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 f