Vectors and systems of linear equations – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(\mathbf a\cdot\mathbf b=3+0+8=11\), \(|\mathbf a|=\sqrt{1+4+4}=3\) and \(|\mathbf b|=\sqrt{9+0+16}=5\) (1). So \(\cos\theta=\frac{11}{15}\) and \(\theta\approx42.8^\circ\) (1).
   2. \(\mathbf a\times\mathbf b=(2\cdot4-2\cdot0,\ 2\cdot3-1\cdot4,\ 1\cdot0-2\cdot3)=(8,2,-6)\) (1). The triangle has half the area of the parallelogram: \(\frac12|\mathbf a\times\mathbf b|=\frac12\sqrt{64+4+36}=\frac12\sqrt{104}\approx5.10\) square units (1).

   In (a), the formula used is \(\cos\theta=\tfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\); a positive dot product means the angle is acute. In (b), check that \(\mathbf a\times\mathbf b\) is perpendicular to both vectors: \((8,2,-6)\cdot(1,2,2)=8+4-12=0\) and \((8,2,-6)\cdot(3,0,4)=24-24=0\).

2. Answer to Question 2 (4 marks)

   Two vectors in the plane are \(\overrightarrow{AB}=(-1,2,0)\) and \(\overrightarrow{AC}=(-1,0,3)\). Their cross product gives a normal: \(\overrightarrow{AB}\times\overrightarrow{AC}=(6,3,2)\) (2). The plane is \(6x+3y+2z=d\), and substituting \(A\) gives \(d=6\), so the plane is \(6x+3y+2z=6\) (1). Check: \(B\) gives \(6\) and \(C\) gives \(6\). The distance from the origin is \(\dfrac{|0-6|}{\sqrt{36+9+4}}=\dfrac6{\sqrt{49}}=\dfrac67\) (1).

   The intercept form \(\tfrac x1+\tfrac y2+\tfrac z3=1\) gives the same plane after multiplying by 6. The normal \((6,3,2)\) has length exactly 7.

3. Answer to Question 3 (4 marks)

   Use row 1 to clear the first column: \(R_2-2R_1\) gives \((0,1,-1\mid-3)\) and \(R_3-R_1\) gives \((0,-2,1\mid4)\) (1). Then \(R_3+2R_2\) gives \((0,0,-1\mid-2)\), so the matrix is in echelon form (1). Back substitution: \(-z=-2\) gives \(z=2\); \(y-z=-3\) gives \(y=-1\); \(x+y+z=2\) gives \(x=1\) (2). Check in the third equation: \(1+1+4=6\).

   The system has a unique solution because elimination produced a pivot in every column. Check the second equation as well: \(2-3+2=1\).

4. Answer to Question 4 (4 marks)

   1. The coefficient matrix \(\begin{pmatrix}1&2\\2&k\end{pmatrix}\) has determinant \(k-4\). There is exactly one solution when it is non-zero, that is \(k\ne4\) (1).
   2. When \(k=4\), the left-hand side of the second equation is \(2x+4y=2(x+2y)\), so it reads \(2(x+2y)=c\) (1). If \(c=6\), it is the first equation multiplied by 2: the same line, so there are infinitely many solutions (1). If \(c\ne6\), the lines are parallel and distinct, so there is no solution (1).

5. Answer to Question 5 (4 marks)

   1. A point of the beam is \((2+t,\ 1+t,\ t)\). Substituting into the plane: \(2+t+2(1+t)+t=4+4t=10\), so \(t=1.5\) (1). The point of impact is \((3.5,\ 2.5,\ 1.5)\) (1). Check: \(3.5+5+1.5=10\).
   2. The angle between a line and a plane is the complement of the angle between the line and the normal \(\mathbf n=(1,2,1)\). So \(\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}=\dfrac4{\sqrt3\sqrt6}\) (1), which gives \(\varphi\approx70.5^\circ\) (1).

Total : 20 marks