Title: Piste- ja ristitulo

Olkoon \(x=([value],[value]), y=([value],[value]), z=([value],[value])\). Määritä

\(\|x-y\|+\|y-z\|=\) ja

\(\|x-z\|=\) .

Mieti, ovatko pisteet \(x,y,z\) samalla suoralla? Miten tämän voi päätellä edellisten laskujen perusteella?

Anna vastaukseksi likiarvo kolmen (3) desimaalin tarkkuudella. Käytä pistettä desimaalierottimena.
Feedback: Mallivastaus: \(\begin{align*} \|x-y\|+\|y-z\| & = \|([value],[value])-([value],[value])\|+\|([value],[value])-([value],[value])\| \\ & = \|([value],[value])\|+\|([value],[value])\| \\ & = \sqrt{([value])^2+([value])^2}+\sqrt{([value])^2+([value])^2} \\ & \approx [value]. \end{align*} \) \(\begin{align*} \|x-z\| & = \|([value],[value])-([value],[value])\| \\ & = \|([value],[value])\| \\ & = \sqrt{([value])^2+([value])^2} \\ & \approx [value]. \end{align*}\) Pisteet ovat samalla suoralla, jos ja vain jos \[ \|x-z\|=\|x-y\|+\|y-z\|. \]

Olkoot \( \mathbf{u} = [value]\mathbf{e}_1 + [value]\mathbf{e}_2 + [value] \mathbf{e}_3, \quad \mathbf{v} = -\mathbf{e}_2 - \mathbf{e}_3 \text{ sekä} \quad \mathbf{w} = \mathbf{e}_1 + [value]\mathbf{e}_2 - \mathbf{e}_3 \).

Tarkista seuraavat kohdat ensin Matlabilla ja kirjoita vastaukset STACKiin vasta sen jälkeen.

\( \textbf{(a)} \) Laske \( \mathbf{u} \times \mathbf{v} \).
Vastaus: \( \mathbf{e}_1 + \) \( \mathbf{e}_2 + \) \( \mathbf{e}_3 \).

\( \textbf{(b)} \) Laske vektorikolmitulo \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \).
Vastaus: \( \mathbf{e}_1 + \) \( \mathbf{e}_2 + \) \( \mathbf{e}_3 \).
Feedback: Mallivastaus: \( \textbf{(a)} \) Karteesisessa koordinaatistossa x- ,y- ja z-akselien suuntaisilla yksikkövektoreilla \( \mathbf{e}_1, \mathbf{e}_2 \text{ ja } \mathbf{e}_3 \) määriteltyjen vektoreiden \( \mathbf{u} \) ja \( \mathbf{v} \) ristitulo voidaan laskea determinantin avulla: \[ \begin{align} \mathbf{u} \times \mathbf{v} & = \text{det} \left[ \begin{matrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ [value] & [value] & [value] \\ 0 & [value] & [value] \end{matrix} \right] \\ & = [value]\mathbf{e}_1 - ([value]) \mathbf{e}_1 + [value] \mathbf{e}_2 - [value] \mathbf{e}_2 + [value] \mathbf{e}_3 - [value]\mathbf{e}_3 \\ & = [value] \mathbf{e}_1 + [value] \mathbf{e}_2 + [value] \mathbf{e}_3. \end{align} \] \( \textbf{(b)} \) Vektorikolmitulo on vektori, joka voidaan laskea kaavalla \[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \mathbf{v}(\mathbf{u} \cdot \mathbf{w}) - \mathbf{w}(\mathbf{u} \cdot \mathbf{v}). \] Nyt \[ \begin{align} \mathbf{u} \cdot \mathbf{w} & = [value] \cdot [value] + [value] \cdot [value] + [value] \cdot ([value]) \\ & = [value] \\ \\ & \text{ja} \\ \\ \mathbf{u} \cdot \mathbf{v} & = [value] \cdot [value] + [value] \cdot ([value]) + [value] \cdot ([value]) \\ & = [value]. \end{align} \] Siten \[ \begin{align} \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) & = \mathbf{v}(\mathbf{u} \cdot \mathbf{w}) - \mathbf{w}(\mathbf{u} \cdot \mathbf{v}) \\ & = [value] \mathbf{v} - ([value]) \mathbf{w} \\ & = [value] \mathbf{e}_1 + ([value]) \mathbf{e}_2 + ([value]) \mathbf{e}_3 - ( [value] \mathbf{e}_1 + ([value]) \mathbf{e}_2 + [value] \mathbf{e}_3 ) \\ & = [value] \mathbf{e}_1 + ([value]) \mathbf{e}_2 + ([value]) \mathbf{e}_3. \end{align} \]

Kolmion kärkipisteet ovat ([value],[value],[value]), ([value],[value],[value]), ([value],[value],[value]). Laske kolmion pinta-ala.
\(\textbf{Vihje:} \) Vektorien \( \mathbf{u} \) ja \( \mathbf{v} \) virittämän suunnikkaan pinta-ala on \( \lVert \mathbf{u} \times \mathbf{v} \rVert \).

