Porisma

Worked solution

Practise five conversions between vector and Cartesian forms

Self-Practice 7:

(a) Convert r=(2,3,−1)+λ(3,4,5)\mathbf r=(2,3,-1)+\lambda(3,4,5), λ real, to Cartesian form.

(b) Convert r=i+j+λ(i+5k)\mathbf r=\mathbf i+\mathbf j+\lambda(\mathbf i+5\mathbf k), λ real, to Cartesian form.

(c) Convert (x−1)/2=y/3=z−4(x-1)/2=y/3=z-4 to a vector equation.

(d) Convert (1−x)/3=(2y−1)/5(1-x)/3=(2y-1)/5, z=4, to a vector equation.

(e) Give vector and Cartesian equations for the line through A with position vector a=i−3j+2k\mathbf a=\mathbf i-3\mathbf j+2\mathbf k, parallel to m=5i+4j−k\mathbf m=5\mathbf i+4\mathbf j-\mathbf k.

CJC · Lecture · Self-Practice 7

(a) Eliminate a parameter with three nonzero coefficients

x=2+3λ, y=3+4λ and z=−1+5λ give (x−2)/3=(y−3)/4=(z+1)/5(x-2)/3=(y-3)/4=(z+1)/5.

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(b) Keep the constant y-coordinate

The vector equation is (1,1,0)+λ(1,0,5)(1,1,0)+\lambda(1,0,5). Thus x=1+λ, y=1, z=5λ, giving x−1=z/5 with y=1.

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(c) Use the common Cartesian value

Set all three expressions to λ: x=1+2λ, y=3λ, z=4+λ. Hence r=(1,0,4)+λ(2,3,1)\mathbf r=(1,0,4)+\lambda(2,3,1), λ real.

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(d) Solve carefully for x and y

For common value λ, 1−x=3λ and 2 y−1=5λ. Thus x=1−3λ, y=1/2+(5/2)λ and z=4. The printed vector equation is r=(1,1/2,4)+λ(−3,5/2,0)\mathbf r=(1,1/2,4)+\lambda(-3,5/2,0), λ real.

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(e) Use the supplied point and direction

r=(1,−3,2)+λ(5,4,−1)\mathbf r=(1,-3,2)+\lambda(5,4,-1), λ real. Its Cartesian equation is (x−1)/5=(y+3)/4=−z+2(x-1)/5=(y+3)/4=-z+2. These are the printed answers; the component conversions above explain them.

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Common mistakes

Distinguish a constant coordinate from a zero coordinate: zero direction in y means y stays at its given value.

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