next up previous
Next: About this document ...

\fbox{ \parbox{16cm}{ 1) The set $x^2+y^2+ z^2-10z=9$\ is a sphere of radius \\ ... ...dd 25 on both sides) to get $x^2+y^2+(z-5)^2=34$. The radius is $\sqrt{34}$. }}

\fbox{ \parbox{16cm}{ 2) $\vec{v},\vec{w}$\ are vectors in space. Then $(\vec{v... ... c) not defined \\ \par a) is correct. We multiply a scalar with a vector. }}

\fbox{ \parbox{16cm}{ 3) If $P,Q,R$\ are three points in space, then \\ a) $\ve... ...{P Q} + \vec{Q R} = \vec{R P}$. \\ \par a) is correct. Just draw a picture. }}

\fbox{ \parbox{16cm}{ 4) If ${\rm proj}_v(w) = {\rm proj}_w(v)$, then \\ a) $\v... ...ot w=0$\ or if $v=w$. Cases $a) and $b) do not apply for example for $v=-w$. }}

\fbox{ \parbox{16cm}{ 5) $\vec{u} \cdot (\vec{v} \times \vec{w})$\ is \\ a) alw... ...v$\ and $w$, the sign changes. The answer can be zero for example of $v=w$. }}

\fbox{ \parbox{16cm}{ 6) Which of the following identities is {\bf not} always t... ...s will be regraded. If you have checked a), then your answer is correct too. }}

\fbox{ \parbox{16cm}{ 7) If $\vec{v} = (1,2,3)$, $\vec{w} = (-2,-4,-6)$, then \\... ...is correct. They are parallel because a scaling with $\lambda=-2$ gives $w$. }}

\fbox{ \parbox{16cm}{ 8) Assume $L$\ is the line $r(t) = P+t \vec{v}$\ with $P=(... ...,8)$. \\ b) $(2,2,2)$. \\ c) $(-2,-2,-2)$. \\ \par b) is not on the line. }}

\fbox{ \parbox{16cm}{ 9) If $\vec{v} = (1,2,3), \vec{w} = (2,2,-2)$, then \\ a)... ... and $\vec{w}$\ are orthogonal. \\ \par c) is correct. They are orthogonal. }}

\fbox{ \parbox{16cm}{ 10) If $\vert\vec{v} + \vec{w}\vert^2 = \vert\vec{v}\vert^... ...w}$\ are orthogonal. \\ \par c) is correct. a) and b) are wrong for $v=-w$. }}



oliver knill 2004-07-08