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Example 0.1. Vector equation of a line.
Find a vector equation for the line through (4, 6 , −3) and parallel to v = 5i − 10 j + 2k.
Solution: With the identifications x 0 = 4, y 0 = 6, z 0 = − 3 , a = 5, b = −10 and c = 2 we obtain from
< x, y, z >=< x(t), y(t), z(t) >=< x 0 + at, y 0 + bt, z 0 + ct >=< x 0 , y 0 , z 0 > +t < a, b, c >,
a vector equation of the line:
< x, y, z >=< 4 , 6 −, 3 > +t < 5 , − 10 , 2 >.
Example 0.2. Vector equation of a line.
Find a vector equation for the line though (2, − 1 , 8) and (5, 6 , −3).
Solution: If we label the points as P 0 (2, − 1 , 8) and P 1 (5, 6 , −3), then a direction vector for the line through P 0 and P 1 is
v =
P 0 P 1 =
OP 1 −
OP 0 =< 5 − 2 , 6 − (−1), − 3 − 8 >=< 3 , 7 , − 11 >.
From the vector equation of that line
r = r 0 + t(r 1 − r 0 ),
(here r =
OP , r 0 =
OP 0 , r 1 =
OP 1 ) a vector equation of the line is < x, y, z >=< 2 , − 1 , 8 > +t < 3 , 7 , − 11 >.
This is one of many possible equations of the line. For example, two alternative equations are
< x, y, z > =< 5 , 6 , − 3 > +t < 3 , 7 , − 11 >, < x, y, z > =< 5 , 6 , − 3 > +t < − 3 , − 7 , 11 >.
Example 0.3. Parametric equation of a line.
Find a parametric equation for the line through (5, 2 , 4) parallel to v = 4i + 7j − 9 k.
Solution:
Make identifications x 0 = 5, y 0 = 2, z 0 = 4, a = 4, b = 7 and c = −9 so that from the parametric equation x(t) = x = x 0 + at, y(t) = y = y 0 + bt, z(t) = z = z 0 + ct,
we get x = 5 + 4t, y = 2 + 7t, z = 4 − 9 t.
Definition 0.1. Two lines L 1 and L 2 with direction vectors v 1 and v 2 , respectively, are i) perpendicular if v 1 · v 2 = 0, and ii) parallel if v 2 = kv 1 for some non-zero scalar k.
Example 0.4. Perpendicular lines.
Determine whether the lines
(0.1) L 1 : x = − 6 − t, y = 20 + 3t, z = 1 + 2t
(0.2) L 2 : x = 5 + 2s y = − 9 − 4 s, z = 1 + 7s
are perpendicular. 1
Solution: Reading off the coefficients of the parameters t and s, we see that
v 1 = −i + 3j + 2k
and v 2 = 2i − 4 j + 7k
are the direction vectors for L 1 and L 2. Because v 1 · v 2 = − 2 − 12 + 14 = 0, we conclude that the lines are perpendicular.
Example 0.5. Parallel lines.
Determine whether the lines
(0.3) L 1 : x = 4 − 2 t, y = 1 + 4t, z = 3 + 10t
L 2 : x = s y = 6 − 2 s, z =
(0.4) − 5 s
are parallel.
Solution: Reading off the coefficients of the parameters t and s, we see that
v 1 = − 2 i + 4j + 10k
and v 2 = i − 2 j − 5 k
are the direction vectors for L 1 and L 2. Because v 1 = − 2 v 2 , we conclude that the lines are parallel.
Example 0.6. Equation of a plane.
Find an equation of the plane that contains the point (4, − 1 , 3) and is perpendicular to the vector n = 2i + 8j − 5 k.
Solution: It follows immediately from the equation of the plane containing P 0 (x 0 , y 0 , z 0 ) and with normal vector n = ai + bj + ck, that is,
a(x − x 0 ) + b(y − y 0 ) + c(z − z 0 ) = 0,
that the identification x 0 = 4, y 0 = − 1 , z 0 = 3, a = 2, b = 8, c = −5 yields
2(x − 4) + 8(y + 1) − 5(z − 3) = 0
Example 0.7. Normal vector a plane.
The graph of a linear equation ax + by + cz + d = 0, with a, b, c NOT all zero, is a plane with normal vector n = ai + bj + ck.
For instance by reading off the coefficients x, y, z in the linear equation 3 x − 4 y + 10z − 8 = 0, we obtain the normal vector n = 3i − 4 j + 10k Of course a non-zero scalar multiple of a normal vector n is still perpendicular to the plane.
Three non collinear points P 1 , P 2 , P 3 also determine a plane S. To obtain an equation of the plane, we need only form two vectors between two pairs of the points. The cross product is a vector normal to the plane containing these vectors. If P (x, y, z) represents any point on the plane and
Example 0.10. Point of intersection.
Find the point of intersection of the plane 3 x− 2 y +z = − 5 and the line x = 1+t, y = −2+2t, z = 4 t.
Solution: If (x 0 , y 0 , z 0 ) denotes the point of intersection, then we must have
3 x 0 − 2 y 0 + z 0 = − 5 , x 0 = 1 + t 0 , y 0 = −2 + 2t, z = 4t 0 ,
for some number t 0. Substituting the latter equations into the equation of the plane gives
3(1 + t 0 ) − 2(−2 + 2t 0 ) + 4t 0 = − 5
or t 0 = − 4.
From the parametric equations for the lines, we then obtain x 0 = − 3 , y 0 = −10 and z 0 = −16. So the point is (− 3 , − 10 , −16).
Example 0.11. Distance D from a point to a plane.
Let P 0 (x 0 , y 0 , z 0 ) be a point on the plane ax + by + cz + d = 0 and let n be a normal vector to the plane.
Solution: If P 1 (x 1 , y 1 , z 1 ) is any point not on the plane, then the distance D from a point to a plane is given by
D =
|ax 1 + by 1 + cz 1 + d| √ a^2 + b^2 + c^2 Homework: prove this formula.
Example 0.12. Distance D from a point to a plane.
Let P 0 (2, 1 , 4) be a point on the plane x − 3 y + z − 6 = 0. Find the distance from P 0 to the plane. Solution: Do this as homework.
Example 0.13. Angle between planes.
The angle between two planes is defined to be the acute angle between their normal vectors.
