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Metu math 115 Anlaytical geometry final exam questions from 2019
Typology: Exams
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2 0 1 8 - 2 0 1 9 F a l l S e m e s t e r
S a t u r d a y , J a n u a r y 1 2
t h
1
2
1
2
Direction vectors of given lines are 𝑢⃗ 1
= ( 3 , 4 , 1 ) and 𝑢⃗
2
= ( 1 , 2 , − 3 ). Since
3
1
4
2
1
− 3
, these vectors are neither
parallel nor coincident, hence ℓ 1
and ℓ
2
either intersect at a point or they are skew. To determine whether they
intersect or not we have to check the following system of equations for a simultaneous solution:
By adding all equations side by side we get 8 𝑡 + 5 = 29 which gives 𝑡 = 3 and substituting this value of t in the first
eqution we obtain s = − 2. Since these values of 𝑡 and 𝑠 satisfy the last two equations as well, 𝑡 = 3 , 𝑠 = − 2 is a
solution of the system. It follows that the lines intersect at the point ( 10 , 11 , 8 ).
The angle between two planes is defined to be the angle between their normal vectors. Normal vectors of the given
planes are 𝑛⃗ 1
= ( 1 , 2 , 2 ) and 𝑛⃗
2
= ( 2 , 3 , 6 ). Then cosine the angle 𝜃 between the given planes is
cos 𝜃 =
1
1
1
2
2
2
𝑥
Rotation around 𝑂𝑥 axis yields the equation (𝑥 − 2 )
2
2
2
= 1 which represents the sphere with center at
( 2 , 0 , 0 ) and radius 1.
𝑦
Rotation around 𝑂𝑦 axis results in the equation (√𝑥
2
2
2
2
= 1 which represents a torus..
𝑥
𝑦
𝑥
𝑦
𝑧
Resulting surface is an annulus in 𝑂𝑥𝑦 plane: 1 ≤ 𝑥
2
2
2
2
1
2
𝒮 is an elliptic paraboloid.
1
2
2
2
1
16
2
Substituting 𝑧 = 4 in 𝑥
2
2
− 𝑧 = 0 , we obtain the
equation
𝑥
2
4
2
= 1 which represents an ellipse. For
the ellipse
𝑥
2
𝑎
2
𝑦
2
𝑏
2
= 1 , the distance between the center
and focus is 𝑓 = √𝑎
2
2
and eccentricity is 𝑒 = 𝑓/𝑎.
Then eccentricity of the given ellipse is √
3 / 2 and focii
are (−√ 3 ,0,4) and (√ 3 ,0,4).