What is the displacement vector FROM the first plane TO the second plane, letting $ˆ i$ $hati$ represent east, $ˆ j$ $hatj$ north, and $ˆ k$ $hatk$ up?

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What is the displacement vector FROM the first plane TO the second plane, letting $ˆ i$ $hati$ represent east, $ˆ j$ $hatj$ north, and $ˆ k$ $hatk$ up?

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Answer 1 · 2018-02-05T07:21:13

(a) ${\to v}_{1 \to 2} = 8686 ˆ i - 4103 ˆ j + 200 ˆ k$ $vecv_(1to2) = 8686hati-4103hatj+ 200hatk$ in meters

(b) $∣ ∣ {\to v}_{1 \to 2} ∣ ∣ = 9608$ $|vecv_(1to2)| = 9608$ meters

Explanation:

We need to determine the standard angle $θ$ $theta$ for the respective airplane; this will make it easy to use the horizontal distance $d$ $d$ to compute the $ˆ i$ $hati$ and $ˆ j$ $hatj$ components of the vector for the plane. The $ˆ k$ $hatk$ is easy because it is given as the as the altitude in meters. For any given airplane, $n$ $n$ , the vector is as follows:

${\to v}_{n} = d_{n} cos (θ_{n}) ˆ i + d_{n} sin (θ_{n}) ˆ j + {altitude}_{n} ˆ k$ $vecv_n = d_ncos(theta_n)hati+d_nsin(theta_n)hatj+"altitude"_nhatk$

For the first airplane convert given heading to a standard angle with the reference point as East.

For airplane 1 we are given $38^{\circ}$ $38^@$ South of West

West is $180^{\circ}$ $180^@$ from East and add $38^{\circ}$ $38^@$ to make it move toward the South:

$θ_{1} = 180^{\circ} + 38^{\circ}$ $theta_1 = 180^@+38^@$

$θ_{1} = 218^{\circ}$ $theta_1 = 218^@$

Convert the distance from km to meters:

$d_{1} = 19.8 km$ $d_1 = 19.8" km"$

$d_{1} = 19800 m$ $d_1 = 19800" m"$

${altitude}_{1} = 900 m$ $"altitude"_1 = 900" m"$

${\to v}_{1} = (19800 cos (218^{\circ}) m) ˆ i + (19800 sin (218^{\circ}) m) ˆ j + (900 m) ˆ k$ $vecv_1 = (19800cos(218^@)" m")hati+(19800sin(218^@)" m")hatj+(900" m")hatk$

${\to v}_{1} = - 15602 ˆ i - 12190 ˆ j + 900 ˆ k$ $vecv_1= -15602hati -12190hatj + 900hatk$ in meters

For the second airplane convert given heading to a standard angle with the reference point as East.

For airplane 2 we are given $23^{\circ}$ $23^@$ West of South

South is $270^{\circ}$ $270^@$ from East and subtract $23^{\circ}$ $23^@$ to make it move toward the West:

$θ_{2} = 270^{\circ} - 23^{\circ}$ $theta_2 = 270^@-23^@$

$θ_{2} = 247^{\circ}$ $theta_2 = 247^@$

Convert the distance from km to meters:

$d_{2} = 17.7 km$ $d_2 = 17.7" km"$

$d_{2} = 17700 m$ $d_2 = 17700" m"$

${altitude}_{2} = 1100 m$ $"altitude"_2 = 1100" m"$

${\to v}_{2} = (17700 cos (247^{\circ}) m) ˆ i + (17700 cos (247^{\circ}) m) ˆ j + (1100 m) ˆ k$ $vecv_2 = (17700cos(247^@)" m")hati+(17700cos(247^@)" m")hatj+(1100" m")hatk$

${\to v}_{2} = - 6916 ˆ i - 16293 ˆ j + 1100 ˆ k$ $vecv_2 = -6916hati -16293hatj + 1100hatk$ in meters

The vector from airplane 1 to airplane 2# is:

${\to v}_{1 \to 2} = {\to v}_{2} - {\to v}_{1}$ $vecv_(1to2) = vecv_2-vecv_1$

${\to v}_{1 \to 2} = (- 6916 - - 15602) ˆ i + (- 16293 - - 12190) ˆ j + (1100 - 900) ˆ k$ $vecv_(1to2) = (-6916- -15602)hati+ (-16293- -12190)hatj+ (1100-900)hatk$

${\to v}_{1 \to 2} = 8686 ˆ i - 4103 ˆ j + 200 ˆ k$ $vecv_(1to2) = 8686hati-4103hatj+ 200hatk$ in meters

The distance between the planes is the magnitude of the vector:

$∣ ∣ {\to v}_{1 \to 2} ∣ ∣ = \sqrt{8686^{2} + {(- 4103)}^{2} + 200^{2}}$ $|vecv_(1to2)| = sqrt(8686^2+ (-4103)^2+ 200^2)$ in meters

$∣ ∣ {\to v}_{1 \to 2} ∣ ∣ = 9608$ $|vecv_(1to2)| = 9608$ meters

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What is the displacement vector FROM the first plane TO the second plane, letting $ˆ i$ $hati$ represent east, $ˆ j$ $hatj$ north, and $ˆ k$ $hatk$ up?

1 Answer

Explanation:

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What is the displacement vector FROM the first plane TO the second plane, letting $ˆ i$ $hati$ represent east, $ˆ j$ $hatj$ north, and $ˆ k$ $hatk$ up?