) with
direction ratios ๐1 , ๐2 , ๐3 โน ๐ 3 = ๐1๐ + ๐2๐ + ๐3๐
Since ๐ฟ โฅ ๐ฟ1
So by condition of perpendicularity
๐ 1 . ๐ 3 = 0 โน 2๐ + 2๐ + 2๐ . ๐1๐ + ๐2๐ + ๐3๐ = 0
2๐1 + 2๐2 + 2๐3 = 0 --------------- (1)
๐ 2 . ๐ 3 = 0 โน 3๐ โ ๐ + 2๐ . ๐1๐ + ๐2๐ + ๐3๐ = 0
3๐1 โ ๐2 โ 2๐3 = 0 -------------- (2)
Now
๐1
2 2
โ1 2
=โ๐2
2 23 2
=
๐3
2 23 โ1
โน
๐1
6=
โ๐2
โ2=
๐3
โ8 โน
๐1
3=
๐2
1=
๐3
โ4
Hence (3,1, โ4) be the direction ratios of required line L
Now equation of required line L passing through the point ๐ 2,0, โ2 is ๐ฅโ2
3=
๐ฆโ0
1=
๐ง+2
โ4
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 12
Find equations of straight line through the given point A and intersecting at right angles the given straight line:
Q#19: ๐ = ๐๐, ๐, โ๐ ๐๐ง๐ ๐ฑ = ๐ โ ๐ญ, ๐ฒ = ๐ + ๐๐ญ, ๐ณ = โ๐ +t.
Solution: Let L be the required line passing through ๐ด = 11,4, โ6 and perpendicular to given line.
Suppose it meets the given line at point B.
Now a point on given line is B 4 โ ๐ก, 7 + 2๐ก, โ1 + ๐ก
๐ด๐ต = 4 โ ๐ก โ 11 ๐ + 7 + 2๐ก โ 4 ๐ + โ1 + ๐ก + 6 ๐
๐ด๐ต = โ๐ก โ 7 ๐ + 2๐ก + 3 ๐ + ๐ก + 5 ๐
Hence direction vector of the given straight line is ๐ = โ๐ + 2๐ + ๐
Since ๐ด๐ต is perpendicular to given line.
By perpendicular condition
๐ด๐ต โฅ ๐ = 0 โน ๐ด๐ต . ๐ = 0
โ๐ก โ 7 ๐ + 2๐ก + 3 ๐ + ๐ก + 5 ๐ . โ๐ + 2๐ + ๐ = 0
โ1 โ๐ก โ 7 + 2 2๐ก + 3 + 1 ๐ก + 5 = 0 โน 7 + ๐ก + 4๐ก + 6 + ๐ก + 5 = 0 โน ๐ = โ๐
Direction vector of required line will become
๐ด๐ต = โ(โ3) โ 7 ๐ + 2(โ3) + 3 ๐ + โ3 + 5 ๐ โน ๐ด๐ต = โ4๐ โ 3๐ + 2๐
Now required equation passing through the point ๐ด = 11,4, โ6 having direction ratios โ4, โ3,2
๐ฅโ11
โ4=
๐ฆโ4
โ3=
๐ง+6
2 OR
๐ฅโ11
4=
๐ฆโ4
3=
๐ง+6
โ2
๐#๐๐: ๐ = ๐, โ๐, ๐ ๐๐ง๐ ๐ฑ
โ๐=
๐ฒ โ ๐
๐=
๐ณ
โ๐
Solution: Given point and line are ๐ด = 5, โ4,4 ๐๐๐ ๐ฅ
โ1=
๐ฆโ1
1=
๐ง
โ2 = ๐ก ๐ ๐๐ฆ
The parametric equations of given line are ๐ฅ = โ๐ก, ๐ฆ = 1 + ๐ก , ๐ง = โ2๐ก
DO YOURSELF AS ABOVE
Q#21: Find the length of the perpendicular from the point ๐(๐ฑ๐, ๐ฒ๐, ๐ณ๐) to the straight line ๐ฑ โ ๐
๐ฅ=
๐ฒ โ ๐
๐ฆ=
๐ณ โ ๐
๐ง , ๐ฐ๐ก๐๐ซ๐ ๐ฅ๐ + ๐ฆ๐ + ๐ง๐ = ๐
Solution: Given point and line are
๐ ๐ฅ1 , ๐ฆ1 , ๐ง1 & ๐ฅโ๐ผ
๐=
๐ฆโ๐ฝ
๐=
๐งโ๐พ
๐
Hence ๐ด = (๐ผ, ๐ฝ, ๐พ) is a point of given line
