Course: Engineering Mathematics I (MATH 1180) Group I: CIV, ELEC, SURV. Lecturer: Mr. Oral Robertson Solutions to Assignment #3 : VECTORS Date: October 15, 2010
MATH 1180: Solutions to Assignment #3
- (a) Prove that the diagonals of a rhombus meet at right angles.
Hint: Let two adjacent sides of the rhombus be represented by the vectors a and b.
First, a rhombus is a parallelogram having all four sides being of equal length.
Consider the rhombus OARB shown below. Its diagonals are OR and AB, having directions obtained from standard vector sum and difference procedure, as OR = a + b and AB = b – a, respectively.
If the diagonals meet at right angles, then the dot product of their directions would equal zero.
ORAB = (a + b )(b–a )
= ab – aa + bb – ba
= bb – aa
= |b|2 – |a|2 , but |b|= |a| ….. equal sides, hence:
ORAB = |a|2 – |a|2
ORAB = 0 ….. proven.
(b)Prove that the altitudes of a triangle are concurrent.
Consider a general triangle ABC as drawn above. Assume that two of the three altitudes, through vertices A and B meet at the point H. So, AH is perpendicular to side BC, while BH is perpendicular to side AC. The third altitude (through C) is concurrent (at H) with the other two altitudes if it can be shown that CH is perpendicular to side AB.
Now, let’s put what we have in vector language:
Firstly, AH is perpendicular to side BC:
AHBC = 0
(h – a)(c – b) = 0
hc – hb – ac + ab = 0……(r1)
Secondly, BH is perpendicular to side AC:
BHAC = 0
(h – b)(c – a) = 0
hc – ha – bc + ba = 0……(r2)
Subtracting the two equations, r1 – r2 , gives:
0 – hb + ha – ac + bc + 0 = 0
h(a – b) – c(a – b) = 0
Factorizing: (h– c)(a – b) = 0
CHBA= 0
CHAB= 0…… hence CH is perpendicular to side AB (proven).
Therefore, the three altitudes of any triangle ABC, are concurrent.■
(c)Show that if R is a point lying on a line segment AB so that it divides AB internally in the ratio :, then the position vector of R is given by
Basically, it is given that , which implies: AR = RB. And since AR and RB are parallel lines then it will also be true that: AR = RB.
And, denoting the position vector of A, B, and R by a, b, and rrespectively then the above equation gives: (r – a) = (b – r)
Expanding:r – a = b – r
r + r = a + b
( + )r= a + b
r = ……. shown.■
- It is known that if four points A, B, C, and D are coplanar then this means that the three displacement vectors AB, AC, and AD (formed from the four points) lie on the same plane and their triple scalar product will be equal to zero.
(i)Show, therefore, that if those four points having position vectors a, b, c, and d, respectively, are indeed coplanar, then:
(a x b + b x c + c x a) . d = a . b x c
With a triple scalar product of zero (as a condition for coplanarity), then:
AB × ACAD= 0
(b – a)×(c – a)(d – a) = 0
And expanding the cross product on the L.H.S. :
(b×c – b×a– a×c + a×a)(d – a) = 0
(b×c + a×b + c×a + 0)(d – a) = 0
(a×b + b×c + c×a)(d – a) = 0
(a×b + b×c + c×a)d – (a×b + b×c + c×a)a = 0
Watching carefully what is required in the proof, we expand the second brackets:
(a×b + b×c + c×a)d – (a×ba + b×ca + c×aa) = 0
And with vectors a×b and c×a being perpendicular to vector a, then we simplify to get:
(a×b + b×c + c×a)d – (0+ b×ca + 0) = 0
(a×b + b×c + c×a)d= b×ca
And the dot product on the R.H.S. is commutative, so that:
(a×b + b×c + c×a)d= ab×c …… shown.■
(ii)Using the expression obtained in part(i), show further that if the four points are coplanar, then :
a x (b – c) . d = b x (d – a) . c
Starting directly from the previous result of part (i):
(a×b + b×c + c×a)d= ab×c
And again watching what is required in the current proof, then, the above equation is manipulated accordingly to give:
(a×b + c×a)d = ab×c – b×cd ….. and since the first dot product on R.H.S. is commutative:
(a×b – a×c)d = b×ca – b×cd
Simple factorization on both sides gives:
a×(b – c)d = b×c(a –d)
Applying the first Triple Vector Product rule on the R.H.S., then we have:
a×(b – c)d = (a –d)×bc
a×(b – c)d = –b× (a –d)c
a×(b – c)d = b × (d – a) c……shown.■
- Two lines L1 and L2 have equations: L1: r = j + (i + 2j + 2k) and L2: r = i – k + (4j + 3k).
