bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

1. i) Find the angle between the vectors $\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ and $\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$.

1. Calculate the Dot Product $\mathbf{A}\cdot \mathbf{B}$$\mathbf{A}\cdot \mathbf{B}$A*B\mathbf{A} \cdot \mathbf{B}$\mathbf{A}\cdot \mathbf{B}$:
   $$\mathbf{A}\cdot \mathbf{B}=(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k})\cdot (\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k})$$$$\mathbf{A}\cdot \mathbf{B}=(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k})\cdot (\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k})$$A*B=(sqrt2i+2j+2k)*(i+sqrt2j+sqrt2k)\mathbf{A} \cdot \mathbf{B} = (\sqrt{2}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{i} + \sqrt{2}\mathbf{j} + \sqrt{2}\mathbf{k})$$\mathbf{A}\cdot \mathbf{B}=(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k})\cdot (\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k})$$
   $$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$$$$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$$=sqrt2*1+2*sqrt2+2*sqrt2= \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2}$$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$$
2. Calculate the Magnitudes of $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$ and $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$:
   • Magnitude of $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$, $|\mathbf{A}|$$|\mathbf{A}|$|A||\mathbf{A}|$|\mathbf{A}|$:$$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}$$$$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}$$|A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2}$$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}$$
   • Magnitude of $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$, $|\mathbf{B}|$$|\mathbf{B}|$|B||\mathbf{B}|$|\mathbf{B}|$:$$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$$$$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$$|B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2}$$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$$
3. Find the Angle $\theta$$\theta$theta\theta$\theta$:
   $$\cos (\theta )=\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$$$\cos (\theta )=\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$cos(theta)=(A*B)/(|A||B|)\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}$$\cos (\theta )=\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$
   $$\theta ={\cos }^{-1}\left(\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right)$$$$\theta ={\cos }^{-1}\left(\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right)$$theta=cos^(-1)((A*B)/(|A||B|))\theta = \cos^{-1}\left(\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}\right)$$\theta ={\cos }^{-1}\left(\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right)$$
   We will calculate $\theta$$\theta$theta\theta$\theta$ after obtaining the dot product and magnitudes.

1. Dot Product $\mathbf{A}\cdot \mathbf{B}$$\mathbf{A}\cdot \mathbf{B}$A*B\mathbf{A} \cdot \mathbf{B}$\mathbf{A}\cdot \mathbf{B}$:
   $$\mathbf{A}\cdot \mathbf{B}=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}=5\sqrt{2}$$$$\mathbf{A}\cdot \mathbf{B}=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}=5\sqrt{2}$$A*B=sqrt2*1+2*sqrt2+2*sqrt2=5sqrt2\mathbf{A} \cdot \mathbf{B} = \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2} = 5\sqrt{2}$$\mathbf{A}\cdot \mathbf{B}=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}=5\sqrt{2}$$
2. Magnitude of $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$, $|\mathbf{A}|$$|\mathbf{A}|$|A||\mathbf{A}|$|\mathbf{A}|$:
   $$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}=\sqrt{10}$$$$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}=\sqrt{10}$$|A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))=sqrt10|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2} = \sqrt{10}$$|\mathbf{A}|=\sqrt{(\sqrt{2}{)}^2+2^2+2^2}=\sqrt{10}$$
3. Magnitude of $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$, $|\mathbf{B}|$$|\mathbf{B}|$|B||\mathbf{B}|$|\mathbf{B}|$:
   $$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}=\sqrt{5}$$$$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}=\sqrt{5}$$|B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))=sqrt5|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{5}$$|\mathbf{B}|=\sqrt{1^2+(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}=\sqrt{5}$$

1. Non-emptiness: $W$$W$WW$W$ must contain the zero vector of ${\mathbb{R}}^3$${\mathbb{R}}^3$R^(3)\mathbb{R}^3${\mathbb{R}}^3$.
2. Closed under addition: For any two vectors $\overrightarrow{u}$$\overrightarrow{u}$vec(u)\vec{u}$\overrightarrow{u}$ and $\overrightarrow{v}$$\overrightarrow{v}$vec(v)\vec{v}$\overrightarrow{v}$ in $W$$W$WW$W$, the sum $\overrightarrow{u}+\overrightarrow{v}$$\overrightarrow{u}+\overrightarrow{v}$vec(u)+ vec(v)\vec{u} + \vec{v}$\overrightarrow{u}+\overrightarrow{v}$ must also be in $W$$W$WW$W$.
3. Closed under scalar multiplication: For any vector $\overrightarrow{u}$$\overrightarrow{u}$vec(u)\vec{u}$\overrightarrow{u}$ in $W$$W$WW$W$ and any scalar $c$$c$cc$c$, the product $c\overrightarrow{u}$$c\overrightarrow{u}$c vec(u)c\vec{u}$c\overrightarrow{u}$ must also be in $W$$W$WW$W$.

