In triangle OAB, OA = 3i + 4k and
OB = i + 2j - 2k. Find OP, where P is the point where the bisector of Angle AOB intersects AB.
Answer is 7i/4 + 5j/4 + k/4
Don't use the matrix method as I haven't learnt it yet and please show working. Thx a lot.
This is discussed here:
http://math.stackexchange.com/questions/1408917/finding-vector-form-of-an-angle-bisector-in-a-triangle
To find the point of intersection, we can find the equation of the bisector line and then solve it with the line AB.
Step 1: Finding the equation of the line AB.
To find the equation of line AB, we need two points on the line. We are given point A(3,0,4) and point B(1,2,-2). Using these points, we can find the direction vector of the line.
Direction vector AB = OB - OA
= (1, 2, -2) - (3, 0, 4)
= (-2, 2, -6)
So, the equation of line AB can be written as:
AB: r = OA + t(AB)
= (3, 0, 4) + t(-2, 2, -6)
= (3 - 2t, 2t, 4 - 6t)
Step 2: Finding the equation of the bisector line.
The bisector line will pass through the origin (0,0,0) and the point P(x, y, z). We need to find the direction vector of this bisector line.
Let the direction vector of the bisector line be (a, b, c).
Since the bisector line is at the angle bisector of AOB, the dot product of the direction vector of AB and the direction vector of the bisector line should be zero.
So, we have the dot product equation:
(-2, 2, -6) • (a, b, c) = 0
Simplifying the dot product equation, we get:
-2a + 2b - 6c = 0
Step 3: Solving the equations to find P.
We have two equations now:
Equation 1: 3 - 2t = ax
Equation 2: 2t = by
Equation 3: 4 - 6t = cz
Equation 4: -2a + 2b - 6c = 0
To find the point P, we need to find the values of x, y, and z.
From Equation 4, we can solve for a in terms of b and c:
-2a = -2b + 6c
a = b - 3c
Substituting the value of a in Equation 1, we get:
3 - 2t = (b - 3c)x
Substituting the value of a in Equation 2, we get:
2t = by
Substituting the value of a in Equation 3, we get:
4 - 6t = (b - 3c)z
From Equation 2, we can solve for t in terms of b and y:
t = y/2b
Substituting the value of t in Equation 1, we get:
3 - 2(y/2b) = (b - 3c)x
2 - (y/b) = (b - 3c)x
2b - (y) = (b - 3c)bx
(y) = (2b - bx) - (3c)bx
y = 2b - bx - 3c^2b
Substituting the value of t in Equation 3, we get:
4 - 6(y/2b) = (b - 3c)z
4 - 3(y/b) = (b - 3c)z
4b - 3y = (b - 3c)bz
4b - 3(2b - bx - 3c^2b) = (b - 3c)bz
4b - 6b + 3bx + 9c^2b = b^2z - 3c^2bz
-2b + 3bx + 9c^2b = b^2z - 3c^2bz
3bx + (3c^2 - 9c^2)b = b^2z + 2bz
3bx - 6c^2b = bz(b + 2)
(3x - 6c^2) = bz(b + 2)
x = (bz(b + 2))/(3 - 6c^2)
Therefore, the point P(x, y, z) is:
P = ((bz(b + 2))/(3 - 6c^2), 2b - bx - 3c^2b, (4b - 3y)/(b - 3c))
For the given answer P = (7/4)i + (5/4)j + (1/4)k:
Comparing the coefficients, we can see that b = 7/4, x = 1/4, and c = 1/4.
Therefore, the point P where the bisector of Angle AOB intersects AB is:
P = (1/4)i + (3/4)j + (1/4)k.