3 – dimensional geometry – day – 4
Equation Of A Line In Three Dimensions
Equation of a line is defined as y= mx+c, where c is the y-intercept and m is the slope. Vectors
can be defined as a quantity possessing both direction and magnitude. Position vectors simply
denote the position or location of a point in the three-dimensional Cartesian system with respect
to a reference origin. Further, we shall study in detail about vectors and Cartesian equation of a
line in three dimensions. It is known that we can uniquely determine a line if:
It passes through a particular point in a specific direction, or
It passes through two unique points
Let us study each case separately and try to determine the equation of a line in both the given
cases.
Equation of a Line passing through a point and parallel to a vector
Let us consider that the position vector of the given point be a⃗ with respect to the origin. The
line passing through point A is given by l and it is parallel to the vector k⃗ as shown below. Let
us choose any random point R on the line l and its position vector with respect to origin of the
rectangular co-ordinate system is given by r⃗ .
Since the line segment, AR¯¯¯¯¯¯¯¯ is parallel to vector k⃗ , therefore for any real number α,
AR¯¯¯¯¯¯¯¯ = α k⃗
Also, AR¯¯¯¯¯¯¯¯=OR¯¯¯¯¯¯¯¯ – OA¯¯¯¯¯¯¯¯
Therefore, α r⃗ = r⃗ – a⃗
From the above equation it can be seen that for different values of α, the above equations give
the position of any arbitrary point R lying on the line passing through point A and parallel to
vector k. Therefore, the vector equation of a line passing through a given point and parallel to a
given vector is given by:
r⃗ = a⃗ + αk⃗
If the three-dimensional co-ordinates of the point ‘A’ are given as (x , y , z ) and the direction
1 1 1
cosines of this point is given as a, b, c then considering the rectangular co-ordinates of point R as
(x, y, z):
Substituting these values in the vector equation of a line passing through a given point and
parallel to a given vector and equating the coefficients of unit vectors i, j and k, we have,
<
Eliminating α we have:
This gives us the Cartesian equation of line.
Equation of a Line passing through two given points
Let us consider that the position vector of the two given points A and B be a⃗ and b⃗ with respect
to the origin. Let us choose any random point R on the line and its position vector with respect to
origin of the rectangular co-ordinate system is given by r⃗ .
Point R lies on the line AB if and only if the vectors AR¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯ are collinear.
Also,
AR¯¯¯¯¯¯¯¯ = r⃗ – a⃗
AB¯¯¯¯¯¯¯¯ = b⃗ – a⃗
Thus R lies on AB only if;
r⃗ –a⃗ =α(b⃗ –a⃗ )
Here α is any real number.
From the above equation it can be seen that for different values of α, the above equation gives
the position of any arbitrary point R lying on the line passing through point A and B. Therefore,
the vector equation of a line passing through two given points is given by:
r⃗ =a⃗ +α(b⃗ –a⃗ )
If the three-dimensional coordinates of the points A and B are given as (x , y , z ) and (x , y , z )
1 1 1 2 2 2
then considering the rectangular co-ordinates of point R as (x, y, z)
Substituting these values in the vector equation of a line passing through two given points and
equating the coefficients of unit vectors i, j and k, we have
Eliminating α we have:
This gives us the Cartesian equation of a line.
The Vector Equation of a Line
You're already familiar with the idea of the equation of a line in two dimensions: the line
with gradient m and intercept c has equation
y=mx+c
When we try to specify a line in three dimensions (or in n dimensions), however, things
get more involved. It can be done without vectors, but vectors provide a really clear and
quick way into the problem.
So, let's work in three dimensions. How much information is needed in order to specify a
straight line? The answer is that we need to know two things: a point through which the
line passes, and the line's direction. Both of those things can be described using
vectors.
Figure 1: straight line through the point A (with position vector a), parallel to the
vector d
Figure 1 shows the straight line through the point A (with position vector a), parallel to
the vector d. How do we get from the origin, O, to some general point P on the line
(where P has position vector r)? One answer is that we first get to the point A, by
travelling along the vector a, and then travel a certain distance in the direction of the
vector d. If the position vector of P is r, this implies that for some value of ,
r=a+ d
So for example, the line through the point 2 0 −1 parallel to the vector −3i+j−k has
equation
r=2i−k+ −3i+j−k
TYPE:1. The Cartesian equation of a line is: 2x – 3 = 3y + 1 = 5 – 6z. Find the vector
equation of a line passing through (7, -5, 0) and parallel to the given line.
Solution : Given equation of the line is: 2x – 3 = 3y + 1 = – 6z + 5
TYPE 2 :
1. Determine the vector equation of the straight line passing through the point with position
vector i − 3j + k and parallel to the vector, 2i + 3j − 4k.
Express the vector equation of the straight line in standard cartesian form.
Solution The vector equation of the straight line is r = i − 3j + k + t(2i + 3j − 4k)
or xi + yj + zk = (1 + 2t)i + (−3 + 3t)j + (1 − 4t)k.
Eliminating t from each component, we obtain the cartesian form of the straight line,
x – 1/ 2 = y + 3 /3 = z – 1/ −4 .
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