1. Non-emptiness:
   The zero vector in ${\mathbb{R}}^3$${\mathbb{R}}^3$R^(3)\mathbb{R}^3${\mathbb{R}}^3$ is $(0,0,0)$$(0,0,0)$(0,0,0)(0, 0, 0)$(0,0,0)$. For this vector, $x=0$$x=0$x=0x = 0$x=0$, $y=0$$y=0$y=0y = 0$y=0$, and $z=0$$z=0$z=0z = 0$z=0$, so $x+y-z=0+0-0=0$$x+y-z=0+0-0=0$x+y-z=0+0-0=0x + y – z = 0 + 0 – 0 = 0$x+y-z=0+0-0=0$. Therefore, the zero vector is in $W$$W$WW$W$, satisfying the non-emptiness condition.
2. Closed under addition:
   Let $\overrightarrow{u}=(x_1,y_1,z_1)$$\overrightarrow{u}=(x_1,y_1,z_1)$vec(u)=(x_(1),y_(1),z_(1))\vec{u} = (x_1, y_1, z_1)$\overrightarrow{u}=(x_1,y_1,z_1)$ and $\overrightarrow{v}=(x_2,y_2,z_2)$$\overrightarrow{v}=(x_2,y_2,z_2)$vec(v)=(x_(2),y_(2),z_(2))\vec{v} = (x_2, y_2, z_2)$\overrightarrow{v}=(x_2,y_2,z_2)$ be any two vectors in $W$$W$WW$W$. This means $x_1+y_1-z_1=0$$x_1+y_1-z_1=0$x_(1)+y_(1)-z_(1)=0x_1 + y_1 – z_1 = 0$x_1+y_1-z_1=0$ and $x_2+y_2-z_2=0$$x_2+y_2-z_2=0$x_(2)+y_(2)-z_(2)=0x_2 + y_2 – z_2 = 0$x_2+y_2-z_2=0$. We need to check if their sum $\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2,z_1+z_2)$$\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2,z_1+z_2)$vec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2),z_(1)+z_(2))\vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2,z_1+z_2)$ also satisfies the condition of $W$$W$WW$W$.
   For $\overrightarrow{u}+\overrightarrow{v}$$\overrightarrow{u}+\overrightarrow{v}$vec(u)+ vec(v)\vec{u} + \vec{v}$\overrightarrow{u}+\overrightarrow{v}$, we have:
   $$(x_1+x_2)+(y_1+y_2)-(z_1+z_2)=(x_1+y_1-z_1)+(x_2+y_2-z_2)=0+0=0$$$$(x_1+x_2)+(y_1+y_2)-(z_1+z_2)=(x_1+y_1-z_1)+(x_2+y_2-z_2)=0+0=0$$(x_(1)+x_(2))+(y_(1)+y_(2))-(z_(1)+z_(2))=(x_(1)+y_(1)-z_(1))+(x_(2)+y_(2)-z_(2))=0+0=0(x_1 + x_2) + (y_1 + y_2) – (z_1 + z_2) = (x_1 + y_1 – z_1) + (x_2 + y_2 – z_2) = 0 + 0 = 0$$(x_1+x_2)+(y_1+y_2)-(z_1+z_2)=(x_1+y_1-z_1)+(x_2+y_2-z_2)=0+0=0$$
   Therefore, $\overrightarrow{u}+\overrightarrow{v}$$\overrightarrow{u}+\overrightarrow{v}$vec(u)+ vec(v)\vec{u} + \vec{v}$\overrightarrow{u}+\overrightarrow{v}$ is in $W$$W$WW$W$, satisfying the closure under addition.
3. Closed under scalar multiplication:
   Let $\overrightarrow{u}=(x,y,z)$$\overrightarrow{u}=(x,y,z)$vec(u)=(x,y,z)\vec{u} = (x, y, z)$\overrightarrow{u}=(x,y,z)$ be any vector in $W$$W$WW$W$ and $c$$c$cc$c$ be any scalar. Since $\overrightarrow{u}$$\overrightarrow{u}$vec(u)\vec{u}$\overrightarrow{u}$ is in $W$$W$WW$W$, $x+y-z=0$$x+y-z=0$x+y-z=0x + y – z = 0$x+y-z=0$. We need to check if $c\overrightarrow{u}=(cx,cy,cz)$$c\overrightarrow{u}=(cx,cy,cz)$c vec(u)=(cx,cy,cz)c\vec{u} = (cx, cy, cz)$c\overrightarrow{u}=(cx,cy,cz)$ also satisfies the condition of $W$$W$WW$W$.
   For $c\overrightarrow{u}$$c\overrightarrow{u}$c vec(u)c\vec{u}$c\overrightarrow{u}$, we have:
   $$cx+cy-cz=c(x+y-z)=c\cdot 0=0$$$$cx+cy-cz=c(x+y-z)=c\cdot 0=0$$cx+cy-cz=c(x+y-z)=c*0=0cx + cy – cz = c(x + y – z) = c \cdot 0 = 0$$cx+cy-cz=c(x+y-z)=c\cdot 0=0$$
   Therefore, $c\overrightarrow{u}$$c\overrightarrow{u}$c vec(u)c\vec{u}$c\overrightarrow{u}$ is in $W$$W$WW$W$, satisfying the closure under scalar multiplication.