\( \mathbf{HUOM!} \) Vastaukseksi hyväksytään vain pinta-alan TARKKA ARVO! Älä anna likiarvoja. Syntaksivihjeitä: Jakolaskun saat /-merkin avulla ja neliöjuuren sqrt()-funktiolla. Esim. \( \dfrac{a}{\sqrt{b}} \) kirjoitetaan a/(sqrt(b)).

Vastaus:
Feedback: \[ \textbf{Mallivastaus:} \] Merkitään kolmion kärkipisteitä vektoreina \[ \mathbf{a} = [value], \quad \mathbf{b} = [value], \quad \mathbf{c} = [value]. \] Näin kolmion sivuina ovat esimerkiksi vektorit \[ \mathbf{u}=\mathbf{b}-\mathbf{a}=[value] \quad \text{ja} \quad \mathbf{v}=\mathbf{c}-\mathbf{a}=[value]. \] Lasketaan ensin \[ \mathbf{u} \times \mathbf{v} = \left| \begin{matrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ [value] & [value] & [value] \\ [value] & [value] & [value] \end{matrix} \right| = \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_1} + \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_2} +\left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_3} = [value] \] Vektorien \( \mathbf{u} \) ja \( \mathbf{v} \) virittämän suunnikkaan pinta-ala on tämän ristitulovektorin pituus \( \lVert \mathbf{u} \times \mathbf{v} \rVert \). Kysytty kolmion pinta-ala on puolet tästä: \[ A = \dfrac{1}{2}\lVert \mathbf{u} \times \mathbf{v} \rVert=\dfrac{\sqrt{([value])^2+([value])^2+([value])^2}}{2} = [value] \approx [value]. \]

The vertices of a triangle are located at ([value],[value],[value]), ([value],[value],[value]) and ([value],[value],[value]). What is the area of the triangle?

\(\textbf{Hint:} \) The area of the parallelogram determined by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \lVert \mathbf{u} \times \mathbf{v} \rVert \).

\( \textbf{Please note!} \) Approximate values are not accepted. Please give the exact value of the area. Use the forward slash ( / ) for division and sqrt() for square root. For example \( \dfrac{a}{\sqrt{b}} \) can be written as \(\mathsf{a/sqrt(b)}\).

Kolmion kärkipisteet ovat ([value],[value],[value]), ([value],[value],[value]) ja ([value],[value],[value]). Laske kolmion pinta-ala.

\(\textbf{Vihje:} \) Vektorien \( \mathbf{u} \) ja \( \mathbf{v} \) virittämän suunnikkaan pinta-ala on \( \lVert \mathbf{u} \times \mathbf{v} \rVert \).

\( \mathbf{HUOM!} \) Vastaukseksi hyväksytään vain pinta-alan TARKKA ARVO! Älä anna likiarvoja. Syntaksivihjeitä: Jakolaskun saat /-merkin avulla ja neliöjuuren sqrt()-funktiolla. Esim. \( \dfrac{a}{\sqrt{b}} \) kirjoitetaan \(\mathsf{a/sqrt(b)}\).

