High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 3 Textbook > Appendix: Review Table for lines and planes

Appendix B: Summary of the course

Line and plane equations, linear independentPrevious Section
Appendix A: Trigonometric FunctionsNext Section
Popular Courses
Find My Course
For the following equations, assume that vectors a⃗\vec a, b⃗\vec b, and c⃗\vec c are defined as:
a⃗=<a1, a2, a3>, b⃗=<b1, b2, b3>, c⃗=<c1, c2, c3>,\vec a=\rowvecth{a_1}{a_2}{a_3},\ \vec b=\rowvecth{b_1}{b_2}{b_3},\ \vec c=\rowvecth{c_1}{c_2}{c_3},
Length of vector a⃗\vec a:
∣a⃗∣=a12+a22+a32|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}
The unit vector corresponding to vector a⃗\vec a is:
a⃗∣a⃗∣=a1ı⃗+a2ȷ⃗+a3k⃗a12+a22+a32\dfrac{\vec a}{|\vec a|}=\dfrac{a_1\vec\imath+a_2\vec\jmath+a_3\vec k}{\sqrt{a_1^2+a_2^2+a_3^2}}
The dot product between vectors a⃗\vec a and b⃗\vec b:
a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta
The angle between the two vectors θ\theta is:
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=a1b1+a2b2+a3b3a12+a22+a32b12+b22+b32\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}
The vector projection of b⃗\vec b onto a⃗\vec a:
projab⃗=a⃗⋅b⃗∣a⃗∣2a⃗proj_a\vec b=\dfrac{\vec a\cdot\vec b}{|\vec a|^2}\vec a
The cross product of a⃗\vec a and b⃗\vec b:
a⃗×b⃗=ı⃗ȷ⃗k⃗a1a2a3b1b2b3=ı⃗(a2b3−a3b2)−ȷ⃗(a1b3−a3b1)+k⃗(a1b2−a2b1)\vec a\times\vec b=\begin{array}{|ccc|} \vec\imath & \vec\jmath & \vec k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{array}=\vec\imath(a_2b_3-a_3b_2)-\vec\jmath(a_1b_3-a_3b_1)+\vec k(a_1b_2-a_2b_1)

PAGE BREAK
The length of the cross product of a⃗\vec a and b⃗\vec b:
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta
Area of a parallelogram formed by a⃗\vec a, b⃗\vec b, and c⃗\vec c:
V=(a⃗×b⃗)⋅c⃗=a⃗⋅(b⃗×c⃗)V=(\vec a\times\vec b)\cdot\vec c=\vec a\cdot(\vec b\times\vec c)

Equation of Lines and Planes:

The scalar format of the equation of line LL passing through point P0(x0,y0,z0)P_0(x_0,y_0,z_0) and parallel to vector a⃗=<a, b, c>\vec a=\rowvecth{a}{b}{c} is:
{x=x0+tay=y0+tbz=z0+tc\begin{cases} x=x_0+ta\\ y=y_0+tb\\ z=z_0+tc \end{cases}
The symmetric format of the above line is:
x−x0a=y−y0b=z−z0c\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}
Equation of a plane passing through point P0(x0,y0,z0)P_0(x_0,y_0,z_0) and normal to vector n⃗=<a, b, c>\vec n=\rowvecth{a}{b}{c} is:
a(x−x0)+b(y−y0)+c(z−z0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0
The angle between two planes with normal vectors n1⃗\vec{n_1} and n2⃗\vec{n_2} is:
cos⁡θ=n1⃗⋅n2⃗∣n1⃗∣∣n2⃗∣\cos\theta=\dfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}
The distance between point P(x1,y1,z1)P(x_1,y_1,z_1) and plane ax+by+cz+d=0ax+by+cz+d=0 is:
D=∣ax1+by1+cz1+d∣a2+b2+c2D=\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

