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A few vector questions
1. Find the point of intersection of the line through the points (2,0,1) and (-1,3,4) and the line through the points (-1,3,0) and (4,-2,5). Calculate the acute angle between the two lines. ans=(1,1,2) 70.5
2. The point A has coordinates (7,-1,3) and the point B has coordinates (10,-2,2). The line l has vector equation r= i + j + k + lambda( 3 i - j + k) where lambda is a real parameter.
a) show that A lies on the line l. ans = (11)^0.5
b)find the length of AB ans=35 degrees
c)find the size of the acute angle between the line l and the line segment AB ans = 1.9
d)hence, calculate the perpendicular distance from B to the line l. soz no ans
3. two submarines are travelling in straight lines. Relative to a fixed origin, the vector equations of 2 lines, l1 and l2, along which they travel are
r= 3i+ 4j -5k + lambda ( i -2j +2k) and
r= 9i+j- 2k + mew (4i +j -k) where lambda and mew are scalars.
a) show that the submarines are moving in perpendicular directions
b) given that l1 and l2 intersect at A, find the position vector of A. The point B has position vector 10j-11k ans =5i- k
c) show that onlyone of the submaries passes through B
d) given that 1 unit on each coordinate axis represents 100m,find in km the distance AB ans=1.5 km
2. The point A has coordinates (7,-1,3) and the point B has coordinates (10,-2,2). The line l has vector equation r= i + j + k + lambda( 3 i - j + k) where lambda is a real parameter.
a) show that A lies on the line l. ans = (11)^0.5
b)find the length of AB ans=35 degrees
c)find the size of the acute angle between the line l and the line segment AB ans = 1.9
d)hence, calculate the perpendicular distance from B to the line l. soz no ans
3. two submarines are travelling in straight lines. Relative to a fixed origin, the vector equations of 2 lines, l1 and l2, along which they travel are
r= 3i+ 4j -5k + lambda ( i -2j +2k) and
r= 9i+j- 2k + mew (4i +j -k) where lambda and mew are scalars.
a) show that the submarines are moving in perpendicular directions
b) given that l1 and l2 intersect at A, find the position vector of A. The point B has position vector 10j-11k ans =5i- k
c) show that onlyone of the submaries passes through B
d) given that 1 unit on each coordinate axis represents 100m,find in km the distance AB ans=1.5 km
1)
A(2,0,1), B(-1,3,4) C(-1,3,0), D(4,-2,5)
Eqn of the line through AB is given by,
r = OA +t(OB - OA)
r = (2,0,1) + t(-3,3,3)
=================
Eqn of the line through CD is given by,
p = OC +s(OD - OC)
p = (-1,3,0) + s(5,-5,5)
==================
intersection of the two lines is gotten by r=p, giving,
(2-3t)i + (0+3t)j + (1+3t)k = (-1+5s)i + (3-5s)j + (0+ 5s)k
equating coefficients,
2 - 3t = -1 + 5s
3t = 3 - 5s
1 + 3t = 5s
solving we get,
s = 2/5, t = 1/3
============
substituting for s or t in either of the eqns for r or p gives the point of intersection as,
(2,0,1) + (1/3)(-3,3,3) = (2,0,1) + (-1,1,1) = (1,1,2)
ans = (1,1,2)
==========
A(2,0,1), B(-1,3,4)
|AB| = √(2+1)² + (0-3)² + (1-4)²
|AB| = √(3² + 3² + 3²)
|AB| = 3√3
========
C(-1,3,0), D(4,-2,5)
|CD| = √(-1-4)² + (3+2)² + (0-5)²
|CD| = √(5² + 5² + 5²)
|CD| = 5√3
========
AB = OB - OA = (-3,3,3)
CD = OD - OC = (5,-5,5)
AB.CD = (-3,3,3).(5,-5,5) = -3.5 + 3.(-5) + 3.5 = -15 -15 + 15 = -15
AB.CD = -15
=========
AB.CD = |AB||CD|.sin@
-15 = 3√3.5√3.sin@
-15 = 45.sin@
sin@ = -1/3
@ = 109.47 deg
smallest angle between lines is 180 - @ = 70.53 deg
=======================================
A(2,0,1), B(-1,3,4) C(-1,3,0), D(4,-2,5)
Eqn of the line through AB is given by,
r = OA +t(OB - OA)
r = (2,0,1) + t(-3,3,3)
=================
Eqn of the line through CD is given by,
p = OC +s(OD - OC)
p = (-1,3,0) + s(5,-5,5)
==================
intersection of the two lines is gotten by r=p, giving,
(2-3t)i + (0+3t)j + (1+3t)k = (-1+5s)i + (3-5s)j + (0+ 5s)k
equating coefficients,
2 - 3t = -1 + 5s
3t = 3 - 5s
1 + 3t = 5s
solving we get,
s = 2/5, t = 1/3
============
substituting for s or t in either of the eqns for r or p gives the point of intersection as,
(2,0,1) + (1/3)(-3,3,3) = (2,0,1) + (-1,1,1) = (1,1,2)
ans = (1,1,2)
==========
A(2,0,1), B(-1,3,4)
|AB| = √(2+1)² + (0-3)² + (1-4)²
|AB| = √(3² + 3² + 3²)
|AB| = 3√3
========
C(-1,3,0), D(4,-2,5)
|CD| = √(-1-4)² + (3+2)² + (0-5)²
|CD| = √(5² + 5² + 5²)
|CD| = 5√3
========
AB = OB - OA = (-3,3,3)
CD = OD - OC = (5,-5,5)
AB.CD = (-3,3,3).(5,-5,5) = -3.5 + 3.(-5) + 3.5 = -15 -15 + 15 = -15
AB.CD = -15
=========
AB.CD = |AB||CD|.sin@
-15 = 3√3.5√3.sin@
-15 = 45.sin@
sin@ = -1/3
@ = 109.47 deg
smallest angle between lines is 180 - @ = 70.53 deg
=======================================
2)
A(7,-1,3), B(10,-2,2)
r= i + j + k + λ( 3 i - j + k)
r = (1+3λ
i + (1-λj + (1+λk
a)
If A is on the line r, then the coords of A satisfy the eqn of r, i.e.
