Complex Trig Root Problem  Solution 
Determine z and three cube roots of z if one cube root is

We can first find z by cubing using De Moivre’s Theorem :
Convert to Polar first: 
Since we saw that z = –8, we could have noticed that the real number cube root of –8 is –2, and then take the conjugate of to get the other root is (from the Complex Conjugate Root Theorem). So the roots are –2, , and .
But we’ll check our answer using the nth root formulas:

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Vector ProblemSolution
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Find x so that the following vectors are orthogonal:
(a)
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(b)
We’ll need to find and make sure it equals 0:
(a)
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(b)
Vector Parametric Problem  Solution 
Find the equation of the plane that passes through the point \(\left( {4,0,1} \right)\) and is perpendicular to the vector \(n=\text{i}3\text{j}+2\text{k}\). 
The equation of this plane is \(\displaystyle x3y+2z=2\). 
Find the equation of the plane that passes through the point \(\left( {3,3,1} \right)\), and is perpendicular to the line \(\displaystyle \frac{{2x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}\).  We know the direction vector of this line is \(\left\langle {3} \right.,\left. {4,2} \right\rangle \) (we have to set each expression to \(t\), and solve back for\(x\), \(y\), and \(z\). Also note that we had to multiply the \(x\) expression by –1).
The equation of this plane is \(\displaystyle 3x+4y+2z=19\). 
Quadratic with Complex Solutions  Solve Using Quadratic Formula 
\({{x}^{2}}2x+2=0\)
\(\begin{array}{l}a=1\\b=2\\c=2\end{array}\) 
\(\require{cancel} \displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{\left( {2} \right)\pm \sqrt{{{{{\left( {2} \right)}}^{2}}4\left( 1 \right)\left( 2 \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\pm \sqrt{{4}}}}{2}=\frac{{2\pm 2i}}{2}\\&=\frac{{\cancel{2}(1\pm i)}}{{\cancel{2}}}=1\pm i\end{align}\) 
\(3{{x}^{2}}2x+2=0\)
\(\begin{array}{l}a=3\\b=2\\c=2\end{array}\) 
\(\displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{\left( {2} \right)\pm \sqrt{{{{{\left( {2} \right)}}^{2}}4\left( 3 \right)\left( 2 \right)}}}}{{2\left( 3 \right)}}\\&=\frac{{2\pm \sqrt{{20}}}}{6}=\frac{{2\pm 2\sqrt{5}\,i}}{6}\\&=\frac{{{}^{1}\cancel{2}(1\pm \sqrt{5}\,i)}}{{{{{\cancel{6}}}^{3}}}}=\frac{1}{3}\pm \frac{{\sqrt{5}i}}{3}\text{ }\text{ or }\text{ }\frac{1}{3}\pm \frac{{\sqrt{5}}}{3}i\text{ }\end{align}\)
