Find the values of k for which the following system of equations has no solutions and the value of k for the system to have an infinite number of solutions.

\[x - 3y + z = 3\]

\[x + 5y - 2z = 1\]

\[16y - 6z = k\]

Given that the system of equations can be solved, find the solutions in the form of a vector equation of a line, r = a + λb , where the components of b  are integers.

The plane \( \div \) is parallel to both the line in part (b) and the line \(\frac{{x - 4}}{3} = \frac{{y - 6}}{{ - 2}} = \frac{{z - 2}}{0}\).

Given that \( \div \) contains the point (1, 2, 0) , show that the Cartesian equation of ÷ is 16x + 24y − 11z = 64 .

The z-axis meets the plane \( \div \) at the point P. Find the coordinates of P.

Find the angle between the line \(\frac{{x - 2}}{3} = \frac{{y + 5}}{4} = \frac{z}{2}\) and the plane \( \div \) .

in augmented matrix form \(\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&1&3 \\
  1&5&{ - 2}&1 \\
  0&{16}&{ - 6}&k
\end{array}} \right|\)

attempt to find a line of zeros     (M1)

\({r_2} - {r_1}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&1&3 \\
  0&8&{ - 3}&{ - 2} \\
  0&{16}&{ - 6}&k
\end{array}} \right|\)     (A1)

\({r_3} - 2{r_2}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&1&3 \\
  0&8&{ - 3}&{ - 2} \\
  0&{0}&{0}&{k + 4}
\end{array}} \right|\)     (A1)

there is an infinite number of solutions when \(k = - 4\)     R1

there is no solution when

\(k \ne - 4,{\text{ }}(k \in \mathbb{R})\)     R1 

Note: Approaches other than using the augmented matrix are acceptable. 

[5 marks]

using \(\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&1&3 \\
  0&8&{ - 3}&{ - 2} \\
  0&{0}&{ 0}&{k + 4}
\end{array}} \right|\) and letting \(\boldsymbol{z} = \lambda \)     (M1)

\(8y - 3\lambda  =  - 2\)

\( \Rightarrow y = \frac{{3\lambda  - 2}}{8}\)     (A1)

\(x - 3y + z = 3\)

\( \Rightarrow x - \left( {\frac{{9\lambda  - 6}}{8}} \right) + \lambda  = 3\)     (M1)

\( \Rightarrow 8x - 9\lambda  + 6 + 8\lambda  = 24\)

\( \Rightarrow x = \frac{{18 + \lambda }}{8}\)     (A1)

\( \Rightarrow \left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\frac{{18}}{8}} \\
  { - \frac{2}{8}} \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  {\frac{1}{8}} \\
  {\frac{3}{8}} \\
  1
\end{array}} \right)\)     (M1)(A1)

\(r = \left( {\begin{array}{*{20}{c}}
  {\frac{9}{4}} \\
  { - \frac{1}{4}} \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  8
\end{array}} \right)\)     A1 

Note: Accept equivalent answers. 

[7 marks]

recognition that \(\left( {\begin{array}{*{20}{c}}
  3 \\
  { - 2} \\
  0
\end{array}} \right)\) is parallel to the plane     (A1)

direction normal of the plane is given by \(\left| {\begin{array}{*{20}{c}}
  \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
  1&3&8 \\
  3&{ - 2}&0
\end{array}} \right|\)     (M1)

= 16i + 24j – 11k     A1

Cartesian equation of the plane is given by 16x + 24y –11z = d and a point which fits this equation is (1, 2, 0)     (M1)

\( \Rightarrow 16 + 48 = d\)

d = 64     A1

hence Cartesian equation of plane is 16x + 24y –11z = 64     AG 

Note: Accept alternative methods using dot product.

[5 marks]

the plane crosses the z-axis when x = y = 0     (M1)

coordinates of P are \(\left( {0,\,0,\, - \frac{{64}}{{11}}} \right)\)     A1 

Note: Award A1 for stating \(z = - \frac{{64}}{{11}}\).

Note: Accept. \(\left( {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  { - \frac{{64}}{{11}}}
\end{array}} \right)\)

[2 marks]

recognition that the angle between the line and the direction normal is given by:

\(\left( {\begin{array}{*{20}{c}}
  3 \\
  4 \\
  2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {16} \\
  {24} \\
  {-11}
\end{array}} \right) = \sqrt {29} \sqrt {953} \cos \theta \) where \(\theta \) is the angle between the line and the normal vector     M1A1

\( \Rightarrow 122 = \sqrt {29} \sqrt {953} \cos \theta \)     (A1)

\( \Rightarrow \theta  = 42.8^\circ {\text{ (0.747 radians)}}\)     (A1)

hence the angle between the line and the plane is 90° – 42.8° = 47.2° (0.824 radians)     A1

[5 marks]  

Note: Accept use of the formula a.b = \(\left| {} \right.\)a\(\left. {} \right|\)\(\left| {} \right.\)b\(\left| {\sin \theta } \right.\) .

Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.

Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.

Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.

Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.

Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.