1. $c_1+c_3=0$$c_1+c_3=0$c_(1)+c_(3)=0c_1 + c_3 = 0$c_1+c_3=0$
2. $c_1+c_2=0$$c_1+c_2=0$c_(1)+c_(2)=0c_1 + c_2 = 0$c_1+c_2=0$
3. $c_2=0$$c_2=0$c_(2)=0c_2 = 0$c_2=0$
4. $c_3=0$$c_3=0$c_(3)=0c_3 = 0$c_3=0$

1. Additivity: $T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$
2. Homogeneity: $T(c\overrightarrow{u})=cT(\overrightarrow{u})$$T(c\overrightarrow{u})=cT(\overrightarrow{u})$T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})$T(c\overrightarrow{u})=cT(\overrightarrow{u})$

1. Additivity:
   Let $\overrightarrow{u}=(x_1,y_1)$$\overrightarrow{u}=(x_1,y_1)$vec(u)=(x_(1),y_(1))\vec{u} = (x_1, y_1)$\overrightarrow{u}=(x_1,y_1)$ and $\overrightarrow{v}=(x_2,y_2)$$\overrightarrow{v}=(x_2,y_2)$vec(v)=(x_(2),y_(2))\vec{v} = (x_2, y_2)$\overrightarrow{v}=(x_2,y_2)$ be any vectors in ${\mathbb{R}}^2$${\mathbb{R}}^2$R^(2)\mathbb{R}^2${\mathbb{R}}^2$. We need to check if $T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$.
   First, calculate $\overrightarrow{u}+\overrightarrow{v}$$\overrightarrow{u}+\overrightarrow{v}$vec(u)+ vec(v)\vec{u} + \vec{v}$\overrightarrow{u}+\overrightarrow{v}$:
   $$\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2)$$$$\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2)$$vec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2))\vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2)$$\overrightarrow{u}+\overrightarrow{v}=(x_1+x_2,y_1+y_2)$$
   Then, apply $T$$T$TT$T$ to $\overrightarrow{u}+\overrightarrow{v}$$\overrightarrow{u}+\overrightarrow{v}$vec(u)+ vec(v)\vec{u} + \vec{v}$\overrightarrow{u}+\overrightarrow{v}$:
   $$T(\overrightarrow{u}+\overrightarrow{v})=T(x_1+x_2,y_1+y_2)=(-(y_1+y_2),x_1+x_2)$$$$T(\overrightarrow{u}+\overrightarrow{v})=T(x_1+x_2,y_1+y_2)=(-(y_1+y_2),x_1+x_2)$$T( vec(u)+ vec(v))=T(x_(1)+x_(2),y_(1)+y_(2))=(-(y_(1)+y_(2)),x_(1)+x_(2))T(\vec{u} + \vec{v}) = T(x_1 + x_2, y_1 + y_2) = (-(y_1 + y_2), x_1 + x_2)$$T(\overrightarrow{u}+\overrightarrow{v})=T(x_1+x_2,y_1+y_2)=(-(y_1+y_2),x_1+x_2)$$
   Now, calculate $T(\overrightarrow{u})$$T(\overrightarrow{u})$T( vec(u))T(\vec{u})$T(\overrightarrow{u})$ and $T(\overrightarrow{v})$$T(\overrightarrow{v})$T( vec(v))T(\vec{v})$T(\overrightarrow{v})$:
   $$T(\overrightarrow{u})=T(x_1,y_1)=(-y_1,x_1)$$$$T(\overrightarrow{u})=T(x_1,y_1)=(-y_1,x_1)$$T( vec(u))=T(x_(1),y_(1))=(-y_(1),x_(1))T(\vec{u}) = T(x_1, y_1) = (-y_1, x_1)$$T(\overrightarrow{u})=T(x_1,y_1)=(-y_1,x_1)$$
   $$T(\overrightarrow{v})=T(x_2,y_2)=(-y_2,x_2)$$$$T(\overrightarrow{v})=T(x_2,y_2)=(-y_2,x_2)$$T( vec(v))=T(x_(2),y_(2))=(-y_(2),x_(2))T(\vec{v}) = T(x_2, y_2) = (-y_2, x_2)$$T(\overrightarrow{v})=T(x_2,y_2)=(-y_2,x_2)$$