Partial derivative of a two variable function such as f(x,y)f(x,y) with respect to xx and yy is defined as:
fx(x,y)=fx=∂f∂x=lim⁡h→0f(x+h,y)−f(x,y)hfy(x,y)=fy=∂f∂y=lim⁡h→0f(x,y+h)−f(x,y)h\begin{aligned} & f_x(x,y)=f_x=\dfrac{\partial f}{\partial x} =\lim_{h\to0}\dfrac{f(x+h,y)-f(x,y)}{h}\\ & f_y(x,y)=f_y=\dfrac{\partial f}{\partial y} =\lim_{h\to0}\dfrac{f(x,y+h)-f(x,y)}{h} \end{aligned}
Higher partial derivative are defined as:
fxx=∂f∂x(fx)=∂∂x(∂f∂x)=∂2f∂x2fxy=∂∂y(∂f∂x)=∂2f∂y∂xfyy=∂f∂y(fy)=∂∂y(∂f∂y)=∂2f∂y2\begin{array}{c} f_{xx} = \dfrac{\partial f}{\partial x}(f_x)=\dfrac{\partial}{\partial x}\bigg(\dfrac{\partial f}{\partial x}\bigg)=\dfrac{\partial^2 f}{\partial x^2}\\[10pt] f_{xy} =\dfrac{\partial}{\partial y}\bigg(\dfrac{\partial f}{\partial x}\bigg)=\dfrac{\partial^2 f}{\partial y\partial x} & \\[10pt] f_{yy} = \dfrac{\partial f}{\partial y}(f_y)=\dfrac{\partial}{\partial y}\bigg(\dfrac{\partial f}{\partial y}\bigg)=\dfrac{\partial^2 f}{\partial y^2}\\ \end{array}
The tangent plane to the two variable function z=f(x,y)z=f(x,y) at point (x0,y0)(x_0,y_0) is:
z=z0+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z=z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)
The value of the above function at point (x0,y0)(x_0,y_0) can be calculated from:
f(x,y)≈f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)f(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Chain rule for functions with multiple variables:


Case 1: if z=f(x,y)z=f(x,y), x=g(t)x=g(t), and y=h(t)y=h(t), then:
dzdt=∂f∂xdxdt+∂f∂ydydt\dfrac{dz}{dt}=\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}
Case 2: if z=f(x,y)z=f(x,y), x=g(t,s)x=g(t,s), y=h(t,s)y=h(t,s), then:
∂z∂t=∂f∂x∂x∂t+∂f∂y∂y∂t∂z∂s=∂f∂x∂x∂s+∂f∂y∂y∂s\begin{aligned} \dfrac{\partial z}{\partial t}&=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial t}\\ \dfrac{\partial z}{\partial s}&=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial s}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial s} \end{aligned}
The partial derivative of function z=f(x,y)z=f(x,y) expressed in the form of F(x,y,z)=0F(x,y,z)=0 is:
∂z∂x=−∂F∂x∂F∂z=−FxFz∂z∂y=−∂F∂y∂F∂z=−FyFz\begin{aligned} \dfrac{\partial z}{\partial x} &=- \dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial z}}=-\dfrac{F_x}{F_z}\\ \dfrac{\partial z}{\partial y} &=- \dfrac{\dfrac{\partial F}{\partial y}}{\dfrac{\partial F}{\partial z}}=-\dfrac{F_y}{F_z} \end{aligned}

PAGE BREAK
The directional derivative of f(x,y)f(x,y) in the direction of unit vector u⃗=<a, b>\vec u=\rowvec{a}{b} is equal to:
Duf(x,y)=fx(x,y)a+fy(x,y)bD_uf(x,y)=f_x(x,y)a+f_y(x,y)b
The maximum value of directional derivative is ∣∇f∣|\nabla f| and it occurs when the unit vector u⃗\vec u is in the direction of ∇f\nabla f.

Critical Points

Critical points of a two variable function are when fx(x0,y0)=0f_x(x_0,y_0)=0 AND fy(x9,y0)f_y(x_9,y_0) or if any of these partial derivatives does not exist. Second Derivative Test is used to classify these critical points:
D=fxx(x0,y0)fxy(x0,y0)fyx(x0,y0)fyy(x0,y0)=fxx(x0,y0)fyy(x0,y0)−[fxy(x0,y0)]2D=\begin{array}{|cc|} f_{xx}(x_0,y_0) & f_{xy}(x_0,y_0)\\ f_{yx}(x_0,y_0) & f_{yy}(x_0,y_0) \end{array}=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-[f_{xy}(x_0,y_0)]^2
i) if D>0D > 0 and fxx(x0,y0)>0f_{xx}(x_0,y_0)>0, then f(x0,y0)f(x_0,y_0) is a local minimum.
ii) if D>0D>0 and fxx(x0,y0)<0f_{xx}(x_0,y_0)<0, then f(x0,y0)f(x_0,y_0) is a local maximum.
iii) if D<0D < 0, then f(x0,y0)f(x_0,y_0) is not a local maximum or minimum. It is called a saddle point.
iv) if D=0D=0, then the test gives no information, which means that (x0,y0)(x_0,y_0) can be a local minimum, local maximum, or a saddle point.
The absolute maximum and absolute minimum values of the continuous function ff over a closed boundary DD:

Step 1: Find the critical points of f over DD.
Step 2: Find the values of ff over the boundaries of DD.
Step 3: The largest value associated with the critical points and the value from Step 2 is the absolute maximum value, while the smallest value from Steps 1 and 2 is the absolute minimum value.
PAGE BREAK

Lagrange Multipliers


Lagrange Multipliers method is used for minimizing or maximizing function ff considering a constraint in the form of an equation or inequality. For minimizing or maximizing the two variable function z=f(x,y)z=f(x,y) subject to the constraint g(x,y)=kg(x,y)=k, we need to solve:
∇f(x,y)=λ∇g(x,y)\nabla f(x,y)=\lambda \nabla g(x,y)
where λ\lambda is a scalar.

This results in:
∂f∂x=λ∂g∂x∂f∂y=λ∂g∂yg(x,y)=k\begin{array}{l} \dfrac{\partial f}{\partial x}=\lambda\dfrac{\partial g}{\partial x}\\[10pt] \dfrac{\partial f}{\partial y}=\lambda\dfrac{\partial g}{\partial y}\\[10pt] g(x,y)=k \end{array}
and for maximizing or minimizing three variable function f(x,y,z)f(x,y,z) subject to the constraint g(x,y,z)=kg(x,y,z)=k, we need to solve:
∂f∂x=λ∂g∂x∂f∂y=λ∂g∂y∂f∂z=λ∂g∂zg(x,y,z)=k\begin{array}{l} \dfrac{\partial f}{\partial x}=\lambda\dfrac{\partial g}{\partial x}\\[10pt] \dfrac{\partial f}{\partial y}=\lambda\dfrac{\partial g}{\partial y}\\[10pt] \dfrac{\partial f}{\partial z}=\lambda\dfrac{\partial g}{\partial z}\\[10pt] g(x,y,z)=k \end{array}


Double integrals over the general region DD defined as D={(x,y)∣a≤x≤b, g1(x)≤y≤g2(x)}D = \{(x, y)|a \le x \le b,\ g_1(x) \le y \le g_2(x) \} is:
∬Df(x,y)dA=∫ab∫g1(x)g2(x)f(x,y)dydx\iint_Df(x,y)dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)dydx
Changing the order of integral requires redefining DD as D={(x,y)∣c≤y≤d, h1(y)≤x≤h2(y)}D = \{(x, y)|c \le y \le d,\ h_1(y) \le x \le h_2(y) \}. In this case, the above double integral is equal to:
∬Df(x,y)dA=∫cd∫h1(y)h2(y)f(x,y)dxdy\iint_Df(x,y)dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)dxdy

Double Integrals in polar Coordinates


Double integrals can be performed in polar coordinates using the following transformations:
x=rcos⁡θy=rsin⁡θr2=x2+y2dA=rdrdθ\begin{array}{c} x=r\cos\theta\\[10pt] y=r\sin\theta\\[10pt] r^2=x^2+y^2\\[10pt] dA=rdrd\theta \end{array}
The general form of a double integral from rectangular coordinates to polar coordinates is:
∬Df(x,y)dA=∫αβ∫abf(rcos⁡θ,rsin⁡θ)rdrdθ\iint_Df(x,y)dA=\int_\alpha^\beta\int_a^bf(r\cos\theta,r\sin\theta)rdrd\theta

Application of Double Integrals


The total mass of a plate with density (mass per unit area) ρ(x,y)\rho(x,y) over area DD is:
m=∬Dρ(x,y)dAm=\iint_D\rho(x,y)dA
The coordinates (xˉ,yˉ)(\bar x,\bar y) of the center of mass is:
xˉ=1m∬Dxρ(x,y)dAyˉ=1m∬Dyρ(x,y)dA\begin{array}{l} \displaystyle\bar x=\dfrac{1}{m}\iint_Dx\rho(x,y)dA\\[10pt] \displaystyle\bar y=\dfrac{1}{m}\iint_Dy\rho(x,y)dA \end{array}
The moment of inertia of plate about xx axis (IxI_x) and about yy axis (IyI_y) is:
Ix=∬Dy2ρ(x,y)dAIy=∬Dx2ρ(x,y)dA\begin{array}{l} \displaystyle I_x=\iint_Dy^2\rho(x,y)dA\\[10pt] \displaystyle I_y=\iint_Dx^2\rho(x,y)dA \end{array}

