u × v = |i  j  k| |1  2  3| |4  5  6| = (-3, 6, -3)

Cross product = |i  j  k| |1  2  3| |4  5  6| = (-3, 6, -3)

Cross product A × B = |i  j  k| |1  2  3| |4  5  6| = (-3, 6, -3).

Distance = √((4-1)² + (5-2)² + (6-3)²) = √(9 + 9 + 9) = 3√3.

Dot product = 1*3 + 2*4 = 3 + 8 = 11

A . B = 2*1 + 3*0 + 4*(-1) = 2 + 0 - 4 = -2.

Direction ratios = (3, 3, 3), hence the line equation is x = 1 + 3t, y = 2 + 3t, z = 3 + 3t.

Magnitude = √(2^2 + (-3)^2 + 6^2) = √(4 + 9 + 36) = √49 = 7.

Magnitude = √(3^2 + 4^2) = √(9 + 16) = √25 = 5

Magnitude |C| = √(6^2 + 8^2 + 10^2) = √(36 + 64 + 100) = √200 = 10√2.

Magnitude of v = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.

Projection = (a · b / |b|^2) * b = (3*1 + 4*0) / 1^2 * (1, 0) = 3 * (1, 0) = (3, 0).

Projection = (A . B / |B|^2) * B = (3*1 + 4*2) / (1^2 + 2^2) * (1, 2) = 11/5 * (1, 2) = (2.2, 4.4).

Resultant = (2 + (-1), 3 + 4) = (1, 7).

Resultant = (1 - 1, 2 - 2) = (0, 0)

Scalar projection = (3*1 + 4*0) / √(1^2) = 3

Scalar projection = (A · B) / |B| = (3*1 + 4*0) / 1 = 3.

Scalar triple product = (1, 2, 3) · ((4, 5, 6) × (7, 8, 9)) = 0.

Scalar triple product = A · (B × C) = 1.

Scalar triple product = a · (b × c). Since b and c are linearly dependent, b × c = 0, hence the scalar triple product is 0.

Unit vector = (3, 4) / √(3^2 + 4^2) = (3/5, 4/5).

Unit vector = (3, 4, 0) / √(3^2 + 4^2) = (3, 4, 0) / 5 = (0.6, 0.8, 0).

Unit vector = (4/5, 3/5) where magnitude = √(4^2 + 3^2) = 5

Unit vector = (4/5, 3/5) since magnitude = 5.

Unit vector = v / |v| = (3, 4) / 5 = (0.6, 0.8)

Unit vector = A / |A| = (3, 4) / 5 = (0.6, 0.8).