Answer: Vastaus:
Feedback: Worked solution: Denote the vertices of the triangle as vectors \[ \mathbf{a} = [value], \quad \mathbf{b} = [value], \quad \mathbf{c} = [value]. \] Now we can denote the edges of the triangle with vectors \[ \mathbf{u}=\mathbf{b}-\mathbf{a}=[value] \quad \text{and} \quad \mathbf{v}=\mathbf{c}-\mathbf{a}=[value]. \] First calculate \[ \mathbf{u} \times \mathbf{v} = \left| \begin{matrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ [value] & [value] & [value] \\ [value] & [value] & [value] \end{matrix} \right| = \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_1} + \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_2} +\left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_3} = [value] \] The area of the parallelogram determined by the vectors \( \mathbf{u} \) ja \( \mathbf{v} \) is \( \lVert \mathbf{u} \times \mathbf{v} \rVert \). The area of the triangle is then half of this. \[ A = \dfrac{1}{2}\lVert \mathbf{u} \times \mathbf{v} \rVert=\dfrac{\sqrt{([value])^2+([value])^2+([value])^2}}{2} = [value] \approx [value]. \] Mallivastaus: Merkitään kolmion kärkipisteitä vektoreina \[ \mathbf{a} = [value], \quad \mathbf{b} = [value], \quad \mathbf{c} = [value]. \] Näin kolmion sivuina ovat esimerkiksi vektorit \[ \mathbf{u}=\mathbf{b}-\mathbf{a}=[value] \quad \text{ja} \quad \mathbf{v}=\mathbf{c}-\mathbf{a}=[value]. \] Lasketaan ensin \[ \mathbf{u} \times \mathbf{v} = \left| \begin{matrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ [value] & [value] & [value] \\ [value] & [value] & [value] \end{matrix} \right| = \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_1} + \left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_2} +\left| \begin{matrix} [value] & [value] \\ [value] & [value] \end{matrix} \right| \mathbf{e_3} = [value] \] Vektorien \( \mathbf{u} \) ja \( \mathbf{v} \) virittämän suunnikkaan pinta-ala on tämän ristitulovektorin pituus \( \lVert \mathbf{u} \times \mathbf{v} \rVert \). Kysytty kolmion pinta-ala on puolet tästä: \[ A = \dfrac{1}{2}\lVert \mathbf{u} \times \mathbf{v} \rVert=\dfrac{\sqrt{([value])^2+([value])^2+([value])^2}}{2} = [value] \approx [value]. \]

Let \({\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) and \({\bf v} = [value]{\bf i} +[value]{\bf j}+[value]{\bf k}\). Calculate \({\bf u}\times {\bf v}\) and the area of the triangle determined by \({\bf u}\) and \({\bf v}\).

\( \textbf{Please note!} \) Approximate values are not accepted. Give the exact value of the cross product and the area. Use the forward slash ( / ) for division and sqrt() for square root. For example \( \dfrac{a}{\sqrt{b}} \) can be written as \(\mathsf{a/sqrt(b)}\). Insert the vector as a list, for example the vector \(2{\bf i}+{\bf j}+3{\bf k}\) should be inserted as \(\mathsf{[2,1,3]}\).

Olkoon \({\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) ja \({\bf v} = [value]{\bf i} +[value]{\bf j}+[value]{\bf k}\). Määritä ristitulo \({\bf u}\times {\bf v}\) ja vektoreiden \({\bf u}\) ja \({\bf v}\) rajaaman kolmion pinta-ala.

\( \textbf{Huom!} \) Anna vastaukseksi tarkka arvo. Likiarvoja ei hyväksytä vastaukseksi. Jakolaskun saat /-merkin avulla ja neliöjuuren sqrt()-funktiolla. Esim. \( \dfrac{a}{\sqrt{b}} \) kirjoitetaan \(\mathsf{a/sqrt(b)}\). Vektorit syötetään listoina, esimerkiksi vektori \(2{\bf i}+{\bf j}+3{\bf k}\) annetaan kirjoittamalla [2,1,3].

\({\bf u}\times {\bf v}=~\)
AreaPinta-ala \(A=~\)
Feedback: Worked solution: \[\begin{array}{ll} \mathbf{u}\times \mathbf{v} &=(u_2v_3-u_3v_2)\mathbf{i}+(u_3v_1-u_1v_3)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k}\\ &=([value]-[value])\mathbf{i}+([value]-[value])\mathbf{j}+([value]-[value])\mathbf{k} \\ &=[value]\mathbf{i}+[value]\mathbf{j}-[value]\mathbf{k} \\ &=([value],[value],[value]). \end{array} \] The area of the triangle determined by vectors \(\mathbf{u}\) ja \(\mathbf{v}\): \[\begin{array}{ll} A &=\frac{1}{2}||\mathbf{u}\times \mathbf{v}|| \\ &=\frac{1}{2} \sqrt{([value])^2+([value])^2+([value])^2} \\ &=\frac{1}{2} \sqrt[value] \\ &=[value]. \end{array} \] Mallivastaus: \[\begin{array}{ll} \mathbf{u}\times \mathbf{v} &=(u_2v_3-u_3v_2)\mathbf{i}+(u_3v_1-u_1v_3)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k}\\ &=([value]-[value])\mathbf{i}+([value]-[value])\mathbf{j}+([value]-[value])\mathbf{k} \\ &=[value]\mathbf{i}+[value]\mathbf{j}-[value]\mathbf{k} \\ &=([value],[value],[value]). \end{array} \] Vektoreiden \(\mathbf{u}\) ja \(\mathbf{v}\) määräämän kolmion pinta-ala: \[\begin{array}{ll} A &=\frac{1}{2}||\mathbf{u}\times \mathbf{v}|| \\ &=\frac{1}{2} \sqrt{([value])^2+([value])^2+([value])^2} \\ &=\frac{1}{2} \sqrt[value] \\ &=[value]. \end{array} \]

