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Vector Projection Question

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Two math students erect a sun "shade" on the beach (similar to a lean too). This "shade" is rectangular in shape with dimensions 1.5 m long and 2 m wide, and it makes an angle of 60° with the ground. When the sun's rays cast down on this "shade" there will be an area of shade made on the ground (also a rectangle). What is the area of shade that the students will have to sit in at 12 noon ? (that is, what is the projection of the shade onto the ground)? (Assume the sun’s rays are shining directly down).

Guest Jul 5, 2015

#2
+19198
+10

Two math students erect a sun "shade" on the beach (similar to a lean too). This "shade" is rectangular in shape with dimensions 1.5 m long and 2 m wide, and it makes an angle of 60° with the ground. When the sun's rays cast down on this "shade" there will be an area of shade made on the ground (also a rectangle). What is the area of shade that the students will have to sit in at 12 noon ? (that is, what is the projection of the shade onto the ground)? (Assume the sun’s rays are shining directly down).

$$\vec{a}=1.5\cdot \begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\qquad \vec{b} = \begin{pmatrix} 0\\2 \end{pmatrix}\\\\ \begin{array}{lll} \text{area}=\left|\vec{a}_{projection}\times\vec{b} \right| & \qquad \vec{a}_{projection} = \left(\vec{a}\cdot \vec{e}_x\right)\cdot \vec{e}_x & \qquad \vec{e}_x = \begin{pmatrix} 1\\0 \end{pmatrix}\\\\ & \qquad \vec{a}_{projection} =\left( 1.5\cdot \begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\cdot \begin{pmatrix} 1\\0 \end{pmatrix}\right)\cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\ & \qquad \vec{a}_{projection} = \left( 1.5\cdot \cos{(60\ensurement{^{\circ}} )} +0\right) \cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\ & \qquad \vec{a}_{projection} = \begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix} \\\\ \text{area}=\left|\begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix}\times\begin{pmatrix} 0\\2 \end{pmatrix} \right| \\\\ \text{area}=2\cdot 1.5\cdot \cos{ (60\ensurement{^{\circ}} ) } \qquad | \qquad \cos{ (60\ensurement{^{\circ}} ) } = \dfrac{1}{2}\\\\ \text{area}=2\cdot 1.5\cdot \dfrac{1}{2}\\\\ \mathbf{\text{area}=1.5 ~\mathrm{m^2}}\\\\ \end{array}$$

heureka  Jul 6, 2015
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#1
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I will make an assumption that the tarp is 1.5 m TALL and 2 m WIDE.

One side of the shadow  will remain the same if the sun is shining directly down: 2m

To calculate the OTHER side of this shadow rectangle,

The hypotenuse of the triangle  is 1.5 m .    We need the cosine of 60 dgrees times the hypotenuse to calculate the other side of the shadow

COS(60) * 1.5 =.75

So you have a rectangle that is 2m x .75m      Area =  2 * .75   = 1.5 sq m

Guest Jul 5, 2015
#2
+19198
+10

Two math students erect a sun "shade" on the beach (similar to a lean too). This "shade" is rectangular in shape with dimensions 1.5 m long and 2 m wide, and it makes an angle of 60° with the ground. When the sun's rays cast down on this "shade" there will be an area of shade made on the ground (also a rectangle). What is the area of shade that the students will have to sit in at 12 noon ? (that is, what is the projection of the shade onto the ground)? (Assume the sun’s rays are shining directly down).

$$\vec{a}=1.5\cdot \begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\qquad \vec{b} = \begin{pmatrix} 0\\2 \end{pmatrix}\\\\ \begin{array}{lll} \text{area}=\left|\vec{a}_{projection}\times\vec{b} \right| & \qquad \vec{a}_{projection} = \left(\vec{a}\cdot \vec{e}_x\right)\cdot \vec{e}_x & \qquad \vec{e}_x = \begin{pmatrix} 1\\0 \end{pmatrix}\\\\ & \qquad \vec{a}_{projection} =\left( 1.5\cdot \begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\cdot \begin{pmatrix} 1\\0 \end{pmatrix}\right)\cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\ & \qquad \vec{a}_{projection} = \left( 1.5\cdot \cos{(60\ensurement{^{\circ}} )} +0\right) \cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\ & \qquad \vec{a}_{projection} = \begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix} \\\\ \text{area}=\left|\begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix}\times\begin{pmatrix} 0\\2 \end{pmatrix} \right| \\\\ \text{area}=2\cdot 1.5\cdot \cos{ (60\ensurement{^{\circ}} ) } \qquad | \qquad \cos{ (60\ensurement{^{\circ}} ) } = \dfrac{1}{2}\\\\ \text{area}=2\cdot 1.5\cdot \dfrac{1}{2}\\\\ \mathbf{\text{area}=1.5 ~\mathrm{m^2}}\\\\ \end{array}$$

heureka  Jul 6, 2015

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