Question

6. An airplane is taking off headed due north with an air speed of 173 miles per hour at an angle of 18° relative to the horizontal. The wind is blowing with a velocity of 42 miles per hour at an angle of S47°E. Find a vector that represents the resultant velocity of the plane relative to the point of takeoff. Let i point east, j point north, and k point up. (Show work)

Answer 1

Answer: The answer is  $u$$plane$ 30.72 $i$ + 193. 18 $j$ + 53. 46 $k .$

Step-by-step explanation:

Component of the plane's velocity in the direction of vector  relative to the air: ;

Component of the plane's velocity in the direction of vector  relative to the air:  .

The direction of the plane's velocity relative to the air is normal to vector . Therefore, the component of the plane's velocity (relative to the air) in that direction will equal zero. Thus

.

Refer to the second diagram,

Component of the velocity of the wind in the direction of vector : ;

Component of the velocity of the wind in the direction of vector : .

Assume that the wind blows horizontally. The direction of the wind will be normal to vector . The component of the velocity of the wind in the direction of vector  will thus equal zero. Therefore,

.

The ground speed of the plane  is the sum of  and .

That is:

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