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QuestionScrivi l'equazione della retta rr che passa per P(1,0,−1)P(1, 0, -1) e è parallela ai piani x+2y+z+3=0x+2y+z+3=0 e 2x+y−z+1=02x+y-z+1=0. [x−1=−y=z+1][x-1=-y=z+1]

Studdy Solution

STEP 1

Assumptions1. The line rr passes through the point (1;0;−1)(1 ;0 ;-1). The line rr is parallel to the planes α\alpha and β\beta
3. The equation of plane α\alpha is x+y+z+3=0x+ y+z+3=0
4. The equation of plane β\beta is x+y−z+1=0 x+y-z+1=0

STEP 2

Since the line rr is parallel to the planes α\alpha and β\beta, the direction vector of the line is perpendicular to the normal vectors of the planes. The normal vector of a plane is the vector of coefficients in the plane equation.For plane α\alpha, the normal vector is nα=(1,2,1)n_{\alpha} = (1,2,1)For plane β\beta, the normal vector is nβ=(2,1,−1)n_{\beta} = (2,1, -1)

STEP 3

The direction vector of the line rr, drd_r, is the cross product of the normal vectors of the planes.
dr=nα×nβd_r = n_{\alpha} \times n_{\beta}

STEP 4

Calculate the cross product of the normal vectors.
dr=(1,2,1)×(2,1,−1)d_r = (1,2,1) \times (2,1, -1)

STEP 5

The cross product of two vectors (a,b,c)(a, b, c) and (d,e,f)(d, e, f) is (bf−ce,cd−af,ae−bd)(bf - ce, cd - af, ae - bd). Apply this to our vectors.
dr=(2∗−1−1∗1,2∗1−1∗2,1∗1−2∗2)d_r = (2*-1 -1*1,2*1 -1*2,1*1 -2*2)

STEP 6

Calculate the direction vector.
dr=(−2−1,2−2,1−4)=(−3,0,−3)d_r = (-2 -1,2 -2,1 -4) = (-3,0, -3)

STEP 7

The parametric equations of a line passing through a point (x0,y0,z0)(x0, y0, z0) with direction vector d=(a,b,c)d = (a, b, c) are x=x0+atx = x0 + at, y=y0+bty = y0 + bt, z=z0+ctz = z0 + ct.Substitute the values of point $$ and the direction vector $d_r$ into these equations.
x=1−3tx =1 -3ty=0y =0z=−1−3tz = -1 -3t

STEP 8

The Cartesian equation of a line in3 space is given by x=x0+atx = x0 + at, y=y0+bty = y0 + bt, z=z0+ctz = z0 + ct.Substitute the values of point $$ and the direction vector $d_r$ into these equations.
x−1=−3tx -1 = -3ty=0y =0z+1=−3tz +1 = -3t

STEP 9

To get a single Cartesian equation, we can equate the expressions for tt in the xx and zz equations.
x−=z+x - = z +So, the Cartesian equations of the line rr are x−=z+x - = z + and y=y =.

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