Relative to the origin O, two points A and B have position vectors a and b respectively. A line, l, passes through A and is parallel to b. It is given that b is a unit vector.
(i) Write down a vector equation of l. Show that the position vector of the point N on l such that the length |ON| is the shortest is given by a-(a.b)b.
(ii) The point M is on AN produced such that kAN = NM, where k is a constant. Given that the position of M is a-5(a.b)b, find k.
It is given that |a|=2 and a.b=1/3.
(iii) Give the geometric meaning of |b x (a-b)| and find its exact value.
(iv) C is a point such that OC bisects the angle AOB. Write down, in terms of a and b, a possible position vector of C.
Any help would be greatly appreciated! Thank you :)
(i) To find a vector equation of line l passing through point A and parallel to vector b, we can use the parametric form of a vector equation. Let's call the position vector of point N as r. Since line l is parallel to vector b, we can express r in terms of parameter t as:
r = a + tb
To find the position vector of point N such that |ON| is the shortest, we need to minimize the distance between the origin O and the point N. The shortest distance between a point and a line is along the perpendicular from the point to the line. So, we want to find the value of t that makes the length of the projection of vector ON onto vector b equal to zero.
Let's calculate the projection of vector ON onto vector b:
Projection of ON onto b = (a + tb) . b
Since b is a unit vector, the magnitude of b is 1. Therefore, the projection simplifies to:
Projection of ON onto b = (a + tb) . b = a.b + t(b . b) = a.b + t(1)
To find the shortest |ON|, we set the projection equal to zero:
a.b + t(1) = 0
Solving for t:
t = -a.b / (1)
Substituting this value of t back into the equation r = a + tb:
r = a - (a.b)b
So, the position vector of the point N on line l such that the length |ON| is the shortest is given by a - (a.b)b.
(ii) Given that the position of point M is a - 5(a.b)b, and we know that kAN = NM, we can determine the value of k by comparing the lengths of vectors AN and NM.
Length of AN = |a - (a.b)b|
Length of NM = |(a - 5(a.b)b) - (a - (a.b)b)| = |4(a.b)b|
Since kAN = NM, we have:
k|a - (a.b)b| = |4(a.b)b|
Dividing both sides by |a - (a.b)b|:
k = |4(a.b)b| / |a - (a.b)b|
Given that |a| = 2 and a.b = 1/3, substituting these values:
k = |4(1/3)b| / |a - (1/3)b|
Since b is a unit vector, its magnitude is 1. Therefore, k simplifies to:
k = (4/3) / |a - (1/3)b|
(iii) The expression |b x (a - b)| represents the magnitude of the cross product between vectors b and (a - b). Geometrically, the magnitude of the cross product represents the area of the parallelogram formed by the two vectors. So, |b x (a - b)| represents the area of the parallelogram formed by vectors b and (a - b).
To find its exact value, we can calculate the cross product:
b x (a - b) = |b||a - b| sin(θ) n
Where θ is the angle between vectors b and (a - b), and n is the unit vector perpendicular to the plane formed by b and (a - b).
Since b is a unit vector, |b| = 1. Similarly, |a - b| represents the magnitude of vector a - b.
Therefore, |b x (a - b)| = |a - b| sin(θ)
(iv) To find a possible position vector of point C, such that OC bisects the angle AOB, we can use the angle bisector theorem. According to the theorem, the position vector of C can be calculated as:
C = (|OB| * A + |OA| * B) / (|OB| + |OA|)
Since |b| = 1, we can substitute |OB| = 1 into the equation:
C = (1 * A + |OA| * B) / (1 + |OA|)
Given that |OA| = |a| = 2, we have:
C = (A + 2B) / 3
Therefore, a possible position vector of C is (A + 2B) / 3, where A and B are the position vectors of points A and B relative to the origin O, respectively.