Quiz 2: Length, Dot Product and Cross Product

This is the length, or magnitude, of the vector .

This is the length, or magnitude, of the vector v.

The difference of two vectors is a vector.

As u is a vector and ||v|| is a scalar (namely, the length of v), this expression does not make sense.

As is a vector, this is minus the length of and so this is a scalar, not a vector.

The parallelogram rule for vector addition shows that when a and b are placed tail to tail, the diagonals of the parallelogram are a + b and a-b. With the vectors a, b as shown here, we see that ||a + b|| > ||a - b||.

Since a + b and a - b are vectors which form the diagonals of a rectangle in this particular case, we see that ||a + b|| = ||a - b||.

Since a + b and a - b are vectors which form the diagonals of a rectangle in this particular case, we see that ||a + b|| = ||a - b||.

The parallelogram rule for vector addition shows that when a and b are placed tail to tail, the diagonals of the parallelogram are a + b and a - b. In this case, after re-drawing b so that its tail is at the tail of a, we see that ||a + b|| < ||a - b||.

Since u⋅v = ||u|| ||v||cosθ, where θ is the angle between the vectors when placed tail to tail, we have u ⋅ v = 2 × 3 × cos45^∘≈ 4.24.

If θ is the required angle, then cosθ = = ≈ 0.577 and hence θ ≈ 0.955 radians.

If the vector equation is simplified, we get Since this holds for all a and b, it will hold when a and b are set equal to 0 in turn. This gives the two conditions α - β - 1 = 0 and 2α + 4 = 0, whose solution is α = -2, β = -3.

The vectors a, -2b, and a- 2b form the sides of a right-angled triangle, with sides of length 8, 6 and hypotenuse of length . Therefore by Pythagoras’ Theorem,

Try drawing a diagram of the vectors and and then use Pythagoras’ Theorem.

The boat’s journey can be represented by the following diagram, where the origin O is taken to be the starting position and Q is the finishing position. The required distance from the starting position is then ||||. Now |||| = 5, and since initially the boat sails in a south-easterly direction, the coordinates of P must be (,-). Since |||| = 3 and Q is due west of P, the coordinates of Q must be ( - 3,-). Hence the required distance from the origin (by Pythagoras’ Theorem) is ≈ 3.6 km.

This is meaningless. The second bracket is a scalar quantity and we can’t take a cross product of a vector with a scalar.

This is meaningless. The cross product is defined between two vectors, not two scalars.

This is a dot product of two vectors and the end quantity is a scalar.

This is a vector since it is a scalar multiple of the vector v × w.

This is meaningless. We can’t add a vector to a scalar.

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u × v = [u₂v₃ - u₃v₂,u₃v₁ - u₁v₃,u₁v₂ - u₂v₁].

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u × v = [u₂v₃ - u₃v₂,u₃v₁ - u₁v₃,u₁v₂ - u₂v₁].

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u × v = [u₂v₃ - u₃v₂,u₃v₁ - u₁v₃,u₁v₂ - u₂v₁].

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u×v = [u₂v₃ -u₃v₂,u₃v₁ -u₁v₃,u₁v₂ -u₂v₁].

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u×v = [u₂v₃-u₃v₂,u₃v₁-u₁v₃,u₁v₂-u₂v₁].

Recall that if u = [u₁,u₂,u₃] and v = [v₁,v₂,v₃] then u × v = [u₂v₃ - u₃v₂,u₃v₁ - u₁v₃,u₁v₂ - u₂v₁].