Given that u is a vector of magnitude 2, v is a vector of magnitude 3 and the
angle between them when placed tail to tail is 45°, what is u · v (to one decimal
place)?
Your answer is correct
Since u · v = |u||v| cos , where is the angle between the vectors when
placed tail to tail, we have .
Not correct. You may try again.
Use the fact that
u · v = |u||v| cos . The answer is 4.2.
Question 2
What is the angle (to two decimal places) between a and b if a · b = 3,
|a| = 2, and |b| = 2.6?
Your answer is correct
If is the required angle, then cos = =
and hence radians.
Not correct. You may try again.
If is the required angle, then cos = =
Question 3
What is a·b if a = 3i-j and b = 2i + j + 4k?
Your answer is correct
a·b = 3 × 2 - 1 × 1 + 0 × 4 = 5.
Not correct. You may try again.
The answer is 5
Question 4
Express u in Cartesian form as ai + bj given that u · u = 12 and u points
towards the north-west.
Not correct. Choice 1 is false.
Not correct. Choice 2 is false.
Not correct. Choice 3 is false.
Your answer is correct
Since u · u = |u|2 = 12, we have |u| = 2. Hence
Question 5
If u = 5j and u · v = 0, what conclusion can be drawn?
Not correct. Choice 1 is false.
Not correct. Choice 2 is false.
Not correct. Choice 3 is false.
Your answer is correct
When the scalar product of two vectors is zero,
we can conclude that either one or both of the vectors is the zero vector, or else
that both the vectors are non-zero and they are mutually perpendicular. In this
case, we are told that u = 5j and so u is non-zero. Hence either v is the zero
vector or v is a non-zero vector perpendicular to u (that is, v points east or
west). If the first statement were amended to include the possibility that
v is the zero vector then it would be the best answer to the question.
Question 6
What is the following expression?
Not correct. Choice 1 is false.
The expression inside the brackets must be a vector in order for the
scalar product to be meaningful, and indeed it is. However, the resulting entire
expression is not a vector.
Not correct. Choice 2 is false.
The expression inside the brackets must be a vector in order for the scalar
product to be meaningful, and indeed it is. The resulting expression is therefore a
scalar.