Question #6201

Find the cosines of the angles that the line r = [-3,2,5] + t[2,2√2] makes with the coordinate axes

Expert's answer

Find the cosines of the angles that the line r = [-3,2,5] + t[2,2–2] makes with the coordinate axes

We'll use the following formula:

cosα = R*i / |R|*|i|,

where i is the direction vector of the coordinate axe and r is the direction vector of the line.

Let's find the direction vector of the line R:

x = -3 + 2t

y = 2 + 2t

z = 5 - 2t

R = dr/dt = (2,2,-2).

|R| = (2²+2²+(-2)²)^(1/2) = (12)^(1/2) = 2*3^(1/2).

Now we can calculate the cosines. Let's begin from the X axe:

ix = (1,0,0), |ix| = 1

R*ix = 2*1+2*0+(-2)*0 = 2

cos(αx) = 2/(2*3^(1/2)) = 1/3^(1/2).

Y axe:

iy = (0,1,0), |iy| = 1

R*iy = 2*0+2*1+(-2)*0 = 2

cos(αy) = 1/3^(1/2).

Z axe:

iz = (0,0,1), |iz| = 1

R*iz = 2*0+2*0+(-2)*1 = -2

cos(αz) = -1/3^(1/2).

We'll use the following formula:

cosα = R*i / |R|*|i|,

where i is the direction vector of the coordinate axe and r is the direction vector of the line.

Let's find the direction vector of the line R:

x = -3 + 2t

y = 2 + 2t

z = 5 - 2t

R = dr/dt = (2,2,-2).

|R| = (2²+2²+(-2)²)^(1/2) = (12)^(1/2) = 2*3^(1/2).

Now we can calculate the cosines. Let's begin from the X axe:

ix = (1,0,0), |ix| = 1

R*ix = 2*1+2*0+(-2)*0 = 2

cos(αx) = 2/(2*3^(1/2)) = 1/3^(1/2).

Y axe:

iy = (0,1,0), |iy| = 1

R*iy = 2*0+2*1+(-2)*0 = 2

cos(αy) = 1/3^(1/2).

Z axe:

iz = (0,0,1), |iz| = 1

R*iz = 2*0+2*0+(-2)*1 = -2

cos(αz) = -1/3^(1/2).

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