RI Vectors Lines and Planes Solns 7A Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 7a (Summary and Tutorial) Topic: Vectors 2 (Lines and Planes) Summary for Lines and Planes Line: , ra b  where r is the position vector of a general point on the line, a is the position vector of a known point on the line and b is the vector that indicate the direction parallel to the line (known as the direction vector) Note : , ra b  is not UNIQUE. Different Forms of Equations of a Line 11 22 33 , ab ab ab          r   Vector Form  11 22 33 , abx ya b z ab               x = a1 +  b1,y = a2 +  b2,z = a3 +  b3  Parametric Form  1 1 xa b   , 2 2 ya b   , 3 3 za b    312 123 zaxa ya bbb    Cartesian Form Plane: dr . n where r is the position vector of a general point on the plane, n is the normal vector to the plane (a vector perpendicular to the plane) d is a scalar constant. Notes: The position vector of any point A on the plane, a will always give the result, a . n d The vector equation of the plane can always be reduced to the form ˆ d Dr . n n . In this form, the normal vector is reduced to a unit vector and |D| is the shortest distance of the plane from the origin.

Different Forms of Equations of a Plane Scalar Product form: rn d. Cartesian form: 12 3nx n y n z d where 1 2 3 ,r n xn yn zn             Vector form: 12 ,,r=a m m     a is the position vector of a point on the plane, m 1 and m2 are non-parallel vectors that are parallel to the plane. Thus, 12mm will be a normal to the plane. We will now summarize the various relationships involving points, lines and planes. Involving Points and Lines  To check whether the point P with position vector p lies on the given line l: , ra b   Let pa b and find a possible value for .  If there is a unique value for  from the three possible equations, then the point P lies on the line.  To find the position vector of the foot of the perpendicular from a point P with position vector p to a given line l : , ra b  Let N be the foot of the perpendicular from the point P to the line, l. Make use of the fact that PN  is perpendicular to l Step 1: Since N lies on l , ON   ab for a unique value of . Step 2: Find PN ON OP       in terms of , where OP  = p is given. Step 3: Solve for the value of  using the fact that 0PN  b Step 4: Using the value of  found above, the position vector of N is given by ab .  To find the perpendicular distance (shortest distance) from a point, P with position vector, p, to a given line l : , ra b  Let N be the foot of the perpendicular from the point P to the line, l Method 1 Step 1 : Find the position vector of the foot of the perpendicular from the point P to the given line l . Step 2 : Perpendicular distance from the point P to the given line l is| PN  |. P N O  l b

Method 2 (Cross Product) Perpendicular distance from the point P to the given line l is ˆbPN AP    Method 3 (Dot Product) Step 1 : Find ˆAN AP b     Step 2 : Using Pythagoras’ Theorem, Perpendicular distance from the point P to the given line l is 22 PN AP AN         To find the position vector of the reflection of a point, P, about a line l : , ra b  Let P’ be the reflection of P about l and that N be foot of the perpendicular from P to l . Method  Find the vector PN  ( Note that PN  = 'NP  )  Position vector of P’ ''OP OP PN NP          2OP PN     ( OR use Ratio Theorem: ' 2 OP OPON      ) Involving Lines  To check if the lines 1 ,l  :r a b  and 2` ,l  :r c d  are parallel If b // d (i.e. b = k d ), then l1 // l2  To find the acute angle between 2 lines, 1 ,l  :r a b  and 2` ,l  :r c d  The acute angle between the lines is determined by the two direction vectors b and d. Method Make use of scalar product of the two direction vectors b and d.  | b  d | = bdcos  where  is the acute angle between the 2 direction vectors, b and d. Therefore cos  b.d bd P p O N l b A a N O l P P’ b d

 To find the position vector of the point of intersection of the two lines 1` ,l   :r a b  and 2` ,l   :r c d  Let P be the point of intersection of the 2 lines l1 and l2 , and p be the position vector of P. Then P lies on both lines and pa b and pc d for some values of  and . Method Step 1 : Since the lines intersect, pa b = cd Step 2 : From Step 1, we will have 3 linear equations in terms of  and . Solve for the values of  and   If there exist unique values for  and , then the position vector of the intersection point is given by ab  or cd  .  If there are no unique values for  and , then the 2 lines are non-intersecting lines. (They can be either parallel or skew lines)  To check if the lines 1 ,l   :r a b  and 2 ,l   :r c d  are skew lines (Skew lines are nonparallel and nonintersecting lines.) Assume that P, the point of intersection of the 2 lines l 1 and l2, and p be the position vector of P exists show that it does not exist. Then P lies on both lines and pa b and pc d  for some values of  and . Method Step 1 : Show that l1 and l2 are not parallel (b  k d ) Step 2 : Then assume ab = cd and show that there is NO unique values for  and  that satisfy the equations formed.  To find the shortest distance between two parallel lines 1 ,l   :r a b  and 2 ,:r c bl     . The shortest distance between two parallel lines can be found by taking any point, says A on 1 ,l   :r a b  and find the shortest distance between point A and 2 ,:r c bl     . OR The shortest distance between two parallel lines can be found by taking any point, says C on 2 ,:r c bl     and find the shortest distance between point C and 1 ,l   :r a b  . (Refer above for the shortest distance between a point and a line) Involving Points, Lines and Planes

 To check whether the point P with position vector p lies on the given plane  : dr . n Method  If dp. n , then the point P lies on the plane. If dp. n , then the point P does not lie on the plane.  To check if the line ,l  :r a b  is parallel to a given plane  : dr . n Method  If 0b. n , then the line is parallel to the plane. (perpendicular to the normal)  To check if the line ,l  :r a b  lies in a given plane  : dr . n Method 1  Show that d(a b) n . for all values of , then the line lies in the plane. Method 2 Step 1 : Show that 0b. n , then the line is parallel to the plane. Step 2 : Show that da . n , then the point with position vector, a lies on the plane. Then conclude the line lies in the plane.  To find the position vector of point of intersection between the line ,l  :r a b  and the plane  : dr . n Let P with position vector p be the point of intersection between the line l and the plane  Then , for some  pa b  and dp. n Method  Since , for some  pa b  and dp. n , then d(a b) n . .  Solve for  and use this value to find the position vector of the point of intersection is given by pa b l n  b l n 

 To find the position vector of the foot of the perpendicular from the point P to the plane  : dr . n Let F be the foot of the perpendicular from the point P to the plane  : dr . n . Method Step 1: Find v