The triple integral of f(x,y,z)f(x,y,z) over region EE defined below:
E={(x,y,z)∣a≤x≤b, h1(x)≤y≤h2(x), z1(x,y)≤z≤z2(x,y)}E =\{ (x,y,z)| a \le x \le b,\ h_1(x) \le y \le h_2(x),\ z_1(x,y) \le z \le z_2(x,y)\}
is:
∭Ef(x,y,z)dV=∫ab∫h1(x)h2(x)∫z1(x,y)z2(x,y)f(x,y,z)dzdydx\iiint_Ef(x,y,z)dV=\int_a^b\int_{h_1(x)}^{h_2(x)}\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)dzdydx

Triple Integrals in Cylindrical Coordinates


Triple integral in cylindrical coordinates is performed by using the following transformations:
x=rcos⁡θy=rsin⁡θz=zr2=x2+y2dxdydz=rdrdθdz\begin{array}{c} x=r\cos\theta\\[10pt] y=r\sin\theta\\[10pt] z=z\\[10pt] r^2=x^2+y^2\\[10pt] dxdydz=rdrd\theta dz \end{array}
The general form of a triple integral using cylindrical coordinates is:
∭Ef(x,y,z)dV=∫αβ∫g1(θ)g2(θ)∫z1(rcos⁡θ,rsin⁡θ)z2(rcos⁡θ,rsin⁡θ)f(rcos⁡θ,rsin⁡θ,z)rdrdθdz\iiint_Ef(x,y,z)dV=\int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)}\int_{z_1(r\cos\theta,r\sin\theta)}^{z_2(r\cos\theta,r\sin\theta)}f(r\cos\theta,r\sin\theta,z)rdrd\theta dz
where, the region DD over which the integral is performed is defined as:
D={(r,θ)∣α≤θ≤β, g1(θ)≤r≤g2(θ)}D=\{(r,\theta)|\alpha\le\theta\le\beta,\ g_1(\theta)\le r\le g_2(\theta)\}

PAGE BREAK

Triple Integrals in Spherical Coordinates


Triple integral in spherical coordinates is performed by using the following transformations:
x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕρ2=x2+y2+z2dV=ρ2sin⁡ϕdϕdρdθ\begin{array}{c} x=\rho\sin\phi\cos\theta\\[10pt] y=\rho\sin\phi\sin\theta\\[10pt] z=\rho\cos\phi\\[10pt] \rho^2=x^2+y^2+z^2\\[10pt] dV=\rho^2\sin\phi d\phi d\rho d\theta \end{array}
The general form of a triple integral using spherical coordinates is:
∭Ef(x,y,z)dV=∫λμ∫αβ∫abf(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ)ρ2sin⁡ϕdρdθdϕ\iiint_Ef(x,y,z)dV=\int_\lambda^\mu\int_\alpha^\beta\int_a^bf(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)\rho^2\sin\phi d\rho d\theta d\phi
where, the region EE over which the integral is performed is defined as:
E={(ρ,θ,ϕ)∣a≤rho≤b, α≤θ≤β, λ≤ϕ≤μ}E=\{(\rho,\theta,\phi)|a\le rho\le b,\ \alpha\le\theta\le\beta,\ \lambda\le\phi\le\mu\}

Application of Triple Integrals


The mass of a solid over region EE with a mass per unit volume ρ(x,y,z)\rho(x,y,z) is:
m=∭Eρ(x,y,z)dVm=\iiint_E\rho(x,y,z)dV
Center of mass (xˉ,yˉ,zˉ)(\bar x,\bar y,\bar z) of the plate is:
xˉ=1m∭Exρ(x,y,z)dVyˉ=1m∭Eyρ(x,y,z)dVzˉ=1m∭Ezρ(x,y,z)dV\begin{aligned} \bar x &=\dfrac{1}{m}\iiint_Ex\rho(x,y,z)dV\\ \bar y &=\dfrac{1}{m}\iiint_Ey\rho(x,y,z)dV\\ \bar z &=\dfrac{1}{m}\iiint_Ez\rho(x,y,z)dV\\ \end{aligned}

PAGE BREAK
The moment of inertia about xx, yy, and zz is:
Ix=∭E(y2+z2)ρ(x,y,z)dVIy=∭E(x2+z2)ρ(x,y,z)dVIz=∭E(x2+y2)ρ(x,y,z)dV\begin{array}{l} \displaystyle I_x=\iiint_E(y^2+z^2)\rho(x,y,z)dV\\[10pt] \displaystyle I_y=\iiint_E(x^2+z^2)\rho(x,y,z)dV\\[10pt] \displaystyle I_z=\iiint_E(x^2+y^2)\rho(x,y,z)dV\\[10pt] \end{array}