12.2 Three-Dimensional Space: Surfaces, Vectors, and Curves (Wk 1)

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State the distance formula in R3\mathbb{R}^3 (Theorem 1) and its special case from the origin.

  • If P=(a1,b1,c1)P=(a_1,b_1,c_1) and Q=(a2,b2,c2)Q=(a_2,b_2,c_2), the distance between PP and QQ is
    ∣P−Q∣=(a2−a1)2+(b2−b1)2+(c2−c1)2|P-Q|=\sqrt{(a_2-a_1)^2+(b_2-b_1)^2+(c_2-c_1)^2}

  • Special case: Q=O=(0,0,0)Q=O=(0,0,0) and P=(a,b,c)P=(a,b,c) gives
    ∣OP∣=a2+b2+c2|OP|=\sqrt{a^2+b^2+c^2}


2
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Give the equations of planes parallel to the coordinate planes.

  • xy-plane: {(x,y,0):x,y∈R}\{(x,y,0):x,y\in\mathbb{R}\}, equation z=0z=0.

  • Plane parallel to the xy-plane: {(x,y,c):x,y∈R}\{(x,y,c):x,y\in\mathbb{R}\}, equation z=cz=c.

  • Similarly:

    • x=ax=a is a plane parallel to the yz-plane.

    • y=by=b is a plane parallel to the xz-plane.


3
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Give the set description and equation of a sphere.

  • A sphere centered at Q=(a,b,c)Q=(a,b,c) with radius RR:
    {(x,y,z):(x−a)2+(y−b)2+(z−c)2=R2}\{(x,y,z):(x-a)^2+(y-b)^2+(z-c)^2=R^2\}

  • Equation: (x−a)2+(y−b)2+(z−c)2=R2(x-a)^2+(y-b)^2+(z-c)^2=R^2.


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Give the equation of a right cylinder in R3\mathbb{R}^3, and compare it with a circle in R2\mathbb{R}^2.

  • A right cylinder in R3\mathbb{R}^3 has radius RR and central axis the vertical line through (a,b,0)(a,b,0).

  • RR is the distance between the point (x,y,z)(x,y,z) and that vertical line (the closest point on the line is (a,b,z)(a,b,z)).

  • Set: {(x,y,z):(x−a)2+(y−b)2=R2}\{(x,y,z):(x-a)^2+(y-b)^2=R^2\}.

  • Equation (for 3D space): (x−a)2+(y−b)2=R2(x-a)^2+(y-b)^2=R^2.

  • Compare: a circle in R2\mathbb{R}^2 with radius RR and center (a,b)(a,b) is {(x,y):(x−a)2+(y−b)2=R2}\{(x,y):(x-a)^2+(y-b)^2=R^2\}. It is the same equation but for 2D space, with only two coordinates.


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How do vectors in 3-space work (components, length, addition, scalar multiplication)?

  • Geometrically it is almost the same as 2D, but pictures are harder to draw.

  • Algebraically there are just 3 coordinates/components.

  • For P=(a1,b1,c1)P=(a_1,b_1,c_1) and Q=(a2,b2,c2)Q=(a_2,b_2,c_2):

    • Vector: PQ→=⟨a2−a1, b2−b1, c2−c1⟩\overrightarrow{PQ}=\langle a_2-a_1,\,b_2-b_1,\,c_2-c_1\rangle (these are the components).

    • Length: ∥PQ→∥=(a2−a1)2+(b2−b1)2+(c2−c1)2\|\overrightarrow{PQ}\|=\sqrt{(a_2-a_1)^2+(b_2-b_1)^2+(c_2-c_1)^2}.

  • For v⃗=⟨v1,v2,v3⟩\vec v=\langle v_1,v_2,v_3\rangle, w⃗=⟨w1,w2,w3⟩\vec w=\langle w_1,w_2,w_3\rangle and λ∈R\lambda\in\mathbb{R}:

    • Addition: v⃗+w⃗=⟨v1+w1, v2+w2, v3+w3⟩\vec v+\vec w=\langle v_1+w_1,\,v_2+w_2,\,v_3+w_3\rangle.

    • Scalar multiplication: λv⃗=⟨λv1,λv2,λv3⟩\lambda\vec v=\langle\lambda v_1,\lambda v_2,\lambda v_3\rangle.


