*Q1* (a) Subtracting twice the second equation from the first gives $y=0$. Substituting for $y$ in the second equation gives $x=0$. Finally we check that $(x,y)=(0,0)$ indeed satisfies both equations. (b) Subtracting twice the second equation from the first gives $y=1$. Substituting for $y$ in the second equation gives $x=1$. Finally we check that $(x,y)=(1,1)$ indeed satisfies both equations. The following plot (done with python) shows that the lines $2x-5y=-3$ (yellow line) and $x-3y=-2$ indeed intersect at $(1,1)$. ![[MA150 Assessment 1 graph.png|400]] (c) Subtracting twice the second equation from the first gives $y=\lambda-2\mu$. Substituting for $y$ in the second equation gives $x=3\lambda-5\mu$. Finally we check that $(x,y)=(\lambda-2\mu,3\lambda-5\mu)$ indeed satisfies both equations. *Q2* (a) Subtracting the second equation from the first gives $3y-2z=0$. So for any $z\in\mathbb{R}$ choose $y=\frac{2}{3}z$. Substituting for $y$ in the first equation gives $x=-\frac{5}{3}z$. Finally check that $(x,y,z)=\left( -\frac{5}{3}t, \frac{2}{3}t,t \right)$ satisfies both equations for all $t\in\mathbb{R}$. Hence the set of solutions corresponds to the line in $\mathbb{R}^{3}$ through the origin parallel to the vector $\left( -\frac{5}{3}, \frac{2}{3},1 \right)^{T}$. (b) Subtracting the second equation from the first gives $3y-2z=-1$. So for any $z\in \mathbb{R}$ choose $y = \frac{2}{3}z-\frac{1}{3}$. Substituting for $y$ in the first equation gives $x+\frac{2}{3}z-\frac{1}{3}+z=2\implies x=\frac{7}{3}-\frac{5}{3}z$. Finally check that indeed $\forall t\in\mathbb{R},\left( \frac{7}{3}-\frac{5}{3}t, \frac{2}{3}t - \frac{1}{3} ,t) \right)$ satisfies both equations. Hence the set of solutions corresponds to line that passes through $\left( \frac{7}{3}, -\frac{1}{3}, 0 \right)$ parallel to the vector $\left( -\frac{5}{3}, \frac{2}{3}, 1 \right)^{T}$. The plot below confirms the lines are parallel: ![[MA150 Assessment 1 graph2.png|400]] *Q3* (a) The equation of $\Pi$ has the form $ax+by+cz=d$ for some $a,b,c,d\in \mathbb{R}$ such that $(a,b,c)\neq \underline{0}$. Given that the points $P,Q,\underline{0}$ lie on the plane, we obtain the following system of linear equations $\begin{align} a+b+c = d \tag{1} \\ a - b + 3c = d \tag{2} \\ 0 = d \tag{3} \end{align}$Adding $(1)$ and $(2)$ gives $2a+4c=0$ (substituting $0$ for $d$ using $(3)$). So for any $c\in\mathbb{R} \setminus \{ 0 \}$, $a=-2c$ and $b=c$ is a solution. Hence the equation of the plane is given by: $ -2x +y +z=0.$ (b) $(1)-(2)$ gives $2b-2c=0 \implies b=c$. Substituting for $b$ in $(1)$ gives $a+2c=d\implies a=d-2c$. So any point in $L_{PQ}$ satisfies $(d-2c)x+cy+cz=d$simultaneously for all $c,d\in \mathbb{R}$ such that $(c,d)\neq(0,0)$ Choosing $(c,d)=(1,0)$ gives: $-2x+y+z=0$. Choosing $(c,d)=(0,1)$ gives: $x=1$. This pair of equations define $L_{PQ}\subset \mathbb{R}^{3}$ since they are independent. (c) $L_{PQ} = \{ (1,1,1)^{T} + \mu (0, 2, -2)^{T }\mid \mu\in \mathbb{R} \}$. *Q4* (a) Length of $\underline{w}=||\underline{w}||=\sqrt{ 2^{2} + 1^{2} }=\sqrt{ 5 }$. So the unit vector in the direction $\underline{w}$ is given by $\underline{\hat{w}}=\left( \frac{2\sqrt{ 5 }}{5} , \frac{\sqrt{ 5 }}{5} \right)^{T}$ Hence the component of $\underline{v}$ in the direction of $\underline{w}$ is given by $\lambda=\underline{v}\cdot \underline{\hat{w}}=\frac{2\sqrt{ 5 }}{5}+\frac{\sqrt{ 5 }}{5}=\frac{3\sqrt{ 5 }}{5}$. The orthogonal projection of $\underline{v}$ in the direction of $\underline{w}$ is given by $\lambda \underline{\hat{w}}=\left( \frac{6}{5}, \frac{3}{5} \right)^{T}$. ![[MA150 Assessment 1 vector.png|400]] (b) Note that $||\underline{w}||=\sqrt{ 1+4+4 }=3$ so $\underline{\hat{w}}=\left( \frac{1}{3}, \frac{-2}{3}, \frac{-2}{3} \right)^{T}$. Now $\underline{v}\cdot \underline{\hat{w}}=\frac{1}{3}-\frac{2}{3}-\frac{2}{3}=-1$. Hence the orthogonal projection of $\underline{v}$ in the direction of $\underline{w}$ is given by $\left( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3} \right)^{T}$.