An ornithologist is in the field studying a formation of geese and is 180 m south, and 385 m west, of where she parked her car. Let u be the vector representing the distance between the...

An ornithologist is in the field studying a formation of geese and is 180 m south, and 385 m west, of where she parked her car.

Let u be the vector representing the distance between the ornithologist and her car, and let v be the vector representing the straight line distance from the ornithologist to the formation.

If the lead goose flies over her car at an altitude 319 m

find |v|

find the angle btw u and v using cos `theta` = u . v / |u||v|

Verify result using right triangle trig

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justaguide | College Teacher | (Level 2) Distinguished Educator

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An ornithologist is in the field studying the formation of a group of geese in flight. She is 180 m south, and 385 m west of where she parked her car. Let the coordinates of her present location be (0, 0, 0). The location of her car is (180, 385, 0)

The vector `vec u` representing the line from where she is standing to her car is [180, 385, 0]. As the lead goose is at an altitude of 319 m above her car, the vector `vec v` representing the line drawn from where she stands to the lead goose is [180, 385, 319]

The magnitude of vector `vec u` is `sqrt(180^2 + 385^2 + 0^2)` = 425

The magnitude of vector `vec v` is `sqrt(180^2 + 385^2 + 319^2)` = 531.4

The dot product of `vec u` and `vec v` is `(180^2 + 385^2 + 319*0)` = 180625

`vec u @ vec v` = `|vec u|*vec v|*cos theta` where `theta` is the angle between the vectors.

`cos theta = 180625/(425*531.4)`

=> `theta = cos^-1(180625/(425*531.4))`

=> `theta = 36.89`

To verify the result, take the right triangle formed by the line joining the ornithologist to her car, the line joining her to the lead goose and the line joining the lead goose to the car. The angle between the line her to the car and the line joining her to the lead goose is `tan^-1(319/425)` = 36.89 degree

The angle between the vectors `vec v` and `vec u` is 36.89 degree

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