Let \( {\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k} \) and \( {\bf v}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k} \). Use \( \theta \) to denote the angle between \( \mathbf{u} \) and \( \mathbf{v} \). Now calculate \( {\bf u}\cdot {\bf v} \) and \( \cos\theta \).

\( \textbf{Please note!} \) Approximate values are not accepted. Give the exact values. Use the forward slash ( / ) for division and sqrt() for square root. For example \( \dfrac{a}{\sqrt{b}} \) can be written as \(\mathsf{a/sqrt(b)}\).

Olkoon \( {\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k} \) ja \( {\bf v}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k} \). Merkitään vektorien välistä kulmaa \( \theta \). Laske nyt \( {\bf u}\cdot {\bf v} \) sekä \( \cos\theta \).

\( \textbf{Huom!} \) Anna vastaukseksi tarkka arvo. Likiarvoja ei hyväksytä vastaukseksi. Jakolaskun saat /-merkin avulla ja neliöjuuren sqrt()-funktiolla. Esim. \( \dfrac{a}{\sqrt{b}} \) kirjoitetaan \(\mathsf{a/sqrt(b)}\).

\( {\bf u}\cdot {\bf v} =~\) |
\(\cos\theta =~\) |
Feedback: Worked solution: The dot product of vectors \({\bf u}=u_1{\bf i}+u_2{\bf j}+u_3{\bf k}\) and \({\bf v}=v_1{\bf i}+v_2{\bf j}+v_3{\bf k}\) is \[ {\bf u}\cdot {\bf v} = u_1v_1+u_2v_2+u_3v_3.\] Therefore the dot product of the two given vectors \({\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) and \({\bf v}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) is \[ {\bf u}\cdot {\bf v} = [value]\cdot [value]+[value]\cdot [value]+[value]\cdot [value]=[value]. \] We denote the angle between \({\bf u}\) and \({\bf v}\) as \(\theta\) so now \[ {\bf u}\cdot {\bf v}=|{\bf u}|\,|{\bf v}|\cos \theta \] from which we may solve the cosine \[ \cos\theta =\frac{{\bf u}\cdot {\bf v}}{|{\bf u}| |{\bf v}|}. \] Now \[ |{\bf u}|=\sqrt{[value]^2+[value]^2+[value]^2}=[value] \] and \[ |{\bf v}|=\sqrt{[value]^2+[value]^2+[value]^2}=[value] \] so \[ \cos \theta=\frac[value]{[value]\cdot [value]}=[value]. \] Mallivastaus: Vektoreiden \({\bf u}=u_1{\bf i}+u_2{\bf j}+u_3{\bf k}\) ja \({\bf v}=v_1{\bf i}+v_2{\bf j}+v_3{\bf k}\) pistetulo on \[ {\bf u}\cdot {\bf v} = u_1v_1+u_2v_2+u_3v_3,\] joten annettujen vektoreiden \({\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) ja \({\bf v}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) pistetulo on \[ {\bf u}\cdot {\bf v} = [value]\cdot [value]+[value]\cdot [value]+[value]\cdot [value]=[value]. \] Jos \(\theta\) on vektoreiden \({\bf u}\) ja \({\bf v}\) välinen kulma, niin silloin \[ {\bf u}\cdot {\bf v}=|{\bf u}|\,|{\bf v}|\cos \theta, \] eli vektoreiden \({\bf u}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) ja \({\bf v}=[value]{\bf i}+[value]{\bf j}+[value]{\bf k}\) välisen kulman kosini on \[ \cos\theta =\frac{{\bf u}\cdot {\bf v}}{|{\bf u}| |{\bf v}|}. \] Nyt \[ |{\bf u}|=\sqrt{[value]^2+[value]^2+[value]^2}=[value], \] ja \[ |{\bf v}|=\sqrt{[value]^2+[value]^2+[value]^2}=[value], \] eli \[ \cos \theta=\frac[value]{[value]\cdot [value]}=[value]. \]

Let \( \mathbf{u}=([value],[value]), \mathbf{v}=([value],[value]) \). Denote the angle between the vectors as \( \theta \). Now calculate the exact value of the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) as well as the angle \( \theta \). Input the angle as radians rounded to 3 decimal places.