(a)Which one of the two lines is either parallel or perpendicular to the x-axis? Explain.
Let v1 and v2 denote the directions of lines L1 and L2 respectively. Therefore, we have:
v1 = i + 2j + 2k and v2 = 4j + 3k.
And noting that a direction of the x-axis is the vector i, then we can now investigate the issues of parallel and perpendicular lines.
Neither v1 nor v2 are scalar multiples of the x-axis direction i, therefore neither L1 nor L2 is parallel to the x-axis.
And for perpendicularity, we apply the dot product as follows:
v1i = (i + 2j + 2k)i = 1 ≠ 0 line L1 is not perpendicular to the x-axis.
v2i = (4j + 3k)i = 0 line L2 is perpendicular to the x-axis.■
(b)Calculate the angle between the two lines.
Using the same notations for the line directions as in part (a), then the required angle, , is given by: cos = = = 0.9333
= cos–1 (0.9333) = 21.0o.
(c)Show that the two lines are skew lines, and find the shortest distance between them.
Let r1 and r2 denote the position vectors of any general point on lines L1 and L2 respectively.
Then: r1 = j + (i + 2j + 2k) and r2 = i – k + (4j + 3k).
If both lines meet, then there will be a unique pair of and values for which r1 = r2. If there isn’t a unique pair of and values for which r1 = r2 , then the lines do not meet and are therefore skew lines. Let’s check to see:
r1 = r2 j + (i + 2j + 2k) = i – k + (4j + 3k)
i + (1 + 2)j + 2k = i + 4j + (–1 + 3)k
Equating alike unit vector coefficients on both sides:i: = 1 …Eqn.1
j: 1 + 2 = 4 …Eqn.2
k: 2 = –1 + 3 …Eqn.3
The procedure from here is to solve the first two equations (simultaneously), and then check if the pair of and values obtained satisfies the third equation.
Solving eqns. (1) and (2) gives: = 1 and = ¾ ;
Substituting both of these values into Eqn.(3) gives:2(1) = –1 + 3(¾)
2 ≠ 5/4 .
So there is no unique pair of and values for which r1 = r2 , so the lines do not meet, and are therefore skew lines. ■
(d)Find out which of the two lines is closer to the origin.
Despite the fancy diagram, we are basically seeking for comparison purposes, the shortest distance of a point (the origin O, in this case) from each of the two given straight lines, L1 and L2 . Let the respective shortest distances from the Origin be denoted by d1 and d2. Points N1 and N2 represent the foot of the perpendicular from O to the lines L1 and L2 , respectively.
And as usual for the shortest point-to-line distance we need just two things the direction of the line, and a displacement vector from the line to the point in space (O, in this case). And the vector protocol is that the angle (), for trigonometric purposes, should be drawn between the two known vectors as done in the diagram above.
For line L1:direction of line is v1=i + 2j + 2k, and a displacement vector is AO = –OA = –j .
So that using the formula d1 =
Now AO× v1 = –j × (i + 2j + 2k) =(i + 2j + 2k) × j
= i×j + 0 + 2k×j = k – 2i
= –2i + k.
d1 =
Similarly, for line L2 , its direction is v2=4j + 3k, and a displacement vector is CO = –OC = –i + k .
And we will use the corresponding formula d2 =
Now CO× v2 = (–i + k) × (4j + 3k) = –4i×j –3i×k + 4k×j + 0
= –4k + 3j – 4i
= –4i+ 3j –4k.
d2 =
And expressing the respective distances conveniently, we have: (units), and we see clearly that d1 < d2 , which implies that line L1 is closer to the origin.
*Alternatively, we need not use the standard cross product–based formula, for the shortest distance between a point and a straight line. Instead, we can use direct right-angle triangle geometry in conjunction with the much more straightforward dot product of the two known vectors (direction of line and displacement vector from line to point). Here’s the technique (with reference to the previous diagram):
From AON1: required distance, d1 = ON1 = AO sin1
But 1 = angle between vectors v1 and AO, therefore:
cos1 = =
sin1 =
d1 = AO sin1 =|AO| sin1 = (1) =
From CON2: required distance, d2 = ON2 = CO sin2
With 1 = angle between vectors v2 and CO, therefore:
cos2 = =
sin2 =
d2 = CO sin2 = |CO| sin2 = =
And the same conclusions are drawn, as before.
(e)Find the equations of the two planes 1 and 2 , where 1 contains the origin O and line L1 , while 2 contains the origin O and the line L2. Hint: To get the normal to a plane, just cross-product any two vectors that lie in the plane.
