If vector v has an initial point at P1 and a terminal point at P2, write v as multiples of the basis vectors That is, write v in the form v = ai + bj.

P1 = (2, −3), P2 = (4, 1) and v = ?

v = 2i + 4j

Step-by-step explanation:

First, we must say what a unit vector is.

In the case of i is (1,0) and j is (0,1), now, they ask us what the vector v is worth, which would be the subtraction from point 1 to point 2, thus:

v = P2 - P1, this because P2 is the end point and point P1 is the start point.

Now to express this in unit vector value, we do the following:

P1 = (2, −3) = 2i - 3j

P2 = (4, 1) = 4i + 1j

Now replacing is in v:

v = P2 - P1 = 4i + 1j - (2i - 3j) = 4i + j - 2i + 3j

v = 2i + 4j

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)