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Define parallel vectors in 3-space and the standard basis i⃗,j⃗,k⃗\vec i,\vec j,\vec k. How do you write any vector in terms of it?

  • Parallel: non-zero v⃗,w⃗\vec v,\vec w are parallel if there exists λ∈R\lambda\in\mathbb{R} with v⃗=λw⃗\vec v=\lambda\vec w.

  • Standard basis in R3\mathbb{R}^3: i⃗=⟨1,0,0⟩\vec i=\langle1,0,0\rangle, j⃗=⟨0,1,0⟩\vec j=\langle0,1,0\rangle, k⃗=⟨0,0,1⟩\vec k=\langle0,0,1\rangle.

  • Any vector in R3\mathbb{R}^3 is a linear combination of i⃗,j⃗,k⃗\vec i,\vec j,\vec k:
    ⟨a,b,c⟩=⟨a,0,0⟩+⟨0,b,0⟩+⟨0,0,c⟩=ai⃗+bj⃗+ck⃗\langle a,b,c\rangle=\langle a,0,0\rangle+\langle0,b,0\rangle+\langle0,0,c\rangle=a\vec i+b\vec j+c\vec k


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How do you describe a general curve in 3D space with a parametric equation?

  • For a point PP on the curve, write r⃗=OP→\vec r=\overrightarrow{OP}.

  • Parametric equation: r⃗=r⃗(t)=⟨x(t),y(t),z(t)⟩\vec r=\vec r(t)=\langle x(t),y(t),z(t)\rangle, or equivalently the system
    x=x(t),y=y(t),z=z(t)x=x(t),\qquad y=y(t),\qquad z=z(t)

  • The notes’ picture marks points on the curve at t=0t=0 and t=1t=1 (position vectors r⃗(0)\vec r(0) and r⃗(1)\vec r(1) from OO).


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Give the equation of a line through a point P=(x0,y0,z0)P=(x_0,y_0,z_0) with direction v⃗=⟨a,b,c⟩\vec v=\langle a,b,c\rangle. Show how it is derived.

  • Derivation: for a point Q=(x,y,z)Q=(x,y,z) on the line, the vector from the base point to QQ is parallel to v⃗\vec v. So there is t∈Rt\in\mathbb{R} with (vector from base point to QQ) =tv⃗=t\vec v. (The notes write this vector as PP0→\overrightarrow{PP_0}, which is presumably the vector between the base point and QQ.) Thus
    ⟨x−x0, y−y0, z−z0⟩=t⟨a,b,c⟩\langle x-x_0,\,y-y_0,\,z-z_0\rangle=t\langle a,b,c\rangle

  • Vector form: ⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨a,b,c⟩\langle x,y,z\rangle=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle.

  • Writing r⃗=OQ→\vec r=\overrightarrow{OQ}: r⃗(t)=OP→+tv⃗\vec r(t)=\overrightarrow{OP}+t\vec v.

  • Parametric (scalar) form:
    x=x0+at,y=y0+bt,z=z0+ctx=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct


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Give the equation of the line through two points P=(a1,b1,c1)P=(a_1,b_1,c_1) and Q=(a2,b2,c2)Q=(a_2,b_2,c_2).

  • Take the direction v⃗=PQ→=⟨a2−a1, b2−b1, c2−c1⟩\vec v=\overrightarrow{PQ}=\langle a_2-a_1,\,b_2-b_1,\,c_2-c_1\rangle.

  • Equation of the line: r⃗(t)=OP→+t PQ→\vec r(t)=\overrightarrow{OP}+t\,\overrightarrow{PQ}.

  • Component form: r⃗(t)=⟨a1,b1,c1⟩+t⟨a2−a1, b2−b1, c2−c1⟩\vec r(t)=\langle a_1,b_1,c_1\rangle+t\langle a_2-a_1,\,b_2-b_1,\,c_2-c_1\rangle.

  • Parametric form:
    x=a1+(a2−a1)t,y=b1+(b2−b1)t,z=c1+(c2−c1)tx=a_1+(a_2-a_1)t,\qquad y=b_1+(b_2-b_1)t,\qquad z=c_1+(c_2-c_1)t