Practice Problems: Review

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\ungraded }[0]{} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Online Problems

Compute the following: $(1,2,3)+(8,3,6)=\answer {(9,5,9)}$ $4(1,-2,4)=\answer {(4,-8,16)}$ $-12((5,2,6)-(8,2,4))=\answer {(36,0,24)}$

Let $h$ be a constant. Compute the following: $(7,2,-1)+(2h,0,h)=\answer {(7+2h,2,h-1)}$ $h(1,8,2)=\answer {(h,8h,2h)}$

For each of the following, determine whether the quantity exists or does not exist.

$(1,8,3,7)+(-1,7,2,7)$

Exists. Does not exist.

$(2,8,3)+(1,7)$

Exists. Does not exist.

$(2,7,3)+1$

Exists. Does not exist.

$(2,8,3)(1,7,3)$

Exists. Does not exist.

$2(7,2,3,7,2)$

Exists. Does not exist.

For points $P_1=(2,-3,7,1)$ and $P_2=(-1,7,2,1)$ , compute the displacement vector $\vec {P_1P_2}$ . $\vec {P_1P_2} = \answer {(-3,10,-5,0)}$

Write the vector $2\mathbf {i}-5\mathbf {j}+2\mathbf {k}$ in $\mathbb {R}^3$ in standard vector notation. $2\mathbf {i}-5\mathbf {j}+2\mathbf {k}=\answer {(2,-5,2)}$

Compute the dot product. $(1,8,3)\cdot (-2,6,0)=\answer {46}$

Compute the dot product. $(1,-5,0,2)\cdot (2,-1,4,1)=\answer {9}$

Compute the dot product. $(1,8,3)\cdot (-3,0,1)=\answer {0}$ What can you conclude about the vectors?

They’re perpendicular. They aren’t perpendicular.

Compute the dot product. $(1,2,3)\cdot (-3,-2,-1)=\answer {-10}$ What can you conclude about the vectors?

They’re perpendicular. They aren’t perpendicular.

Compute the dot product. $(1,8,3,6)\cdot (3,-3,-1,4)=\answer {-0}$ What can you conclude about the vectors?

They’re perpendicular. They aren’t perpendicular.

For each expression, determine whether it exists or does not exist.

$(2,8,3)\cdot (1,8,2,4)$

Exists. Does not exist.

$(2,8,3,1,8)\cdot (1,8,2,4,2)$

Exists. Does not exist.

Compute the angle between the vectors $(2,8)$ and $(-8,2)$ in degrees. $\theta = \answer {90}^\circ$

Compute the angle between the vectors $(2,1,3,1,1)$ and $(-3,1,1,1,-2)$ in degrees. $\theta = \answer {104.48}^\circ$ (Give your answer as a positive number to two decimal places.

Suppose you have vectors $\vec {v}$ and $\vec {w}$ such that $\|\vec {v}\|=6$ and $\|\vec {w}\|=2$ , and the angle between $\vec {v}$ and $\vec {w}$ is $\frac {\pi }{4}$ radians. Compute the dot product of $\vec {v}$ and $\vec {w}$ . $\vec {v}\cdot \vec {w} = \answer {6\sqrt {2}}$

Compute the projection of $\vec {v}=(1,7,3)$ onto $\vec {w}=(-3,-2,1)$ . $\textrm {proj}_{\vec {w}}(\vec {v}) = \answer {(7/2,7/3,-7/6)}$

Compute the projection of the vector $\vec {v}=(1,2,3)$ onto the vector $\vec {w} = (-3,2,-1)$ . $\textrm {proj}_{\vec {w}}(\vec {v}) = \answer {(0,0,0)}$

Why does your answer make sense?

The vectors are parallel. The vectors are perpendicular. They are the same length.

Compute the cross product. $(1,7,3)\times (2,8,-1) = \answer {(-31,7,-6)}$

Compute the cross product. $(-1,7,2)\times (2,8,3) = \answer {(5,7,-22)}$

For each of the following, determine whether the expression exists or does not exist.

$(-1,7,3)\times (1,7,2)$

Exists. Does not exist.

$(-1,7,3,8)\times (1,7,2,0)$

Exists. Does not exist.

$(2,0,0)\times (1,7,2,0)$

Exists. Does not exist.

$(0,0,0)\times (0,0,0)$

Exists. Does not exist.

Compute the area of the parallelogram determined by $(1,6)$ and $(1,0)$ . $\textrm {Area} = \answer {6}$

Compute the volume of the parallelepiped determined by $(1,6,2)$ , $(-1,2,0)$ , and $(0,3,1)$ . $\textrm {Volume} = \answer {2}$

Suppose $\vec {v}$ and $\vec {w}$ are unit vectors in the $xy$ -plane, and we know that they are perpendicular. What is $\vec {v}\times \vec {w}$ ?

$(0,0,1)$ $(0,0,-1)$ Not enough information.

Find a parametrization $\vec {x}(t)$ of the line parallel to the vector $(1,6,3)$ and through the point $(1,3,2)$ , such that $\vec {x}(0) = (1,3,2)$ . $\vec {x}(t) = \answer {(t+1,6t+3,3t+2)}$

Find a parametrization $\vec {x}(s,t)$ of the plane containing vectors $(1,6,2)$ and $(1,3,2)$ , and passing through the point $(1,0,0)$ , such that $\vec {x}(0,0) = (1,0,0)$ and $\vec {x}(1,0) = (1,6,2)$ . $\vec {x}(s,t) = \answer {(s+t+1, 6s+3t, 2s+2t)}$

Give an equation which describes the plane perpendicular to the vector $(1,7,3)$ and through the point $(-2,4,1)$ . $0=\answer {(x+2)+7(y-4)+3(z-1)}$

Written Problems

For any vector $\vec {v}$ in $\mathbb {R}^n$ , prove that $\vec {v}\cdot \vec {v}=\|\vec {v}\|^2$ .

Prove that vectors $\vec {v}$ and $\vec {w}$ in $\mathbb {R}^n$ are perpendicular if and only if $\textrm {proj}_{\vec {w}}(\vec {v})$ is the zero vector.

For any vector $\vec {v}$ in $\mathbb {R}^3$ , prove that $\vec {v}\times \vec {v}$ is the zero vector.