Direction vector of given line is ๐ = ๐๐ + ๐๐ + ๐๐
Now ๐ด๐ = ๐ฅ1 โ ๐ผ ๐ + ๐ฆ1 โ ๐ฝ ๐ + ๐ง1 โ ๐พ ๐
Let d be the required distance then by using formula
๐ = ๐ด๐ ร๐
๐ โ โ โ โ ๐ด
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 13
โด ๐ด๐ ร ๐ = ๐ ๐ ๐
๐ฅ1 โ ๐ผ ๐ฆ1 โ ๐ฝ ๐ง1 โ ๐พ๐ ๐ ๐
= ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ ๐ โ ๐ ๐ฅ1 โ ๐ผ โ ๐ ๐ง1 โ ๐พ ๐ + ๐ ๐ฅ1 โ ๐ผ โ ๐ ๐ฆ1 โ ๐ฝ ๐
๐ด๐ ร ๐ = ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ ๐ + ๐ ๐ง1 โ ๐พ โ ๐ ๐ฅ1 โ ๐ผ ๐ + ๐ ๐ฅ1 โ ๐ผ โ ๐ ๐ฆ1 โ ๐ฝ ๐
๐ด๐ ร ๐ = ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ 2 + ๐ ๐ง1 โ ๐พ โ ๐ ๐ฅ1 โ ๐ผ
2 + ๐ ๐ฅ1 โ ๐ผ โ ๐ ๐ฆ1 โ ๐ฝ 2
๐ด๐ ร ๐ = ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ 2 & ๐ = ๐2 + ๐2 + ๐2 = 1 = 1
Putting in equation (A)
๐ = ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ
2
1 โน ๐ = ๐ ๐ฆ1 โ ๐ฝ โ ๐ ๐ง1 โ ๐พ
2 ๐๐๐๐ข๐๐๐๐ ๐๐๐ ๐ก๐๐๐๐
Q#22: Find equations of the perpendicular from the point ๐(๐, ๐, ๐) to the straight line ๐ฑ
๐=
๐ฒ โ ๐
๐=
๐ณ โ ๐
๐
Also obtain its length and coordinates of the foot of the perpendicular.
Solution:
Given point ๐(1,6,3) and equation of straight line is
๐ฅ
1=
๐ฆ โ1
2=
๐ง โ2
3
Let ๐ด๐ be the length of perpendicular from point A to given line
๐ฅ
1=
๐ฆโ1
2=
๐งโ2
3 = ๐ก
So parametric equations of given line are
๐ฅ = ๐ก
๐ฆ = 1 + 2๐ก ๐ง = 2 + 3๐ก
Any point on this line is ๐ก, 1 + 2๐ก, 2 + 3๐ก
So coordinates of point P are ๐ ๐ก, 1 + 2๐ก, 2 + 3๐ก
Direction ratios of line ๐ด๐ are = ๐ก, 1 + 2๐ก, 2 + 3๐ก โ 1,6,3
๐ด๐ = ๐ก โ 1 ๐ + 2๐ก โ 5 ๐ + 3๐ก โ 1 ๐
& direction vector of given line is ๐ = ๐ + 2๐ + 3๐
๐๐๐๐๐ ๐ด๐ โฅ ๐ โน ๐ด๐ . ๐ = 0
๐ก โ 1 ๐ + 2๐ก โ 5 ๐ + 3๐ก โ 1 ๐ . ๐ + 2๐ + 3๐ = 0
1 ๐ก โ 1 + 2 2๐ก โ 5 + 3 3๐ก โ 1 = 0 โน 14๐ก โ 14 = 0 โน ๐ = ๐
So coordinates of point P are ๐ 1,3,5
Length of perpendicular = ๐ด๐ = 1 โ 1 2 + 3 โ 1 2 + 5 โ 3 2 = 0 + 9 + 4 = 13
Now equation of perpendicular ๐ด๐ is
๐ฅโ1
1โ1=
๐ฆโ6
3โ6=
๐งโ3
5โ3 โน
๐ฅโ1
0=
๐ฆโ6
โ3=
๐งโ3
2 ๐๐๐๐ข๐๐๐๐ ๐๐๐ข๐๐ก๐๐๐ ๐๐ ๐๐๐๐
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 14
Q#23: Find necessary and sufficient condition that the point ๐ ๐ฑ๐, ๐ฒ๐, ๐ณ๐ , ๐(๐ฑ๐, ๐ฒ๐, ๐ณ๐) and ๐(๐ฑ๐, ๐ฒ๐, ๐ณ๐) are collinear.
