Question

Let vectors A =(2,1,−4), B=(−3,0,1), and C=(−1,−1,2)

What is the angle θAB between vector A and B?

Answer 1

The angle θAB between vectors A and B is calculated using the dot product and magnitudes of both vectors, then the inverse cosine function to determine the angle in degrees.

Step-by-step explanation:

To find the angle θAB between vectors A and B, we must use the dot product of vectors A and B along with the magnitude of both vectors. The dot product A·B is given by the sum of the products of the respective components of vectors A and B:

A·B = Ax × Bx + Ay × By + Az × Bz = 2×(-3) + 1×(0) + (-4)×1 = -6 + 0 - 4 = -10

The magnitudes of vectors A and B are given by |A| = √(Ax² + Ay² + Az²) and |B| = √(Bx² + By² + Bz²), respectively. Thus:

|A| = √(2² + 1² + (-4)²) = √(4 + 1 + 16) = √21

|B| = √((-3)² + 0² + 1²) = √(9 + 0 + 1) = √10

Now we can calculate the angle using the formula:

cos(θ) = (A·B) / (|A| × |B|)

cos(θ) = -10 / (√21 × √10)

θ = cos⁻¹(-10 / (√21 × √10))

To find the angle θ, we take the inverse cosine of the result. Remember, the angle will be between 0° and 180°. The final step is to carry out the calculation on a calculator to determine the angle between vectors A and B.

Answer 2

the answer: to find the angle between the two vectors, the following method can be used: first, computing the scalar product between the two vectors A and B, after, their length. the next is to use the main formula cos T = A*B / //A// //B//, where A*B is scalar product T= teta Practice: A*B = (2, 1,-4) * (-3, 0,1) =(2x-3)+(1x0)+(-4x1)=-6-4= -10 //A// = sqrt( 2² + 1² +4²) = sqrt(4+1+16) = sqrt(21)=4.58 //B// = sqrt(10)=3.16 so, cosT = -10 / 3.16*4.58 = -10/14.48 = -0.69, cosT = -0.69, therefore, T= arccos(-0.69)=46,33°