logo

Plane Equations in 3D Geometry

   

Added on  2020-05-08

31 Pages991 Words345 Views
 | 
 | 
 | 
CALCULUS Student id [Pick the date]
Plane Equations in 3D Geometry_1

Question 1 Let Vector1=aVector2=vPoint located on the line (a+tv) which is near to the origin in such as way that a┴v.a┴vonly when the dot product would be equal to zero. a.v=0Line equation ¿a+tvThe distance (t,a+tv) from origin (0, 0) would be determined as shown below:¿√(a−c)2+¿¿a=a,b=a+tv,c=0,d=0¿√(t−0)2+¿¿¿√t2+¿¿Now, letf(x)=√t2+¿¿1
Plane Equations in 3D Geometry_2

It is essential to minimize the square of the distance and hence, g(x)=t2+¿The next step is to determine the first derivative of f(x).ddxf(x)=ddx¿¿t+v(a+tv)√t2+¿¿¿Further, find the first derivative of g(x)ddxg(x)=ddxt2+¿¿2t+¿Let the value of denominator close to zero. t2+¿0 (derivative is defined) t=0,a+tv=0∨t=−avThis is termed as critical point of the and therefore, point (−av,0¿is not close to the origin. 2
Plane Equations in 3D Geometry_3

Question 2 All unit vectors which are perpendicular to the vector (512)=?Let the vector is a = 5i +12 j Two vectors would be termed as perpendicular to each other when their dot product would be equal to zero. cosθ=a.b|a||b|θ=90°(perpendicular)a.b|a||b|=cos90=0a.b=0Assume that vector (ai+bj) is perpendicular to (5i +12 j). Now, ¿(ai+bj).(5i+12j)5a+12b=0.............(1)3
Plane Equations in 3D Geometry_4

In order to determine the unit vector which would be perpendicular to the (5i +12 j) hit and trial method needs to be applied. ata=12,b=−5(5∗12)+(12∗−5)=0satisfiedu1=12i−5j√(12)2+(−5)2=(12i13−5j13)Ata=−12,b=5¿u2=−12i+5j√(−12)2+(5)2=(−12i13+5j13)Hence, the two unit vectors i.e. (12i13−5j13) and (−12i13+5j13) are perpendicular to (512).Question 3 Angle between any of the two line segment from carbon atom to hydrogen atom =? 4
Plane Equations in 3D Geometry_5

LetabethevectoralongAC∧bbethevectoralongCD.AC=(12−0)i+(12−0)j+(12−1)kAC=a=12i+12j−12kSimilarly CD=b=12i−12j+12k5
Plane Equations in 3D Geometry_6

Now cosθ=a.b|a||b|cosθ=(12i+12j−12k).(12i−12j+12k)√(12)2+(12)2+(−12)2√(12)2+(−12)2+(12)2¿14−14−14√34√34¿−1434¿−412cosθ=−13θ=cos−1(−13)θ=121.6°Hence, the angle would be 121.60.6
Plane Equations in 3D Geometry_7

Question 4 Given figureLet the center of the systemO(0,0,0)And the corner would be A(±1,±1,±1)Hence, the by co-ordinate system as per the figure is highlighted below:OA=(−1i+1j+1k)OB=(−1i+1j−1k)OA.OB=(−1i+1j+1k).(−1i+1j−1k)¿1+1−1=1Hence, |OA|=√(−1)2+(1)2+(1)2=√3|OB|=√(−1)2+(1)2+(−1)2=√37
Plane Equations in 3D Geometry_8

End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
Stage 2 Specialist Mathematics
|4
|756
|89

HNCB036 Applied Mathematics For Engineers
|27
|566
|122