Olkoon \( \mathbf{u}=([value],[value]), \mathbf{v}=([value],[value]) \). Merkitään vektoreiden välistä kulmaa \( \theta \). Laske vektoreiden välisen pistetulon tarkka arvo sekä kulman \( \theta \) arvo radiaaneina kolmen desimaalin tarkkuudella.

\( \textbf{Huom!} \) Käytä desimaalierottimena pistettä pilkun sijaan.

\( {\bf u}\cdot {\bf v} =~\) |
\( \theta =~\) |
Feedback: Worked solution: The dot product of two vectors \( \mathbf{u}=(u_1,u_2) \) and \( \mathbf{v}=(v_1,v_2) \) is \[ \mathbf{u}\cdot \mathbf{v} = u_1v_1+u_2v_2 \] so the dot product between the given vectors \( \mathbf{u}=([value],[value]) \) and \( \mathbf{v}=([value],[value]) \) can be calculated as \[ \mathbf{u}\cdot \mathbf{v} = [value]\cdot [value]+[value]\cdot [value]=[value]. \] If we denote the angle between the vectors as \( \theta \) we may write \[ \mathbf{u}\cdot \mathbf{v}=|\mathbf{u}|\,|\mathbf{v}|\cos \theta \] and solve for the angle: \[ \theta = \cos^{-1} \frac{\mathbf{u}\cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}. \] Now \[ |\mathbf{u}|=\sqrt{[value]^2+[value]^2}=[value], \] and \[ |\mathbf{v}|=\sqrt{[value]^2+[value]^2}=[value], \] so \[ \theta = \cos^{-1} \frac[value]{[value]\cdot [value]}\approx [value] \ \text{(rad)}. \] Mallivastaus: Vektoreiden \( \mathbf{u}=(u_1,u_2) \) ja \( \mathbf{v}=(v_1,v_2) \) pistetulo on \[ \mathbf{u}\cdot \mathbf{v} = u_1v_1+u_2v_2, \] joten annettujen vektoreiden \( \mathbf{u}=([value],[value]) \) ja \( \mathbf{v}=([value],[value]) \) pistetulo on \[ \mathbf{u}\cdot \mathbf{v} = [value]\cdot [value]+[value]\cdot [value]=[value]. \] Jos \( \theta \) on vektoreiden \( \mathbf{u} \) ja \( \mathbf{v} \) välinen kulma, niin silloin \[ \mathbf{u}\cdot \mathbf{v}=|\mathbf{u}|\,|\mathbf{v}|\cos \theta, \] eli vektoreiden \( \mathbf{u}=([value],[value]) \) ja \(\mathbf{v}=([value],[value]) \) välinen kulma on \[ \theta = \cos^{-1} \frac{\mathbf{u}\cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}. \] Nyt \[ |\mathbf{u}|=\sqrt{[value]^2+[value]^2}=[value], \] ja \[ |\mathbf{v}|=\sqrt{[value]^2+[value]^2}=[value], \] eli \[ \theta = \cos^{-1} \frac[value]{[value]\cdot [value]}\approx [value] \ \text{(rad)}. \]

Let \( {\bf a} = ([value], [value]) \). Find a vector \({\bf b}\neq {\bf 0}\) such that it is perpendicular to \(\mathbf{a}\).

Olkoon vektori \( {\bf a} = ([value], [value]) \). Anna vektori \({\bf b}\neq {\bf 0}\) siten, että vektorit ovat kohtisuorassa toisiaan vastaan.

\({\bf b} = \Big(\)\(,\) \(\Big)\)
Feedback: Worked solution: Two vectors are said to be perpendicular to each other when their dot product is zero. If \( {\bf a} = ([value], [value]) \) then for example the vector \({\bf b} = ([value], [value]) \) is perpendicular to \( \mathbf{a} \): \[ {\bf a} \cdot {\bf b} = [value]\cdot[value]+[value]\cdot ([value])=0.\] Vector \({\bf 0}\) is said to be perpendicular to all other vectors. Mallivastaus: Vektorit ovat kohtisuorassa toisiaan vastaan, kun niiden välinen pistetulo on nolla. Siis jos vektori \( {\bf a} = ([value], [value]) \), niin esimerkiksi vektori \({\bf b} = ([value], [value]) \) toteuttaa tämän: \[ {\bf a} \cdot {\bf b} = [value]\cdot[value]+[value]\cdot ([value])=0.\] Jos vektori \({\bf b}\) on nollavektori, sanotaan että se on kohtisuorassa kaikkia muita vektoreita vastaan.

