To determine the direction of u × v we have the right hand rule; Fingers point the
first vector, Palm points the second vector, tumb gives the direction of the cross product.
Theorem 3.5.1 (Relationship involving Cross Product and Dot Product:) If u,v and w are vectors in 3-space, then
(a) u · (u × v) = 0 (that is u × v and u are orthogonal)
(b) v · (u × v) = 0 (that is u × v and v are orthogonal)
(c) ||u × v||2 = ||u||2 ||v||2 − (u · v)2 (Lagrange’s Identity)
(d) u × (v × w) = (u · w)v − (u · v)w
(e) (u × v) × w = (u · w)v − (v · w)u
Proof: Write u = (u1 , u2 , u3 ), v = (v1 , v2 , v3 ) and apply the definitions of dot product and cross product.
Importance:
• From part (a) and (b) we see that if one needs to find a vector that is orthogonal
to both vectors u and v then u × v will do the work.
• We will see that this property can be used to write a plane equation.
i j k
Example: Let u = (1, 0, 2), v = (−1, 1, 3) then u×v = 1 0 2 = −2i−5j+k =
−1 1 3
(−2, −5, 1)
(u × v) · u = (−2, −5, 1) · (1, 0, 2) = 0, (u × v) · v = (−2, −5, 1) · (−1, 1, 3) = 0.
Theorem 3.5.2 (Properties of Cross Product): If u,v and w are any vectors
in 3-space and k is any scalar, then
(a) u × v = −(v × u)
(b) u × (v + w) = (u × v) + (u × w)
(c) (u + v) × w = (u × w) + (v × w)
(d) k(u × v) = (ku) × v = u × (kv)
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(e) u × 0 = 0 × u = 0
(f) u × u = 0
Geometric Interpretation of Cross Product and Determinants:
Theorem 3.5.3 (Area of a Parallelogram) If u and v are vectors in 3-space,
then ||u × v|| is equal to the area of the parallelogram determined by u and v.
Why ||u × v|| = ||u||||v|| sin (θ)?
From Theorem 3.5.1 we have Lagrange’s Identity ||u × v||2 = ||u||2 ||v||2 − (u · v)2 and
u·v
we have cos (θ) = ||u||||v||
⇒ u · v = ||u||||v|| cos (θ). Then
||u × v||2 =
=
=
=
||u||2 ||v||2 − (u · v)2
||u||2 ||v||2 − ||u||2 ||v||2 cos2 (θ)
||u||2 ||v||2 (1 − cos2 (θ))
||u||2 ||v||2 sin2 (θ)
⇒ ||u × v|| = ||u||||v|| sin (θ)
Theorem 3.5.4:
u1 u2
(a) The absolute value of the determinant det
is equal to the area of the parv1 v2
allelogram in 2-space determined by the vectors u = (u1 , u2 ) and v = (v1 , v2 ).


u1 u2 u3
(b) The absolute value of the determinant det  v1 v2 v3  is equal to the volume
w1 w2 w3
of the parallelopiped in 3-space determined by the vectors u = (u − 1, u2 , u3 ) and
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v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 ).
u1 u2 u3 Definition: v1 v2 v3 = u·(v×w) is called the scalar triple product of u,v and w.
w1 w2 w3 Theorem 3.5.5: If the vectors u = (u1 , u2 , u3 ), v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 )
have the same initial point, then they lie in the same plane if and only if
u1 u2 u3 u · (v × w) = v1 v2 v3 = 0
w 1 w 2 w 3 Section 3.4, The Geometry of Linear Systems
We will write equation of a line and a plane:
For line we will either use a point and a vector or 2 points to write a line equation.
Point + vector: In this case we want to write the equation of a line passing through
a point say x0 and in the direction of a vector say v
How to write this line L?
Let x be a general point on the line that we want to write the equation for. Then the
vector x~0 x = x − x0 must be parallel to the vector v. This means x − x0 = tv for some
scalar t and this equation describes the line
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Definition: The vector equation of the line that passes through the point x0 and
parallel to the vector v is given by
x = x0 + tv
Example: Find a vector equation of the line that passes through P0 (1, 2, −3) and
parallel to v = (4, −5, 1)
Solution: (x, y, z) = (1, 2, −3) + t(4, −5, 1) is the vector equation of the line.
• Expressing each coordinate of the points on the line gives the parametric equations
of the line, that is
x = 1 + 4t, y = 2 − 5t, z = −3 + t are the parametric equations of the line described
in the above example.
In general,
Definition: If x0 and v are vectors in Rn , and if v is nonzero then we define the line
through x0 that is parallel to v to be the set of all vectors x in Rn that are expressible in
the form
x = x0 + tv, −∞ < t < ∞
point+ vector in R2 : We can also find a line equation with a given point and a
vector perpendicular to the line. Let n = (a, b) be perpendicular to the line we want to
write the equation for, n is called normal vector of the line. Let x0 = (x0 , y0 ) be a given
point on the line then for a general point on the line say x = (x, y) we can form a vector
x~0 x = (x − x0 , y − y0 ) which is perpendicular to n. Hence the equation of the line is
given by
n · x~0 x = a(x − x0 ) + b(y − y0 ) = 0 ⇒ ax + by = c point-normal equation of the line
2 points case: In this case we will write the equation of a line that passes through
two points P0 , P1 .
