THREE DIMENSIONAL GEOMETRY

Chapter * For various activities in three dimensional geometry, one may refer to the Book “A Hand Book for designing Mathematics Laboratory in Schools” , NCERT, Leonhard Euler ( - ) Note that a given line in space can be extended in two opposite directions and so it has two sets of direction cosines. In order to have a unique set of direction cosines for a given line in space, we must take the given line as a directed line. These unique direction cosines are denoted by l , m and n . Remark If the given line in space does not pass through the origin, then, in order to find its direction cosines, we draw a line through the origin and parallel to the given line.

Now take one of the directed lines from the origin and find its direction cosines as two parallel line have same set of direction cosines. Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line. If l , m , n are direction cosines and a , b , c are direction ratios of a line, then a = λ l , b = λ m and c = λ n , for any nonzero λ ∈ R . A Note Some authors also call direction ratios as direction numbers.

Let a , b , c be direction ratios of a line and let l , m and n be the direction cosines ( d . c ’s) of the line. Then a = m b = n c = (say), k being a constant. Therefore l = ak , m = bk , n = ck But l + m + n = Therefore k ( a + b + c ) = or k = ± Fig .

Hence, from ( ), the d . c .’ s of the line are =± = ± = ± where, depending on the desired sign of k , either a positive or a negative sign is to be taken for l , m and n . For any line, if a , b , c are direction ratios of a line, then ka , kb , kc ; k ≠ is also a set of direction ratios. So, any two sets of direction ratios of a line are also proportional.

Also, for any line there are infinitely many sets of direction ratios. . . Direction cosines of a line passing through two points Since one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points P( x , y , z ) and Q( x , y , z ) as follows (Fig .

(a)). Fig . Let l , m , n be the direction cosines of the line PQ and let it makes angles α , β and γ with the x , y and z -axis, respectively. Draw perpendiculars from P and Q to XY-plane to meet at R and S.

Draw a perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠ PQN= γ (Fig . (b). Therefore, cos γ = NQ Similarly cos α = and cos β= Hence, the direction cosines of the line segment joining the points P( x , y , z ) and Q( x , y , z ) are where PQ = A Note The direction ratios of the line segment joining P( x , y , z ) and Q( x , y , z ) may be taken as x – x , y – y , z – z or x – x , y – y , z – z Example If a line makes angle °, ° and ° with the positive direction of x , y and z -axis respectively, find its direction cosines.

Solution Let the d . c .' s of the lines be l , m , n . Then l = cos = , m = cos = , n = cos = . Example If a line has direction ratios , – , – , determine its direction cosines.

Solution Direction cosines are ) ) or Example Find the direction cosines of the line passing through the two points (– , , – ) and ( , , ). Solution We know the direction cosines of the line passing through two points P( x , y , z ) and Q( x , y , z ) are given by where PQ = Here P is (– , , – ) and Q is ( , , ). So PQ = ( ( )) ) ( )) Thus, the direction cosines of the line joining two points is Example Find the direction cosines of x , y and z -axis. Solution The x -axis makes angles °, ° and ° respectively with x , y and z -axis.

Therefore, the direction cosines of x -axis are cos °, cos °, cos ° i.e., , , . Similarly, direction cosines of y -axis and z -axis are , , and , , respectively. Example Show that the points A ( , , – ), B ( , – , ) and C ( , , – ) are collinear. Solution Direction ratios of line joining A and B are – , – – , + i.e., – , – , .

The direction ratios of line joining B and C are – , + , – – , i.e., , , – . It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

EXERCISE . . If a line makes angles °, °, ° with the x , y and z -axes respectively, find its direction cosines. .

Find the direction cosines of a line which makes equal angles with the coordinate axes. . If a line has the direction ratios – , , – , then what are its direction cosines ? .

Show that the points ( , , ), (– , – , ), ( , , ) are collinear. . Find the direction cosines of the sides of the triangle whose vertices are ( , , – ), (– , , ) and (– , – , – ). .

Equation of a Line in Space We have studied equation of lines in two dimensions in Class XI, we shall now study the vector and cartesian equations of a line in space. A line is uniquely determined if (i) it passes through a given point and has given direction, or (ii) it passes through two given points. . .

Equation of a line through a given point and parallel to given vector Let be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector . Let be the position vector of an arbitrary point P on the line (Fig . ).

Then AP is parallel to the vector , i.e., AP = λ , where λ is some real number. But AP = OP – OA i.e. λ = Conversely, for each value of the parameter λ , this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by » r a + Remark If ai bj ck , then a , b , c are direction ratios of the line and conversely, if a , b , c are direction ratios of a line, then ai bj ck will be the parallel to the line.

Here, b should not be confused with | |. Derivation of cartesian form from vector form Let the coordinates of the given point A be ( x , y , z ) and the direction ratios of the line be a , b , c . Consider the coordinates of any point P be ( x , y , z ). Then xi yj zk ; x i y j a i b j c k Substituting these values in ( ) and equating the coefficients of ˆ , i j and k ˆ , we get x = x + λ a ; y = y + λ b ; z = z + λ c ...

( ) These are parametric equations of the line. Eliminating the parameter λ from ( ), we get x – x y – y z – z ... ( ) This is the Cartesian equation of the line. A Note If l , m , n are the direction cosines of the line, the equation of the line is x – x y – y z – z Example Find the vector and the Cartesian equations of the line through the point ( , , – ) and which is parallel to the vector Solution We have Fig .

Therefore, the vector equation of the line is ( Now, is the position vector of any point P( x , y , z ) on the line. Therefore, xi y j z k ( ) ( ) ( ) + − −λ Eliminating λ , we get x − = − which is the equation of the line in Cartesian form. . Angle between Two Lines Let L and L be two lines passing through the origin and with direction ratios a , b , c and a , b , c , respectively.

Let P be a point on L and Q be a point on L . Consider the directed lines OP and OQ as given in Fig . . Let θ be the acute angle between OP and OQ.

Now recall that the directed line segments OP and OQ are vectors with components a , b , c and a , b , c , respectively. Therefore, the angle θ between them is given by cos θ = a a c c The angle between the lines in terms of sin θ is given by sin θ = cos θ