OABCDE is a right triangular prism with OA = 2units , OD= 2units and AB = 4 units. The unit vectors i,j and k are parallel to OA,OD and OC respectively.
a) Find the unit vector parallel to DB.
are the two triangular bases OAB and CDE?
and the three lengths OC, AD and BE?
the bases are OAB and OCB. and the lengths are OD , AD and BE
Sorry. In a prism, the bases are parallel and thus do not intersect. Points O and B cannot possibly be in both bases.
To find the unit vector parallel to DB, we need to first find the vector that represents the line segment DB.
We are given that the vector i is parallel to OA, j is parallel to OD, and k is parallel to OC.
From the given information, we can see that D is the origin (0,0,0) and B has coordinates (4,0,0) since AB = 4 units and A shares the same x-coordinate as D.
The vector DB can be found by subtracting the coordinates of D from the coordinates of B:
DB = B - D = (4,0,0) - (0,0,0) = (4,0,0)
Since we are only interested in the direction of DB, we can convert DB into a unit vector by dividing each component by its magnitude.
The magnitude of DB is found using the Pythagorean theorem:
|DB| = sqrt(4^2 + 0^2 + 0^2) = sqrt(16) = 4
Now, we can find the unit vector by dividing each component of DB by its magnitude:
unit vector DB = (4/4, 0/4, 0/4) = (1, 0, 0)
Therefore, the unit vector parallel to DB is (1, 0, 0).