First step is to form the vector between P0 and P1 , P0~P1 = P1 − P0 . Then choose one
of the points, say P0 to use the line equation we had above, that is
x = P0 + tP0~P1 = P0 + t(P1 − P0 ) = (1 − t)P0 + tP1 , −∞ < t < ∞
Example: Find the vector and parametric equations of the line determined by the
points (1, −1, 1) and (2, 1, 1).
Solution: vector equation of the line : x = (1, −1, 1) + t(1, 2, 0),
parametric equations of the line: x = 1 + t, y = −1 + 2t, z = 1, −∞ < t < ∞
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Plane equation: For plane equation we will use either a point and a special vector
or a point and 2 vectors.
point + special vector: special vector is the vector that is perpendicular to every
vector in the plane. This vector is called the normal vector of the plane and is denoted by n.
Take P = (x0 , y0 , z0 ) be a given point, take Q = (x, y, z) a general point on the
plane. Assume n = (A, B, C) is given then v = P~Q = (x − x0 , y − y0 , z − z0 )⊥n =
(A, B, C) ⇒ n · v = 0
A(x − x0 ) + B(y − y0 ) + C(z − z0 ) = 0 is the point-normal equation of the plane
If we expand parantheses we get
Ax + By + Cz = D general equation of the plane
Example: Find the general equation of the plane that passes through the point
P (−1, −1, −1) and has normal n = (3, 2, 1).
Solution: 3(x+1)+2(y +1)+1(z +1) = 0 is the point-normal equation of the plane,
3x + 2y + z = −6 is the general equation of the plane.
Point + 2 vectors: Assume v1 , v2 are given vectors parallel to the plane and
v1 6= tv2 . Let x0 be a given point on the plane. We will create a parallelogram with
adjacent sides t1 v1 and t2 v2 with −∞ < t1 < ∞ and −∞ < t2 < ∞
⇒ x~0 x = t1 v1 + t2 v2 ⇒ x = x0 + t1 v1 + t2 v2 is the vector equation of the plane
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Expressing each coordinate of the general point on the plane, we will get the parametric
equation of the plane, that is say x = (x, y, z), x0 = (x0 , y0 , z0 ), v1 = (a1 , b1 , c1 ),
v2 = (a2 , b2 , c2 )
x = x0 +t1 a1 +t2 a2 , y = y0 +t1 b1 +t2 b2 , z = z0 +t1 c1 +t2 c2 are parametric equations of the plane
In general,
Definition: If x0 , v1 , v2 are vectors in Rn , and if v1 and v2 are nonzero that are not
scalar multiples of one another, then we define the plane through x0 that is parallel to v1
and v2 to be the set of vectors x in Rn that are expressible in the form
x = x0 + t1 v1 + t2 v2 , −∞ < t1 < ∞, −∞ < t2 < ∞
Example: Find parametric equations of the plane that is parallel to the plane
3x + 2y − z = 1 and passes through the point P(1,1,1).
Solution: 3x+2y-z=4 is the general equation of the plane and y = t1 , z = t2 , x =
+ t32 + 43 are parametric equations of the plane (Note: Parametric equations are not
unique, one may get different parametric equations).
−2
t
3 1
Theorem 3.3.4:
(a) In R2 the distance D between the point P0 (x0 , y0 ) and the line ax+by+c=0 is
D=
|ax0 + by0 + c|
√
a2 + b 2
(b) In R3 the distance D between the point P0 (x0 , y0 , z0 ) and the plane ax+by+cz+d=0
is
|ax0 + by0 + cz0 + d|
√
D=
a2 + b 2 + c 2
Example: Find the distance between the point (−3, 1) and the line 4x + 3y + 4 = 0.
Solution: 1st way: we can use the formula given in Theorem 3.3.4, and get D = 1.
2nd way: using projections, Take a point on the line, (−1, 0) then form a vector between
(−1, 0) and (−3, 1), denoted by v = (−2, 1) and note that normal vector of the line is
(4, 4). Then D = ||projn v|| = 1.
Example: Find the distance between the point (3, 1, −2) and the plane x + 2y − 2z = 4.
Solution: 1st way: we can use the formula given in Theorem 3.3.4 and get D = 53 .
2nd way: using projections; (0, 1, −1) is a point on the plane form the vector between
(0, 1, −1) and (3, 1, −2) denoted by v = (3, 0, −1). Note that n = (1, 2, −2) is normal
vector of the plane. Then D = ||projn v|| = 53 .
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Example: Find the distance between the parallel planes 2x − y − z = 5 and
−4x + 2y + 2z = 12.
Solution: Take a point in one of the planes and apply the distance formula between
a point and a plane. Or one can find two points one from each plane and form a vector
in between and the take projection on the corresponding normal vector. In either case
D = √116
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