Solution: Given points are ๐ ๐ฅ1 , ๐ฆ1 , ๐ง1 , ๐(๐ฅ2 , ๐ฆ2 , ๐ง2) & ๐ (๐ฅ3 , ๐ฆ3 , ๐ง3)
Suppose that the points P,Q &R are collinear.
Now equation of line through ๐ ๐ฅ1 , ๐ฆ1 , ๐ง1 & ๐ ๐ฅ2 , ๐ฆ2 , ๐ง2 ๐๐ ๐ฅโ๐ฅ1
๐ฅ2โ๐ฅ1=
๐ฆโ๐ฆ1
๐ฆ2โ๐ฆ1=
๐งโ๐ง1
๐ง2โ๐ง1
Since points P,Q &R are collinear. So point ๐ (๐ฅ3 , ๐ฆ3 , ๐ง3) lies on line.
๐ฅ3โ๐ฅ1
๐ฅ2โ๐ฅ1=
๐ฆ3โ๐ฆ1
๐ฆ2โ๐ฆ1=
๐ง3โ๐ง1
๐ง2โ๐ง1 = ๐ก ๐ ๐๐ฆ
๐ฅ3 โ ๐ฅ1 = ๐ก ๐ฅ2 โ ๐ฅ1
โน ๐ฆ3 โ ๐ฆ1 = ๐ก ๐ฆ2 โ ๐ฆ1
๐ง3 โ ๐ง1 = ๐ก ๐ง2 โ ๐ง1 ๐๐
๐ฅ3 โ ๐ฅ1 โ ๐ก๐ฅ2 + ๐ก๐ฅ1 = 0๐ฆ3 โ ๐ฆ1 โ ๐ก๐ฆ2 + ๐ก๐ฆ1 = 0๐ง3 โ ๐ง1 โ ๐ก๐ง2 + ๐ก๐ง1 = 0
โน ๐ก โ 1 ๐ฅ1 โ ๐ก๐ฅ2 + ๐ฅ3 = 0 ๐ก โ 1 ๐ฆ1 โ ๐ก๐ฆ2 + ๐ฆ3 = 0 ๐ก โ 1 ๐ง1 โ ๐ก๐ง2 + ๐ง3 = 0
Eliminating ๐ก โ 1 , โ๐ก, 1 from above equations
๐ฅ1 ๐ฅ2 ๐ฅ3๐ฆ1 ๐ฆ2 ๐ฆ3๐ง1 ๐ง2 ๐ง3
= 0 ๐๐
๐ฅ1 ๐ฆ1 ๐ง1๐ฅ2 ๐ฆ2 ๐ง2๐ฅ3 ๐ฆ3 ๐ง3
= 0 Which is necessary condition for three points ๐, ๐ &๐ to be collinear.
Q#24: If ๐ฅ๐, ๐ฆ๐, ๐ง๐; ๐ฅ๐, ๐ฆ๐, ๐ง๐ and ๐ฅ๐, ๐ฆ๐, ๐ง๐ are direction cosines of three mutual perpendicular lines, prove that the lines whose direction cosines are proportional to ๐ฅ๐ + ๐ฅ๐ + ๐ฅ๐, ๐ฆ๐ + ๐ฆ๐ + ๐ฆ๐, ๐ง๐ + ๐ง๐ + ๐ง๐ makes congruent angles with them.
