LECTURE 5 OF 8 Topic 5: VECTORS 5.5 Application of Vectors In Geometry.

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LECTURE 5 OF 8 Topic 5: VECTORS 5.5 Application of Vectors In Geometry

LEARNING OUTCOMES find a vector equation of the straight line in the form of where t is a scalar. (b) find the parametric and cartesian equations of a straight line.

The Equation of a Line In Space A line in space is a straight line. Suppose that R (x,y,z) is a point which is free to move on a line containing a fixed point A (a1,a 2,a3). If is a direction vector of the line NOTE :

, t is a scalar A vector equation of a line is : NOTE : To find a vector equation of a line, we have to determine a position vector of one point on the line and any vector which is parallel to the line, then use the above equation .

In the component form, it can be written as Equating the respective components, we obtain, x = a1 + tv1 y = a2 + tv2 z = a3 + tv3 The parametric equations of the line

x = a1 + tv1 y = a2 + tv2 z = a3 + tv3 tv1 = x - a1 tv2 = y – a2 tv3 = z – a3 Isolating t in each of these equations gives, The Cartesian equations of the line

Find the vector equation,the parametric Example 1 Find the vector equation,the parametric equations and the Cartesian equation of the line through (1,-2,3) in the direction of Solution the vector equation is

The parametric equations are x = 1 + 4t, y = -2 + 5t, z = 3 - 6t The Cartesian equations are

Example 2 Find a vector equation for a line passing through a point with position vector and is parallel to the line with cartesian equation Hence find a point on the line when y = 0

Solution The vector equation is A general point on the line has coordinates x = 5 + 3t, y = -2 + 4t, z = -4 + 5t If y = 0 0 = -2 + 4t t =

when t = The point on the line where y = 0 is

Example 3 a) Find parametric equations for the line L passing through the points A (2,4,-1) and B (5,0,7). b) Where does the line intersect the xy-plane ?

Solution a) The vector equation is The parametric equations x = 2 + 3t, y = 4 - 4t, z = -1 + 8t

b) The line intersects the xy-plane at the point where z = 0 So, when z = 0 0 = -1 + 8t t = When t = x = 2 + 3( ) = 2 + =

y = 4 - 4( ) = 4 - = The line intersects the xy-plane at

Conclusion the vector equation of the line : To find the equation, determine

The vector equation of the line : The parametric equations of the line : x = a1 + tv1 y = a2 + tv2 z = a3 + tv3 The Cartesian equations of the line