Olkoon \( {\bf u}=[value]{\bf e_1}+[value]{\bf e_2}+[value]{\bf e_3}+[value]{\bf e_4} \) ja \( {\bf v}=[value]{\bf e_1}+[value]{\bf e_2}+[value]{\bf e_3}+[value]{\bf e_4} \). Laske

\( {\bf u}\cdot {\bf v} = \) ja

vektoreiden välisen kulman \(\theta\) kosinin tarkka arvo

\(\cos\theta=\) .

Neliöjuuri kirjoitetaan esim. muodossa sqrt(5), kertolasku esim. 3*5 ja jakolasku esim. 4/9.
Feedback: Mallivastaus: Vektoreiden \({\bf u}=u_1{\bf e_1}+u_2{\bf e_2}+u_3{\bf e_3}+u_4{\bf e_4}\) ja \({\bf v}=v_1{\bf e_1}+v_2{\bf e_2}+v_3{\bf e_3}+v_4{\bf e_4}\) pistetulo on \[{\bf u}\cdot {\bf v} = u_1v_1+u_2v_2+u_3v_3+u_4v_4,\] joten annettujen vektoreiden \( {\bf u}=[value]{\bf e_1}+[value]{\bf e_2}+[value]{\bf e_3}+[value]{\bf e_4} \) ja \( {\bf v}=[value]{\bf e_1}+[value]{\bf e_2}+[value]{\bf e_3}+[value]{\bf e_4} \) pistetulo on \[{\bf u}\cdot {\bf v}= [value]\cdot [value]+[value]\cdot [value]+[value]\cdot [value] +[value]\cdot[value]=[value].\] Jos \(\theta\) on vektoreiden \({\bf u}\) ja \({\bf v}\) välinen kulma, niin silloin \[{\bf u}\cdot {\bf v}=|{\bf u}|\,|{\bf v}|\cos \theta,\] eli vektoreiden \({\bf u} \) ja \({\bf v}\) välisen kulman kosini on \[\cos\theta =\frac{{\bf u}\cdot {\bf v}}{|{\bf u}| |{\bf v}|}.\] Nyt \[|{\bf u}|=\sqrt{[value]^2+[value]^2+[value]^2+[value]^2}=[value], \] ja \[|{\bf v}|=\sqrt{[value]^2+[value]^2+[value]^2+[value]^2}=[value],\] eli \[\cos \theta=\frac[value]{[value]\cdot [value]}=[value].\]

Olkoon \( u=([value],[value]), v=([value],[value]) \).

Määritä pistetulo

\( u\cdot v = \) ja

vektoreiden \( u \) ja \( v \) välinen kulma \( \theta \) (radiaaneina)

\( \theta \approx \) .

Anna kulman likiarvo kolmen desimaalin tarkkuudella. Käytä pistettä desimaalierottimena.
Feedback: Mallivastaus: Vektoreiden \( u=(u_1,u_2) \) ja \( v=(v_1,v_2) \) pistetulo on \[ u\cdot v = u_1v_1+u_2v_2, \] joten annettujen vektoreiden \( u=([value],[value]) \) ja \( v=([value],[value]) \) pistetulo on \[u\cdot v = [value]\cdot [value]+[value]\cdot [value]=[value]. \] Jos \( \theta \) on vektoreiden \( u \) ja \( v \) välinen kulma, niin silloin \[u\cdot v=|u|\,|v|\cos \theta, \] eli vektoreiden \( u=([value],[value]) \) ja \(v=([value],[value]) \) välinen kulma on \[\theta = \cos^{-1} \frac{u\cdot v}{|u|\,|v|}.\] Nyt \[|u|=\sqrt{[value]^2+[value]^2}=[value],\] ja \[|v|=\sqrt{[value]^2+[value]^2}=[value],\] eli \[ \theta = \cos^{-1} \left( \frac[value]{[value]\cdot [value]} \right) \approx [value] \ \text{(rad)}.\]

Tarkastellaan seuraavia vektoreita:

\(\vec{v}_1=([value],[value],[value]), \vec{v}_2=([value],[value],[value]), \vec{w}_1=([value],[value],[value],[value]), \vec{w}_2=([value],[value],[value],[value])\) ja \(\vec{w}_3=([value],[value],[value],[value])\)

a) Määritä seuraavat pistetulot

\(\vec{v}_1 \cdot \vec{v}_2 =\).