(1+3λi + (1-λj + (1+λk = 7i -j + 3k
equating coeffts,
1+3λ = 7
1-λ = -1
1+λ = 3
solving for λ, we get,
λ = 2
Hence A lies on the line l
==================
b)
|AB| = √(10 - 7)² + (-2 + 1)² + (2 - 3)²
|AB| = √(9 + 1 + 1)
|AB| = √11
========
c)
get two points, C and D,say, on the line l by letting λ=0 and λ=1,say.
λ = 0 -> C = (1,1,1)
λ = 1 -> D = (4,0,2)
|CD| = √(3² + 1² + 1²) = √11
|CD| = √11
=========
AB = OB - OA = (3,-1,-1)
CD = OD - OC = (3,-1,1)
AB.CD = (3,-1,-1).(3,-1,1)
AB.CD = 9 + 1 - 1 = 9
AB.CD = 9
========
AB.CD = |AB||CD|.cos@
9 = √11√11.cos@
cos@ = 9/11
@ = 35 deg
=========
d)
I'm not sure if this is the right method, but it gets the answer. Perhaps someone else can give a better solution.
Let P be the point on the line l such that BP is perpindicular to l.
P lies on l, so p is given by,
P = (1,1,1) + λ(3,-1,1) = (1+3λ, 1-λ, 1+λ
A(7,-1,3), B(10,-2,2)
BP = OP - OB = (-9 +3λ, 3-λ, -1+λ
AP = OP - OA = (-6+3λ, 2-λ, -2+λ
If BP is perpindicular to AP then BP.AP = 0
BP.AP = (-9 +3λ, 3-λ, -1+λ.(-6+3λ, 2-λ, -2+λ = 0
54 - 27λ - 18λ + 9λ² + 6 - 5λ + λ² + 2 - 3λ + λ² = 0
11λ² -53λ + 62 = 0
(11λ - 31)(λ - 2) = 0
λ = 31/11, λ = 2
=============
The solution, λ=2, gives the point A so ignore this solution
For λ = 31/11,
P = (1,1,1) + (31/11)(3,-1,1)
P = (9.4545, -1.8182, 3.8182)
B(10,-2,2)
|BP| = √(10 - 9.4545)² + (-2 + 1.8182)² + (2 - 3.8182)²
|BP| = √(0.2975 + 0.03306 + 3.3056)
|BP| = √3.6363
|BP| = 1.907
==========
A(7,-1,3), B(10,-2,2)
r= i + j + k + λ( 3 i - j + k)
r = (1+3λ
i + (1-λj + (1+λka)
If A is on the line r, then the coords of A satisfy the eqn of r, i.e.
(1+3λ
i + (1-λj + (1+λk = 7i -j + 3kequating coeffts,
1+3λ = 7
1-λ = -1
1+λ = 3
solving for λ, we get,
λ = 2
Hence A lies on the line l
==================
b)
|AB| = √(10 - 7)² + (-2 + 1)² + (2 - 3)²
|AB| = √(9 + 1 + 1)
|AB| = √11
========
c)
get two points, C and D,say, on the line l by letting λ=0 and λ=1,say.
λ = 0 -> C = (1,1,1)
λ = 1 -> D = (4,0,2)
|CD| = √(3² + 1² + 1²) = √11
|CD| = √11
=========
AB = OB - OA = (3,-1,-1)
CD = OD - OC = (3,-1,1)
AB.CD = (3,-1,-1).(3,-1,1)
AB.CD = 9 + 1 - 1 = 9
AB.CD = 9
========
AB.CD = |AB||CD|.cos@
9 = √11√11.cos@
cos@ = 9/11
@ = 35 deg
=========
d)
I'm not sure if this is the right method, but it gets the answer. Perhaps someone else can give a better solution.
Let P be the point on the line l such that BP is perpindicular to l.
P lies on l, so p is given by,
P = (1,1,1) + λ(3,-1,1) = (1+3λ, 1-λ, 1+λ
A(7,-1,3), B(10,-2,2)
BP = OP - OB = (-9 +3λ, 3-λ, -1+λ
AP = OP - OA = (-6+3λ, 2-λ, -2+λ
If BP is perpindicular to AP then BP.AP = 0
BP.AP = (-9 +3λ, 3-λ, -1+λ
.(-6+3λ, 2-λ, -2+λ = 054 - 27λ - 18λ + 9λ² + 6 - 5λ + λ² + 2 - 3λ + λ² = 0
11λ² -53λ + 62 = 0
(11λ - 31)(λ - 2) = 0
λ = 31/11, λ = 2
=============
The solution, λ=2, gives the point A so ignore this solution
For λ = 31/11,
P = (1,1,1) + (31/11)(3,-1,1)
P = (9.4545, -1.8182, 3.8182)
B(10,-2,2)
|BP| = √(10 - 9.4545)² + (-2 + 1.8182)² + (2 - 3.8182)²
|BP| = √(0.2975 + 0.03306 + 3.3056)
|BP| = √3.6363
|BP| = 1.907
==========
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