answer:To find the cross product of two vectors A and B using trigonometry, we can use the formula: A x B = |A||B|sin(θ)n where |A| and |B| are the magnitudes of vectors A and B, θ is the angle between the two vectors, and n is the unit vector perpendicular to both A and B. First, let's find the magnitudes of vectors A and B: |A| = √(2^2 + 3^2 + (-1)^2) = √(4 + 9 + 1) = √14 |B| = √(1^2 + (-5)^2 + 4^2) = √(1 + 25 + 16) = √42 Next, we need to find the angle θ between the two vectors. We can use the dot product formula: A · B = |A||B|cos(θ) The dot product of A and B is: A · B = (2)(1) + (3)(-5) + (-1)(4) = 2 - 15 - 4 = -17 Now, we can find the cosine of the angle between the two vectors: cos(θ) = (A · B) / (|A||B|) = (-17) / (√14√42) Now, we can find the sine of the angle using the Pythagorean identity: sin^2(θ) + cos^2(θ) = 1 sin^2(θ) = 1 - cos^2(θ) sin(θ) = √(1 - (-17/√(14*42))^2) Now, we can find the cross product: A x B = |A||B|sin(θ)n To find the unit vector n, we can use the right-hand rule. The cross product of A and B will be a vector perpendicular to both A and B. We can find the components of the cross product by taking the determinant of the following matrix: | i j k | | 2 3 -1 | | 1 -5 4 | A x B = (3 * 4 - (-1) * (-5))i - (2 * 4 - (-1) * 1)j + (2 * (-5) - 3 * 1)k A x B = (12 - 5)i - (8 + 1)j + (-10 - 3)k A x B = 7i - 9j - 13k So, the cross product of A and B is: A x B = <7, -9, -13>

answer:To find the cross product of two vectors A and B, we can use the formula: |A x B| = |A| * |B| * sin(θ) where |A| and |B| are the magnitudes of vectors A and B, and θ is the angle between them. First, we need to find the angle between the two vectors. Since vector A is at 30 degrees to the x-axis and vector B is at 45 degrees to the x-axis, the angle between them is: θ = 45 - 30 = 15 degrees Now we can find the magnitude of the cross product: |A x B| = |A| * |B| * sin(θ) |A x B| = 5 * 7 * sin(15) To find sin(15), we can use the half-angle formula: sin(15) = sqrt((1 - cos(30))/2) cos(30) = sqrt(3)/2, so: sin(15) = sqrt((1 - sqrt(3)/2)/2) Now, we can plug this back into the formula for the magnitude of the cross product: |A x B| = 5 * 7 * sqrt((1 - sqrt(3)/2)/2) |A x B| ≈ 5 * 7 * 0.2588 |A x B| ≈ 9.03 The magnitude of the cross product of vector A and B is approximately 9.03. The direction of the cross product is perpendicular to the plane formed by vectors A and B. Since we are working in two dimensions, the direction of the cross product will be either into or out of the plane (positive or negative z-direction). To determine the direction, we can use the right-hand rule. In this case, the cross product will be in the positive z-direction. So, the cross product of vector A and B has a magnitude of approximately 9.03 and is directed in the positive z-direction.

answer:To find the cross product of two vectors in 3-dimensional space, we can use the following formula: 𝐀 × 𝐁 = (𝐀y * 𝐁z - 𝐀z * 𝐁y, 𝐀z * 𝐁x - 𝐀x * 𝐁z, 𝐀x * 𝐁y - 𝐀y * 𝐁x) Plugging in the values for 𝐀 and 𝐁: 𝐀 × 𝐁 = ((-2) * (-2) - (4) * (5), (4) * (1) - (3) * (-2), (3) * (5) - (-2) * (1)) Calculating the values: 𝐀 × 𝐁 = (4 - 20, 4 + 6, 15 + 2) 𝐀 × 𝐁 = (-16, 10, 17) So, the cross product of 𝐀 and 𝐁 is (-16, 10, 17).

answer:To find the cross product of two vectors A and B using the formula |A||B|sinθ, we first need to find the angle θ between the two vectors. However, the formula you provided is used to find the magnitude of the cross product, not the cross product itself. The cross product of two vectors A and B can be found using the determinant method: A x B = (A_y * B_z - A_z * B_y)i - (A_x * B_z - A_z * B_x)j + (A_x * B_y - A_y * B_x)k Given the vectors A = 3i + 2j + k and B = 2i - j + 2k, we can find the cross product as follows: A x B = ((2 * 2) - (1 * (-1)))i - ((3 * 2) - (1 * 2))j + ((3 * (-1)) - (2 * 2))k A x B = (4 + 1)i - (6 - 2)j + (-3 - 4)k A x B = 5i - 4j - 7k So, the cross product of the two vectors A and B is 5i - 4j - 7k.