\(\vec{w}_1 \cdot \vec{w}_2 =\).

\(\vec{w}_1 \cdot \vec{w}_3 =\).

b) Laske seuraavat pistetuloon liittyvät laskut.

\(\vec{w}_2 \cdot \vec{w}_1 =\)  .

\(([value] \vec{w}_1) \cdot ([value]\vec{w}_2) = \).

\((\vec{w}_3 + \vec{w}_2)  \cdot \vec{w}_1 = \) .
Feedback: Ratkaisu a) Pistetulot lasketaan pistetulon määritelmää käyttäen. Kannattaa pitää mielessä, että pistetulon tulos on aina reaaliluku, ei vektori. \(\vec{v}_1 \cdot \vec{v}_2 = [value] \cdot [value] + [value] \cdot [value] + [value] \cdot [value] = [value]\). \(\vec{w}_1 \cdot \vec{w}_2 = [value] \cdot [value] + [value] \cdot [value] + [value] \cdot [value] + [value] \cdot [value] = [value]\). \(\vec{w}_1 \cdot \vec{w}_3 = [value] \cdot [value] + [value] \cdot [value] + [value] \cdot [value] + [value] \cdot [value] = [value]\). b) Tehtävän pistetulot voidaan laskea suoraan, mutta niissä kannattaa käyttää pistetulon laskusääntöjä. Pistetulo on vaihdannainen, joten \(\vec{w}_2 \cdot \vec{w}_1 = \vec{w}_1 \cdot \vec{w}_2 = [value]\). Toista laskua varten käytetään laskusääntöä, jonka mukaan vektoreille \(\vec{a}\) ja \(\vec{b}\) ja reaaliluvulle \(c\) pätee \((c \vec{a}) \cdot \vec{b} = c (\vec{a} \cdot \vec{b})\). Siis \(([value] \vec{w}_1) \cdot ([value]\vec{w}_2) = [value] (\vec{w}_1 \cdot ([value]\vec{w}_2)) = [value] \cdot (([value] \vec{w}_2) \cdot \vec{w}_1) =[value] \cdot [value] (\vec{w}_2 \cdot \vec{w}_1) = [value] \cdot [value] = [value]\). Kolmanteen laskuun käytetään pistetulon vaihdannaisuutta ja osittelulakia: \((\vec{w}_3 + \vec{w}_2) \cdot (\vec{w}_1) = (\vec{w}_2 + \vec{w}_3) \cdot (\vec{w}_1) = \vec{w}_1 \cdot (\vec{w}_2 + \vec{w}_3) = \vec{w}_1 \cdot \vec{w}_2 + \vec{w}_1 \cdot \vec{w}_3 = [value]+[value] = [value]\).

Olkoon \(\mathbf{u} = [value]\) ja \(\mathbf{v} = [value]\).

Laske

\(\mathbf{u}\cdot \mathbf{v} = \)

\((2\mathbf{u})\cdot \mathbf{v} = \)

\(2(\mathbf{u}\cdot \mathbf{v}) = \)

\((c\mathbf{u})\cdot \mathbf{v} = c(\mathbf{u}\cdot \mathbf{v})\)
Feedback: Mallivastaus: \(\mathbf{u} \cdot \mathbf{v} = [value]\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right)+[value]\cdot[value] = [value]\) \((2\mathbf{u}) \cdot \mathbf{v} = (2\cdot[value])\cdot\left([value]\right)+(2\left([value]\right))\cdot\left([value]\right)+(2\cdot[value])\cdot[value] = [value]\) \(2(\mathbf{u} \cdot \mathbf{v}) = 2([value]\cdot\left([value]\right))+2([value]\cdot\left([value]\right))+2([value]\cdot[value]) = [value]\) \((c\mathbf{u})\cdot \mathbf{v} = c\mathbf{u}_1\mathbf{v}_1 + c\mathbf{u}_2\mathbf{v}_2 + \ldots + c\mathbf{u}_n\mathbf{v}_n = c(\mathbf{u}_1\mathbf{v}_1 + \mathbf{u}_2 \mathbf{v}_2 + \ldots + \mathbf{u}_n \mathbf{v}_n) = c(\mathbf{u}\cdot \mathbf{v})\)

Olkoon \(\mathbf{u} = \begin{bmatrix}[value] \\ [value] \\[value]\end{bmatrix}\) ja \(\mathbf{v} = \begin{bmatrix}[value] \\ [value] \\[value]\end{bmatrix}\).

