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c. If this equation represents the same line as the equation in part a., it is necessary
for the two lines to have the same direction and contain the same set of points.
The line has Q 14 as its direction vector.
3 , 4R
3
The two lines will have the same direction vectors because
2 Q14 , 4R 17, 62.
3
The two lines have the same direction, and if these lines have a point in
common, then the equations represent the same line. The easiest approach is
to substitute 115, 72 into the first equation to see if this point is on the line.
Substituting gives 115, 72 11, 52 s17, 62 or 15 1 7s and
7 5 6s. Since the solution to both of these equations is s 2, the
point 115, 72 is on the line, and the two equations represent the same line.
r! 115, 72 t Q 14 3 , 4R, tR, CHAPTER 8
In the next example, vector properties will be used to determine equations for
lines that involve perpendicularity.
EXAMPLE 4
Selecting a strategy to determine the vector equation of a
perpendicular line
Determine a vector equation for the line that is perpendicular to
r ! 14, 12 s13, 22, sR, and passes through point P16, 52.
Solution
The direction vector for the given line is m ! 13, 22, and this line is drawn
through 14, 12, as shown in red in the diagram. A sketch of the required line,
passing through 16, 52 and perpendicular to the given line, is drawn in blue.
6
y
P(6, 5)
4
m = (– 3, 2)
2 (4, 1)
x
–4 –2 0 2 4 6 8
–2
r = (4, 1) + s(–3, 2)
–4
–6
Let the direction vector for the required blue line be v ! 1a, b2. Since the direction
vector of the given line is perpendicular to that of the required line,
Therefore,
or 3a 2b 0.
The simplest integer values for a and b, which satisfy this equation, are a 2 and
b 3. This gives the direction vector and the required vector equation for
the perpendicular line is r ! 12, 32
16, 52 t12, 32, tR.
1a, b2 # 13, 22 0
v ! # m
! 0.
NEL
431