a and b are perpendicular vectors.a=(2i+3j) and b=ti+dj(d>0 and i and j are unit vectors) and |b|=1
Find t and d?
I know we have to use the cross product here.
a.b=|a||b|*cos(π/2)
(2i+3j)(ti+dj)= √13*1*0
2t+3j=√13*0
How do we find t and d?
As you say, we need
a•b = 0
2t+3d = 0
but b is also a unit vector, so
t^2+d^2 = 1
Solving those two equations gives the values of t and d. Pick the solution where d>0.
Thanks Steve!
To find t and d, we need to equate the dot product of a and b to zero and use the fact that |b| equals 1.
The dot product of a and b is given by:
a · b = 2t + 3d
Since a and b are perpendicular vectors, their dot product is equal to zero:
2t + 3d = 0
Now let's consider the magnitude of b, which is given by:
|b| = 1
The magnitude of b is the square root of the sum of the squares of its components:
√(t^2 + d^2) = 1
Squaring both sides of this equation, we get:
t^2 + d^2 = 1
Now we have a system of two equations:
2t + 3d = 0 ---(1)
t^2 + d^2 = 1 ---(2)
To solve this system, we can use substitution or elimination.
Using substitution, we isolate one variable from one equation and substitute it into the other equation:
From equation (1), we can solve for t:
t = -1.5d
Now substitute this value of t into equation (2):
(-1.5d)^2 + d^2 = 1
2.25d^2 + d^2 = 1
3.25d^2 = 1
d^2 = 1/3.25
d = √(1/3.25)
Therefore, d is equal to the square root of 1 divided by 3.25. Taking the square root, we get the value of d. Similarly, substituting this value of d in equation (1), we can find the value of t.