Solution: Suppose that ๐ฟ1 , ๐ฟ2 & ๐ฟ3 are given line such that
Direction cosines of ๐ฟ1 are ๐1 , ๐1 , ๐1
Direction cosines of ๐ฟ2 are ๐2 , ๐2 , ๐2
Direction cosines of ๐ฟ3 are ๐3 , ๐3 , ๐3
Since lines ๐ฟ1 , ๐ฟ2 & ๐ฟ3 are mutually perpendicular
So ๐1๐2 + ๐1๐2 + ๐1๐2 = 0๐2๐3 + ๐2๐3 + ๐2๐3 = 0๐1๐3 + ๐1๐3 + ๐1๐3 = 0
Let L be the line having direction ratios ๐1 + ๐2 + ๐3 , ๐1 + ๐2 + ๐3 & ๐1 + ๐2 + ๐3
Let ๐ผ be the angle between ๐ฟ & ๐ฟ1 then
cos ๐ผ =๐1 ๐1+๐2+๐3 +๐1 ๐1+๐2+๐3 +๐1 ๐1+๐2+๐3
๐12+๐1
2+๐12 . ๐1+๐2+๐3
2+ ๐1+๐2+๐3 2+ ๐1+๐2+๐3
2
= ๐1
2+๐12+๐1
2 + ๐1๐2+๐1๐2+๐1๐2 + ๐1๐3+๐1๐3+๐1๐3
๐12+๐1
2+๐12 + ๐2
2+๐22+๐2
2 + ๐32+๐3
2+๐32 +2 ๐1๐2+๐1๐2+๐1๐2
cos ๐ผ =1+0+0
1+1+1+0+0+0 โน cos ๐ผ =
1
3 โน ๐ผ = cosโ1
1
3 โ โ โ โ โ (1)
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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal ([email protected]) Mardawal Naushehra ,KHUSHAB Page 15
Let ๐ฝ be the angle between ๐ฟ & ๐ฟ2 then
cos ๐ฝ =๐2 ๐1 + ๐2 + ๐3 + ๐2 ๐1 + ๐2 + ๐3 + ๐2 ๐1 + ๐2 + ๐3
๐22 + ๐2
2 + ๐22. ๐1 + ๐2 + ๐3
2 + ๐1 + ๐2 + ๐3 2 + ๐1 + ๐2 + ๐3
2
= ๐1๐2 + ๐1๐2 + ๐1๐2 + ๐2
2 + ๐22 + ๐2
2 + + ๐2๐3 + ๐2๐3 + ๐2๐3
๐12 + ๐1
2 + ๐12 + ๐2
2 + ๐22 + ๐2
2 + ๐32 + ๐3
2 + ๐32 + 2 ๐1๐2 + ๐1๐2 + ๐1๐2
cos ๐ฝ =0+1+0
1+1+1+0+0+0 โน cos ๐ฝ =
1
3 โน ๐ฝ = cosโ1
1
3 โ โ โ โ โ (2)
Let ๐พ be the angle between ๐ฟ & ๐ฟ1 then
cos ๐พ =๐3 ๐1 + ๐2 + ๐3 + ๐3 ๐1 + ๐2 + ๐3 + ๐3 ๐1 + ๐2 + ๐3
๐32 + ๐3
2 + ๐32 . ๐1 + ๐2 + ๐3
2 + ๐1 + ๐2 + ๐3 2 + ๐1 + ๐2 + ๐3
2
= ๐1๐3 + ๐1๐3 + ๐1๐3 + ๐2๐3 + ๐2๐3 + ๐2๐3 + ๐3
2 + ๐32 + ๐3
2
๐12 + ๐1
2 + ๐12 + ๐2
2 + ๐22 + ๐2
2 + ๐32 + ๐3
2 + ๐32 + 2 ๐1๐2 + ๐1๐2 + ๐1๐2
cos ๐พ =1+0+0
1+1+1+0+0+0 โน cos ๐พ =
1
3 โน ๐พ = cosโ1
1
3 โ โ โ โ 3
From (1) (2) & (3) it is proved that โน ๐ผ โ ๐ฝ โ ๐พ ๐๐๐๐ข๐๐๐๐ ๐๐๐ ๐ข๐๐ก
Q#25: A variable line in two adjacent positions has direction cosines ๐, ๐, ๐ and ๐ + ๐น๐, ๐ + ๐น๐, ๐ + ๐น๐. Show that measure of the small angle ๐ ๐ between the two positions is given by ๐น๐ฝ ๐ = ๐น๐ ๐ + ๐น๐ ๐ + ๐น๐ ๐. Solution:
Let OA and OB be the two adjacent positions of the line. Let PQ be the arc of the circle with centre at O and radius 1.
Then the coordinates of the points.
P & Q are ๐ ๐, ๐, ๐ & ๐ ๐ + ๐ฟ๐, ๐ + ๐ฟ๐, ๐ + ๐ฟ๐ .
Let ๐ฟ๐ be the angle between two positions of line.
Now ๐ฟ๐ = ๐๐๐๐ ๐๐
So ๐ฟ๐ = ๐๐
๐ฟ๐ = ๐ + ๐ฟ๐ โ ๐ 2 + ๐ + ๐ฟ๐ โ ๐ 2 + ๐ + ๐ฟ๐ โ ๐ 2
๐ฟ๐ = ๐ฟ๐2 + ๐ฟ๐2 + ๐ฟ๐2
โน ๐ฟ๐2 = ๐ฟ๐2 + ๐ฟ๐2 + ๐ฟ๐2
Checked by: Sir Hameed ullah ( hameedmath2017 @ gmail.com)
Specially thanks to my Respected Teachers
Prof. Muhammad Ali Malik (M.phill physics and Publisher of (www.Houseofphy.blogspot.com)
Muhammad Umar Asghar sb (M.Sc Mathematics)
Hameed Ullah sb ( M.Sc Mathematics)
http://www.houseofphy.blogspot.com/