   Finally, add $T(\overrightarrow{u})$$T(\overrightarrow{u})$T( vec(u))T(\vec{u})$T(\overrightarrow{u})$ and $T(\overrightarrow{v})$$T(\overrightarrow{v})$T( vec(v))T(\vec{v})$T(\overrightarrow{v})$:
   $$T(\overrightarrow{u})+T(\overrightarrow{v})=(-y_1,x_1)+(-y_2,x_2)=(-(y_1+y_2),x_1+x_2)$$$$T(\overrightarrow{u})+T(\overrightarrow{v})=(-y_1,x_1)+(-y_2,x_2)=(-(y_1+y_2),x_1+x_2)$$T( vec(u))+T( vec(v))=(-y_(1),x_(1))+(-y_(2),x_(2))=(-(y_(1)+y_(2)),x_(1)+x_(2))T(\vec{u}) + T(\vec{v}) = (-y_1, x_1) + (-y_2, x_2) = (-(y_1 + y_2), x_1 + x_2)$$T(\overrightarrow{u})+T(\overrightarrow{v})=(-y_1,x_1)+(-y_2,x_2)=(-(y_1+y_2),x_1+x_2)$$
   Since $T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$, the additivity property is satisfied.
2. Homogeneity:
   Let $\overrightarrow{u}=(x,y)$$\overrightarrow{u}=(x,y)$vec(u)=(x,y)\vec{u} = (x, y)$\overrightarrow{u}=(x,y)$ be any vector in ${\mathbb{R}}^2$${\mathbb{R}}^2$R^(2)\mathbb{R}^2${\mathbb{R}}^2$ and $c$$c$cc$c$ be any scalar. We need to check if $T(c\overrightarrow{u})=cT(\overrightarrow{u})$$T(c\overrightarrow{u})=cT(\overrightarrow{u})$T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})$T(c\overrightarrow{u})=cT(\overrightarrow{u})$.
   First, calculate $c\overrightarrow{u}$$c\overrightarrow{u}$c vec(u)c\vec{u}$c\overrightarrow{u}$:
   $$c\overrightarrow{u}=c(x,y)=(cx,cy)$$$$c\overrightarrow{u}=c(x,y)=(cx,cy)$$c vec(u)=c(x,y)=(cx,cy)c\vec{u} = c(x, y) = (cx, cy)$$c\overrightarrow{u}=c(x,y)=(cx,cy)$$
   Then, apply $T$$T$TT$T$ to $c\overrightarrow{u}$$c\overrightarrow{u}$c vec(u)c\vec{u}$c\overrightarrow{u}$:
   $$T(c\overrightarrow{u})=T(cx,cy)=(-cy,cx)$$$$T(c\overrightarrow{u})=T(cx,cy)=(-cy,cx)$$T(c vec(u))=T(cx,cy)=(-cy,cx)T(c\vec{u}) = T(cx, cy) = (-cy, cx)$$T(c\overrightarrow{u})=T(cx,cy)=(-cy,cx)$$
   Now, calculate $T(\overrightarrow{u})$$T(\overrightarrow{u})$T( vec(u))T(\vec{u})$T(\overrightarrow{u})$ and then multiply by $c$$c$cc$c$:
   $$T(\overrightarrow{u})=T(x,y)=(-y,x)$$$$T(\overrightarrow{u})=T(x,y)=(-y,x)$$T( vec(u))=T(x,y)=(-y,x)T(\vec{u}) = T(x, y) = (-y, x)$$T(\overrightarrow{u})=T(x,y)=(-y,x)$$
   $$cT(\overrightarrow{u})=c(-y,x)=(-cy,cx)$$$$cT(\overrightarrow{u})=c(-y,x)=(-cy,cx)$$cT( vec(u))=c(-y,x)=(-cy,cx)cT(\vec{u}) = c(-y, x) = (-cy, cx)$$cT(\overrightarrow{u})=c(-y,x)=(-cy,cx)$$
   Since $T(c\overrightarrow{u})=cT(\overrightarrow{u})$$T(c\overrightarrow{u})=cT(\overrightarrow{u})$T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})$T(c\overrightarrow{u})=cT(\overrightarrow{u})$, the homogeneity property is satisfied.