Navigation A plane is flying with a bearing of 302degree. Its speed with respect

Question

Navigation A plane is flying with a bearing of 302degree. Its speed with respect to the air is 900 kilometers per hour. The wind at the plane's altitude is from the southwest at 100 kilometers per hour (see figure). What is the true direction of the plane, and what is its speed with respect to the ground? Find the magnitude and direction of the resultant of the vectors: Calculate the dot product between FI and F2 Calculate the cross product between F1 and F3 Find the equation of the plane: The plane passes through (1,2, 3), (3, 2, 1), and ( - 1, - 2, 2). The plane passes through the point (1,2, 3) and is parallel to the yz-plane. Calculate the angle between the plane from part a and b. The position vector r describes the path of an object moving in space. Find the magnitude and direction of velocity and acceleration of the object after 5 s.

Explanation / Answer

2. Fx = 2Cos33 + 2.5Cos110 + 3Cos(360-125) = -0.898 Fy = 2Sin33 + 2.5Sin110 + 3Sin(360-125) = 0.981 Resultant force F = sqrt(-0.898^2 + 0.981^2) = 1.33 Direction of resultant force = atan(Fy/Fx) = atan(0.981/(-0.898)) = 132.5 deg a) F1.F2 = |2*3*Cos(125+33)| = 5.563 b) F1XF3 = 2*2.5*Sin(110-33) = 4.8718

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