IDZ Ryabushko 2.1 Option 10

No. 1. Let the vectors $a = \alpha\cdot m + \beta\cdot n$, $b = \gamma\cdot m + \delta\cdot n$, $|m| = k$, $|n| = \ell$, $(m;n) = \varphi$. You need to find: a) $(\lambda\cdot a + \mu\cdot b)\cdot(\nu\cdot a + \tau\cdot b)$; b) projection of the vector $\nu\cdot a + \tau\cdot b$ onto the vector $b$; c) $\cos(a + \tau\cdot b)$.

Known: $\alpha = 5$, $\beta = -3$, $\gamma = 4$, $\delta = 2$, $k = 4$, $\ell = 1$, $\varphi = \frac {2\pi}{3}$, $\lambda = 2$, $\mu = -\frac{1}{2}$, $\nu = 3$, $\tau = 0$.

а) Note that the quotation marks $\lambda\cdot a + \mu\cdot b$: $$ \begin{aligned} \lambda\cdot a + \mu\cdot b &= \lambda\cdot(\alpha \cdot m + \beta\cdot n) + \mu\cdot(\gamma\cdot m + \delta\cdot n) \ &= (\lambda\cdot\alpha + \mu\cdot\gamma)\cdot m + (\ lambda\cdot\beta + \mu\cdot\delta)\cdot n. \end{aligned} $$ Language, Language Language Language $\nu\cdot a + \tau\cdot b$: $$ \begin{aligned} \nu\cdot a + \tau\cdot b & = \nu\ cdot(\alpha\cdot m + \beta\cdot n) + \tau\cdot(\gamma\cdot m + \delta\cdot n) \ &= (\nu\cdot\alpha + \tau\cdot\gamma); \cdot m + (\nu\cdot\beta + \tau\cdot\delta)\cdot n. \end{aligned} $$ Similarly aligned to aligned $(\lambda\cdot a + \mu\cdot b)\cdot(\nu\cdot a + \tau\ cdot b)$:$$\begin{aligned} &(\lambda\cdot a + \mu\cdot b)\cdot(\nu\cdot a + \tau\cdot b) \ &= ((\lambda\cdot\alpha + \mu\cdot\gamma)\cdot m + (\lambda\cdot\beta + \mu\cdot\delta)\cdot n)\cdot((\nu\cdot\alpha + \tau\cdot\gamma)\cdot m + (\nu\cdot\beta + \tau\cdot\delta)\cdot n) \ &= (\lambda\cdot\alpha + \mu\cdot\gamma)\cdot(\nu\cdot\alpha + \tau\cdot\gamma)\cdot| m|^2 + (\lambda\cdot\beta + \mu\cdot\delta)\cdot(\nu\cdot\beta + \tau\cdot\delta)\cdot|n|^2 \ &\quad + ( (\lambda\cdot\alpha + \mu\cdot\gamma)\cdot(\nu\cdot\beta + \tau\cdot\delta) + (\lambda\cdot\beta + \mu\cdot\delta)\cdot (\nu\cdot\alpha + \tau\cdot\gamma))\cdot(m;n) \ &= (10\cdot3 - \tfrac{1}{2}\cdot4)\cdot16 + (-15\cdot0 + 1\cdot0)\cdot1 + ((10\cdot0 - \tfrac{1}{2}\cdot3)\cdot(-3) + (5\cdot4 + (-3)\cdot2))\cdot\cos\ frac{2\pi}{3}\&= 155 - 23\sqrt{3}. \end{aligned} $$

б) Проекция вектора $\nu\cdot a + \tau\cdot b$ на вектор $b$ равна: $$ \begin{aligned} &\operatorname{proj}_b(\nu\cdot a + \tau\cdot b) \ &= \frac{(\nu\cdot a + \tau\cdot b)\cdot b}{|b|^2}\cdot b \ &= \frac{(\nu\cdot(\alpha\cdot m + \beta\cdot n) + \tau\cdot(\gamma\cdot m + \delta\cdot n))\cdot(\gamma\cdot m + \delta\cdot n)}{\gamma^2+\delta^2}\cdot (\gamma\cdot m + \delta\cdot n) \ &= \frac{(3\cdot 5 + 0)\cdot 4 + (0 - \tfrac{1}{2}\cdot 2)\cdot (-3)}{4^2+2^2}\cdot (4\cdot m + 2\cdot n) \ &= \frac{58}{20}\cdot (4\cdot m + 2\cdot n) \ &= \frac{29}{10}\cdot (2\cdot m + n). \end{aligned} $$