Laske \(\mathbf{w}=\mathbf{u}\times \mathbf{v}\)= |

Laske \(\mathbf{w}\cdot \mathbf{u}\) =

Laske \(\mathbf{w}\cdot \mathbf{v}\) =

Laske \(\mathbf{u}\cdot \mathbf{v}\) =
Feedback: Mallivastaus: \(\mathbf{w}=\mathbf{u}\times \mathbf{v}=\begin{bmatrix} [value]\cdot\left([value]\right)-\left([value]\right)\cdot\left([value]\right) \\ [value]\cdot\left([value]\right)-\left([value]\right)\cdot\left([value]\right) \\ [value]\cdot\left([value]\right)-\left([value]\right)\cdot\left([value]\right) \end{bmatrix} = [value]\) \(\mathbf{w}\cdot \mathbf{u} = [value]\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right) = [value]\) \(\mathbf{w}\cdot \mathbf{v} = [value]\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right) = [value]\) \(\mathbf{u}\cdot \mathbf{v} = [value]\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right)+\left([value]\right)\cdot\left([value]\right) = [value]\)

Olkoot \(x\) ja \(y\) reaalilukuja. Olkoot edelleen \(\vec a=(x,[value],y)\), \(\vec b=(-1,[value],x)\) ja \(\vec c=(0,[value],1)\) vektoriavaruuden \(\mathbb{R}^3\) vektoreita.

Määrää reaaliluvut \(x\) ja \(y\) siten, että vektori \(\vec a\) on kohtisuorassa molempia vektoreita \(\vec b\) ja \(\vec c\) vastaan.

Let \(x\) and \(y\) be real numbers. Furthermore, let \(\vec a=(x,[value],y)\), \(\vec b=(-1,[value],x)\) and \(\vec c=(0,[value],1)\) be vectors in the vector space \(\mathbb{R}^3\).

Find the real numbers \(x\) and \(y\) such that the vector \(\vec a\) is perpendicular (orthogonal) to both of the vectors \(\vec b\) and \(\vec c\).

\(x=\), \(y=\) .
Feedback: Malliratkaisu: Model solution: \[\left\{ \begin{align*} \vec a\cdot\vec b &=(x,[value],y) \cdot (-1,[value],x)=[value] \\ \vec a\cdot\vec c&=(x,[value],y) \cdot (0,[value],1)=[value]+y=0 \end{align*}\right.\] Alemmasta yhtälöstä saadaan \(y=[value]\), joka sijoitettuna ylempään yhtälöön antaa \(x=[value]\). TAI Vektori \(\vec{b}\times\vec{c}\) on kohtisuorassa molempia vektoreita \(\vec b\) ja \(\vec c\) vastaan. Siis \[ \begin{align*} \vec{b}\times\vec{c}&=t\vec{a}\,\,\textrm{jollakin}\,\,t\in\mathbb{R}, t\ne0\\ \\ \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -1 & [value] & x \\ 0 & [value] & 1 \end{array}\right| &=t(x,[value],y)\\ \\ ([value],1,[value]) &=(tx,[value],ty). \end{align*} \] Saadaan yhtälöryhmä \[ \left\{ \begin{array}{rcl} [value] &=& tx \\ 1 &=& [value] \\ [value] &=& ty, \end{array} \right. \] jonka ratkaisu on \(t=[value]\), \(y=[value]\), \(x=[value]\). From the lowest equation \(y=[value]\), which gives when placed in the highest equation \(x=[value]\). OR The vector \(\vec{b}\times\vec{c}\) is perpendicular (orthogonal) to both of the vectors \(\vec b\) and \(\vec c\). Therefore \[ \begin{align*} \vec{b}\times\vec{c}&=t\vec{a}\,\,\textrm{for some}\,\,t\in\mathbb{R}, t\ne0\\ \\ \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -1 & [value] & x \\ 0 & [value] & 1 \end{array}\right| &=t(x,[value],y)\\ \\ ([value],1,[value]) &=(tx,[value],ty). \end{align*} \] The system of equations will be obtained \[ \left\{ \begin{array}{rcl} [value] &=& tx \\ 1 &=& [value] \\ [value] &=& ty \end{array} \right. \] the solution of which is \(t=[value]\), \(y=[value]\), \(x=[value]\).