With reference to the same diagram drawn at the beginning of part (d) above, we see that two vectors lying in plane 1 (same as plane AON1) would be v1and AO , while similarly in plane 2 we have the vectors v2 andCO. Hence, denoting the respective normal to these two planes as n1 and n2, then we may write:
n1= AO × v1 = = –j × (i + 2j + 2k) = –2i + k…. already calculated from part (d).
n2= CO × v2 = = (–i + k) × (4j + 3k) = –4i+ 3j –4k…. was also evaluated from part (d).
And using each normal with a known point on the respective plane (use point O, for simplicity), we obtain the standard vector equations of the planes as follows:
Plane 1 : rn1 = On1 r(–2i + k) = O(–2i + k) r(–2i + k) = 0. Ans.
Plane 2 : rn2 = On2 r(–4i+ 3j –4k) = O(–4i+ 3j –4k) r–4i+ 3j –4k) = 0. Ans.
- Two planes 1 and 2 have equations: 1: x + y + z = 9 and 2: r.(4i –5j –2k) = –15.
(a)Find the angle between the two given planes.
The angle () between two planes 1 and 2 may be taken as the angle between their respective normals, n1 and n2 .
Here, n1 = i + j + k and n2 = 4i –5j –2k; so that:
cos = =
= cos–1(–0.2582) = 105.0o …Ans.
- Please note that in calculating the angle between two planes, the supplement of any angle obtained is also a correct answer. So that, in this case, 180o – 105.0o = 75.0o is also correct. However, this (supplement rule) is not so for the angle between two straight lines, due to the direction specificity of each line. In other words the angle between two lines is unique, while the angle between two planes (that are not perpendicular) is not unique but instead has two answers.
(b)Find the equation of the line of intersection between the planes: Use both the “n1×n2” method and the “free variable” method, and verify that the two answers obtained represent the exact (same) straight line.
(i) “n1×n2” method:
Remember that any line or vector lying in (or is parallel to) a plane is automatically perpendicular to the normal of the plane. Therefore since the required line (of intersection) is contained in each plane, then its direction (v) will be perpendicular to both normals (n1 and n2).
And as such we may write, on a direction basis, that:v= n1 × n2 .
v= (i + j + k)× (4i –5j –2k)
= = i(–2+5) –j(–2 – 4) + k(–5 – 4) = 3i + 6j – 9k
And now to get a point on the line of intersection, we first lay out the Cartesian equations of both planes: 1: x + y + z = 9 …….. Eqn.(1)
2: 4x –5y –2z = –15.…….. Eqn.(2)
Putting any numerical the value of one variable (say x = 0), into both equations, give:
y + z = 9, …….. Eqn.(3)
–5y –2z = –15…….. Eqn.(4)
Solving simultaneously, 2× Eqn.(3) + Eqn.(4) gives: –3y = 3 y = –1
And Eqn.(3) implies: z = 9 – (– 1) = 10. So, a point on the line of intersection is (0, –1, 10).
Therefore, a valid equation of the line of intersection is:
r = .
(ii) “free variable” method:
This method involves solving the two Cartesian equations simultaneously. The system will have two equations but three variables (x, y, and z), so one of these variables can be designated to be a free variable, in the solution.
1: x + y + z = 9…….. Eqn.(1)
2: 4x –5y –2z = –15.…….. Eqn.(2)
If we plan on making x the free variable, then we can start by eliminating either y or z. Let’s eliminate y, as follows: 5×Eqn.(1) + Eqn.(2) gives: 9x + 3z = 30 z = 10 – 3x.
Now, for y in terms of x:Eqn.(1) x + y + (10 – 3x) = 9 y = 9 – 10 + 2x y = –1 + 2x .
And, of course, x = x. So we have expressed all [three] variables in terms of the free variable x.
Therefore, letting x = , then y = –1 + 2, and z = 10 – 3 …… and these are the parametric equations of the line of intersection ( a fully valid form of the answer).
For the vector equation: recall that r= xi + yj + zk
= i + (–1 + 2)j + (10 – 3)k
r= –j + 10k + (i + 2j – 3k)
or: r = .
And comparing the two line equations from the two different methods, we see that both lines contain the same point (0, –1, 10), and are also parallel…… directions are scalar multiples of each other. Hence, with any two parallel lines containing a known common point, we conclude that those two lines are the exact same line.
(c)Obtain the equation of a new plane that is parallel to 1 and passes through the point (1, 5, 4).
Parallel planes have ‘equal’ normals. Hence normal (n) of new plane = normal of plane 1 .