в) Infinite matrix $a + \tau\cdot b$: $$ \begin{aligned} a + \tau\cdot b &= \alpha\cdot m + \beta\cdot n + \tau\cdot(\gamma \cdot m + \delta\cdot n) \ &= (5 + 0)\cdot m + (-3 + 0)\cdot n + 0\cdot m + 2\cdot n \ &= 5\cdot m - \cdot n. \end{aligned} $$ Let's aligned the range of $a$ and $a + \tau\cdot b$ begin: $$ \begin{aligned} \cos(a + \tau\cdot b) &= \fra c{ a\cdot(a + \tau\cdot b)}{|a|\cdot|a + \tau\cdot b|} \ &= \frac{(\alpha\cdot m + \beta\cdot n)\cdot (5\cdot m - n)}{\sqrt{\alpha^2+\beta^2}\cdot\sqrt{5^2+1^2}} \ &= \frac{(5\cdot 3 - \cdot(-3))\cdot4}{\sqrt{5^2+3^2}\cdot\sqrt{5^2+1^2}} \ &= \frac{64}{65}. \end{aligned} $$

No. 2. Using the coordinates of points $A$, $B$ and $C$ for the indicated vectors, you need to find: a) the modulus of the vector $a$; b) scalar product of vectors $a$ and $b$; c) projection of vector $c$ onto vector $d$; d) coordinates of the point $M$ dividing the segment $\ell$ in relation $\alpha$.

Дано: $A(0; 2; 5)$, $B(2;-3;4)$, $C(3;2;-5)$, $a = \overrightarrow{AB}$, $b = \overrightarrow{BC}$, $c = \overrightarrow{AC}$, $d = (1, 1, 1)$, $\ell = AB$, $\alpha = \frac{1}{3}$.

a) Vector $a$ has coordinates $(2-0,-3-2,4-5) = (2,-5,-1)$, so its modulus is equal to $\sqrt{2^2+5^2 +1^2} = \sqrt{30}$.

b) The scalar product of vectors $a$ and $b$ is equal to: $$ \begin{aligned} a\cdot b &= (2,-5,-1)\cdot(1,4,-9) \ &= 2 \cdot 1 + (-5)\cdot 4 + (-1)\cdot(-9) \ &= -33. \end{aligned} $$

c) Projection of vector $c$ onto vector $d$ equal to: $$ \begin{aligned} &\operatorname{proj}_d c \ &= \frac{c\cdot d}{|d|^2}\cdot d \ &= \frac{(3\cdot 1 + 2\cdot 1 + (-5)\cdot 1)}{1^2+1^2+1^2}\cdot (1,1,1) \ & = 0\cdot (1,1,1) \ &= (0,0,0). \end{aligned} $$

d) The coordinates of the point $M$ can be found using the parametric equation of the line passing through the points $A$ and $B$: $$ M = A

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5-4) = (2,-5,-1)$, so its modulus is equal to: $$ |a| = \sqrt{2^2 + (-5)^2 + (-1)^2} = \sqrt{30}. $$b) The scalar product of vectors $a$ and $b$ is equal to: $$ \begin{aligned} a\cdot b &= (2,-5,-1)\cdot(3,-5,-9) \ &= 2\cdot3 + (-5)\cdot(-5) + (-1)\cdot(-9) \ &= 35. \end{aligned} $$c) Projection of vector $c$ onto vector $d $ is equal to: $$ \begin{aligned} &\operatorname{proj}_d c \ &= \frac{c\cdot d}{|d|^2}\cdot d \ &= \frac{(0-3+ 2)\cdot 1 + (2+(-3)-5)\cdot 1 + (5+4+(-5))\cdot 1}{1^2+1^2+1^2}\cdot( 1,1,1) \ &= \frac{2}{3}\cdot(1,1,1). \end{aligned} $$d) The coordinates of the point $M$ can be found using the parametric equation of the line passing through the points $A$ and $B$: $$ \begin{aligned} x(t) &= 0 + 2t, \ y(t) &= 2 + (-3-2)t, \ z(t) &= 5 + (4-5)t. \end{aligned} $$ To find the coordinates of the point $M$ dividing the segment $AB$ in the relation $\alpha = \frac{1}{3}$, you can substitute $t = \alpha$ into the parametric equation: $$ \begin{aligned} x(\alpha) &= 2\cdot\frac{1}{3} = \frac{2}{3},\ y(\alpha) &= 2 + (-3-2)\ cdot\frac{1}{3} = -\frac{1}{3}, \z(\alpha) &= 5 + (4-5)\cdot\frac{1}{3} = \frac{13 }{3}. \end{aligned} $$ So the coordinates of point $M$ are equal to: $$ M\left(\frac{2}{3}, -\frac{1}{3}, \frac{13}{3}\ right). $$

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