Download Suggested problems

Suggested problems - solutions
Orthogonal vectors
P1: Identify which pairs of vectors are orthogonal.
(a) < 1, 3 > and < −3, −1 >
< 1, 3 > · < −3, −1 >= 1(−3) + 3(−1) = −6 6= 0
Not orthogonal.
(b) < 1, 3 > and < 10, − 10
3 >
< 1, 3 > · < 10, −
10
10
>= 1(10) + 3(− ) = 0
3
3
Orthogonal.
(c) < −2, 4, 1 > and < 4, 1, −2 >
< −2, 4, 1 > · < 4, 1, −2 >= −2(4) + 4(1) + 1(−2) = −6 6= 0
Not orthogonal.
(d) < x, y > and < −y, x >
< x, y > · < −y, x >= −xy + xy = 0
Orthogonal for any values of x and y.
(e) < a, b, a + b > and < 1, 1, −1 >
< a, b, a + b > · < 1, 1, −1 >= a + b − (a + b) = 0
Orthogonal for any values of a and b.
P2: Find a vector which is orthogonal to v =< −1, 4 >.
You can do this by inspection easily enough (< 4, 1 > is the obvious answer), but the formal
way is to set up an equation; let w =< x, y >, and we want
< −1, 4 > · < x, y >= −x + 4y = 0
That’s one equation with two unknowns, so we have a free parameter - let y = t, and then
x = 4y = 4t.
Any vector in the form < 4t, t > is orthogonal to < −1, 4 >. Picking various values for t, such
as t = 1, would generate an infinite number of such vectors.
P3: Find a unit vector which is orthogonal to v =< 3, 7 >.
The key word here is unit vector - normalize your answer.
Inspection immediately gives a vector of < −7, 3 > (or you could write the equation as before; a
parameterized answer is x = t, y = − 73 t). Then, normalize the answer; a unit vector orthogonal
to < 3, 7 > is
1
1
−7
3
q
< −7, 3 >= √ < −7, 3 >=< √ , √ >
58
58 58
(−7)2 + 32
Although there are an infinite number of vectors orthogonal to < 3, 7 >, there are only two unit
−3 >.
vectors that are: the one given above and the one in the opposite direction, < √7 , √
58 58
Additional problems of this type at COW:
3.Calculus Book
> 2.Vectors and
> 2.Dot Product
> 5.Unit Vector
III
Analytic Geometry
in the Plane
Orthogonal to a Given Vector (2-d)
P4: Find two vectors which are orthogonal to v =< 1, −4, 7 >, and are not scalar multiples of each
other.
Let w =< x, y, z >, and we want
< 1, −4, 7 > · < x, y, z >= x − 4y + 7z = 0
That gives us two free parameters to work with - let y and z be arbitrary, and then x = 4y − 7z.
Pick values for y and z; say y = 1, z = 1, and so x = −3. The vector < −3, 1, 1 > is orthogonal
to < 1, −4, 7 >.
Or, let y = 1, z = 2, and so x = −10. The vector < −10, 1, 2 > is orthogonal to < 1, −4, 7 >.
Note that < −3, 1, 1 > and < −10, 1, 2 > are not scalar multiples of each other.
Your answers will probably vary, as there are an infinite number of choices.
P5: Find a unit vector which is orthogonal to v =< 8, 1, 1 >.
First, find any vector orthogonal to < 8, 1, 1 > using 8x + y + z = 0. Letting x = 1 and y = 1
gives z = −9, and so < 1, 1, −9 > is orthogonal to < 8, 1, 1 >. Normalizing gives
1
√ < 1, 1, −9 >≈< .1098, .1098, −.9879 >
83
In the 2D case, we saw an infinite number of orthogonal vectors, but they all boiled down to two
unit vectors when normalized, and one of those was just the negative of the other. In 2D, all the
vectors orthogonal to a given vector are simply scalar multiples of each other - they may have
different magnitudes, but they all point in the same (or opposite) direction.
In 3D, as in this problem, that is no longer the case. There are an infinite number of vectors
orthogonal to < 8, 1, 1 >, all headed off in an infinite number of directions. Choosing different
parameters would lead to a different unit vector as well; for example, letting x = −1, y = 1 gives
z = 7 and the vector < −1, 1, 7 > normalizes to
1
√ < −1, 1, 7 >≈< −.1400, .1400, .9802 >
51
Additional problems of this type at COW:
3.Calculus Book
> 2.Vectors and
> 3.Dot Product
> 5.Unit Vector
III
Analytic Geometry
in Three Dimensions
Orthogonal to a Given Vector (3-d)
P6: Prove the following:
For vectors u, v, and w, suppose that u and w are orthogonal, and v and w are
orthogonal. Prove that the vector sum u + v is orthogonal to w as well.
Suppose that u and w are orthogonal, so u · w = 0, and v and w are orthogonal, so v · w = 0.
Then
(u + v) · w = u · w + v · w
= 0+0=0
as well, and u + v is orthogonal to w.
P7: Verify that the set
{< 1, −1, 0 >, < 2, 2, 5 >, < −5, −5, 4 >}
is an orthogonal set. Construct an orthonormal set based on this set.
An orthogonal set requires that the vectors be pairwise orthogonal: the dot product of any
pairing must be zero. You can quickly verify that
< 1, −1, 0 > · < 2, 2, 5 >= 0
< 1, −1, 0 > · < −5, −5, 4 >= 0
< 2, 2, 5 > · < −5, −5, 4 >= 0
all the possible pairings.
Then, to construct an orthonormal set, normalize the vectors:
1
1
2
2
5
5
5
4
{< √ , − √ , 0 >, < √ , √ , √ >, < − √ , − √ , √ >}
2
2
33 33 33
66
66 66