That is, n = n1 n = i + j + k
And using known point on the plane as A(1, 5, 4), then required equation of new plane takes the form: rn = OAn r(i + j + k) =(i + 5j + 4k)(i + j + k)
r(i + j + k) = 1+ 5 + 4
r(i + j + k) = 10Ans.
(d)Obtain the equation of a straight line that is perpendicular to 1 and passes through (1, 5, 4).
Any straight line perpendicular to a plane has its direction parallel to the normal of the plane.
Hence, direction (v) of the line equals the normal (n1) of plane 1.
So, v = n1 = i + j + k.
And using known point on line as A(1, 5, 4), then the required vector equation of the line takes the standard form: r = OA + v
r = (i + 5j + 4k) + (i + j + k)Ans.
(e)What is the distance from the point (1, 5, 4) to the plane 2 ?
Now, a known point on the plane 2 is P(0, 3, 0)……. or any point that satisfies 4x –5y –2z = –15.
Hence a displacement vector from point in space A(1, 5, 4) to a point P(0, 0, 9) on the plane, is given by:
AP = OP – OA = 3j – (i + 5j + 4k) = –i – 2j – 4k
Now, n2 = 4i –5j –2k ….. normal to plane2 .
AP ň2 =
Therefore, using absolute value of the AP ň2result, then required distance: AN = units. ■
- Please note that in these “shortest distances” questions, you may label known or obtained points (eg. those on a given plane, line, or just a given point in space) anyhow you wish to aid in the organization of your solution process. In the preceding calculation, we chose to label the point on the plane as P and the given point in space as A just to be consistent with the earlier labeling as A(1, 5, 4) in parts (c ) and (d) of the question. This is very important when applying any [vector] distance formulae, the key is to know exactly what each vector label in the formula represents (especially the displacement vector which is a common ‘theme’ in these types of calculations.)
(f)Obtain the equation of the straight line that is parallel to both planes and passes through O.
Since any line parallel to (or lying in) a plane is perpendicular to the normal of the plane, then being parallel to both planes means that this straight line is perpendicular to both normals. Hence, its direction (v) is the cross product of both normals (n1 and n2).
So, v= n1 × n2 = (i + j + k)× (4i –5j –2k) = 3i + 6j – 9k …… already obtained from part (b).
Using point O(0, 0, 0) on the line, then its equation takes the form: r = O +v
r = (3i + 6j – 9k)or, simplifying the direction of the line, we get: r = (i + 2j – 3k).■
- Triangle ABC contains the points A(2,– 1, 4), B(3, 1, 6) , and C(0, –3, 3).
(a)Find the size of the angle at A.
Let the angle at A be denoted by , as depicted
in the diagram on the right.
So applying the dot product formula, then:
cos =
Now,
AB= OB – OA
=(3i + j + 6k)– (2i – j + 4k)
= i + 2j + 2k
AC= OC – OA
=(–3j + 3k)– (2i – j + 4k)
= –2i – 2j – k
|AB|= |i + 2j + 2k| = = 3
|AC|= |–2i – 2j – k| = = 3
ABAC= (i + 2j + 2k)(–2i – 2j – k)
= –2 – 4 – 2
= –8
Hence,cos = = –0.8889
Therefore, = cos–1 (–0.8889) = 152.7o Ans.
(b)Find the area of triangle ABC.
Area of ABC = ½ |AB AC|
Now, AB AC= (i + 2j + 2k)(–2i – 2j – k)
= = i(–2+ 4) –j(–1+4) + k(–2+4)= 2i– 3j +2k
Required Area = ½ |2i– 3j +2k| = ½= units2. ■
(c)Find the length of each of the three sides of the triangle. What kind of triangle is it?
For the length of the three sides AB, AC, and BC:
AB = |AB| = 3 units
AC = |AC| = 3 units
BC = |BC|
BC = OC – OB = (–3j + 3k) – (3i + j + 6k) = –3i – 4j – 3k
BC = |BC|= = units.
And with two of the three sides having equal lengths, the triangle is isosceles.■
(d)Find the equation of the plane ABC.
As usual, the normal (n) to any plane ABC is given by: n = AB AC.
And from part (b), we had the cross product AB AC = 2i– 3j +2k.
So using the normal,n = 2i – 3j + 2k, with any one of the three known points lying on the plane ABC, say C(0, –3, 3), then the required plane equation is given by: rn = OCn.
r(2i – 3j + 2k) =(–3j + 3k)(2i – 3j + 2k)
r(2i – 3j + 2k) =9 + 6
r(2i – 3j + 2k) =15.■
(e)If the point O is used as the vertex of a tetrahedron OABC (with ABC being the base), find the height (h) of the tetrahedron. Hint: This is just the distance of O from the plane ABC.
Note: The triple scalar product used must comprise three displacement vectors concurrent at any one of the four vertices of the tetrahedron. Specifically, we use any two base vectors (AB and AC in this case) and the third vector being the ‘slant height’ vector (AO in this case)….. all three of these vectors being concurrent through the vertex A.
(f)Useeither the pyramid formulaor one based on the triple scalar product to find the volume of the tetrahedron OABC.
Pyramid formula: V= ⅓ Area of Base x perpendicular height
= ⅓(Area of ABC) x h
And using the previous results from parts (b) and (e), then:
V= = units3.Ans.
Or, using the Triple Scalar Product (T.S.P.) route, then:
Volume, V = ⅙(AB× AC) ·AO
This form is chosen because we already have vector AB× AC from earlier calculations.
Therefore, with AB× AC =2i– 3j +2k…… from part(b),
and AO = – OA = –(2i –j + 4k) = –2i + j – 4k,
then, the triple scalar product:
(AB× AC) ·AO= (2i– 3j +2k)·(–2i + j – 4k)
= –4 – 3 – 8
= –15
Hence, V = ⅙–15 = units3.Ans.
- Note: The triple scalar product used should comprise threedisplacement vectors concurrent at any one of the four vertices of the tetrahedron. Specifically, we use any two base vectors (AB and AC in this case) and the third vector being the ‘slant height’ vector (AO in this case)….. all three of these vectors being concurrent through the vertex A.
- In finding the volume of the tetrahedron OABC using vectors (that is, one-sixth of the T.S.P.), it does not matter which two of any three defining vectors are used to represent the base. As such, we could have used faces OAB, OAC, or OBC (instead of ABC) to denote the base of the tetrahedron. These respective choices would mean that the volume of the tetrahedron is calculated by:
V = ⅙(OA × OB) · OC,V = ⅙(OA × OC) · OB, or V = ⅙(OB × OC) · OA , respectively.
- (a) Calculate the distance between:
(i)the two parallel lines having equations: r = j + (i + j – k) and x – 5 = y + 4 = 2 – z.
The equation of the second line in the standard Cartesian form is:
.
This means that a known point on the second line is C(5, –4, 2). Its direction is obviously i + j – k, just like the first line. A known point on the first line (based on its vector equation) is A(0, 1, 0).
These known points are illustrated in the diagram below.
The required distance between two parallel lines is essentially a “distance from a point to a line” problem. This is because all points on one line is equidistant from the other line, and vice versa. So, in the present context, if you observe the diagram carefully, the required distance between the two lines is equivalent to calculating the distance from the point A(0, 1, 0) to the second line; or the distance from the point C(5, –4, 2) to the first line.
So applying the point-to-line formula, using A(0, 1, 0) as the point in space and r = OC + v as the vector equation ofthe [second] line, then the required distance (CN, or D)is given by:
D =
Here, AC is just the displacement vector from one line to the other parallel line (in the context of the overall problem), while v is the direction of the line(s). In the pseudo-context hinted above, AC is the familiar displacement vector from the point (A) to the [second] line.
Now, AC= OC – OA
=(5i – 4j + 2k) – ( j )
AC = 5i – 5j + 2k
AndACv= (5i – 5j + 2k)(i + j – k)
=
= i(5 – 2) –j(–5 – 2) + k(5+5)
ACv = 3i+ 7j +10k
|ACv| = |3i+ 7j +10k| = =
And, | v| = |i+ j –k| = =
D = = or, D = 7.257 units. Ans.
Alternatively, if you are not too fond of the cross product formulae, then you may use the dot product via some trigonometry on ANC.
Observe, required distance CN is given by:CN = AC sin
But, = angle between displacement vector AC and the line direction v.
Hence: cos =
sin=
Now, AC = |AC|= |5i – 5j + 2k|= units
Therefore: CN = AC sin =
=
= ■
(ii)the two parallel planes having equations: r.(3i + 4k) = 5and 3x + 4z = –10.
The distance between two parallel planes is [logically] obtained by the difference in their numerical distances from any fixed reference point (usually the origin, for simplicity). Recall, that for any plane having equation r.n= d, its numerical distance from the origin is given by d/|n|. Note that the vector equation of the second plane is r.(3i + 4k) = –10, although its n and d values could easily have been read off from the given Cartesian equation (without converting to its vector equation).
Anyway, the numerical distance of the first plane from O is given by:
D1 = = 1 unit
Similarly, the numerical distance of the second plane from O is given by:
D2 = = –2 units …… do not take absolute value, as yet!