Topics

Three planes, and systems containing an unknown constant

Every equation is a plane

$ax + by + cz = d$ is the equation of a plane in three-dimensional space.
A solution of the system is a point lying on all three planes at once.
The nature of the solution is exactly the geometry of how the three planes are arranged, which is why the geometric picture is worth knowing rather than just the algebraic signatures.

The configurations

Configuration	Solutions	How to recognise it
Meet at a point	Unique	three independent equations
Meet along a line ("sheaf")	Infinitely many, 1 parameter	one equation is a combination of the other two
All three coincide	Infinitely many, 2 parameters	two equations are multiples of the third
Triangular prism	None	each pair meets in a line, but the three lines are parallel and distinct
Two parallel, one crossing	None	two equations have proportional coefficients but non-proportional constants
All three parallel	None	all three have proportional coefficients, different constants

The prism is the subtle case. Every pair of equations is perfectly consistent — you can solve any two of them — but no point satisfies all three. It is the geometric picture behind "the pieces all agree, but not together".

Reading the coefficients

Proportional coefficient rows mean parallel planes:
- $2x - y + z = 4$ and $4x - 2y + 2z = 9$ have coefficients in the ratio $1:2$ but constants $4$ and $9$ , which is not $1:2$ — parallel and distinct, no solution.
- If the constant had been $8$ , the whole rows would be proportional — the same plane.
Check the constant separately from the coefficients. The coefficients decide whether the planes are parallel; the constant decides whether they coincide.

Systems containing an unknown constant

A standard Excellence question: for what values of $k$ does the system have a unique solution, infinitely many, or none?
The strategy is always the same:
- Eliminate normally, carrying $k$ along as an ordinary symbol.
- Reduce to a single equation of the form $(\text{expression in } k)\,z = (\text{expression in } k)$ .
- Then read off the three cases:

Final equation $Az = B$	Condition	Outcome
$A \neq 0$	whatever $B$ is	Unique solution
$A = 0$ and $B = 0$		Infinitely many
$A = 0$ and $B \neq 0$		No solution

Do not divide by an expression containing $k$ . Dividing by $(k^2 - 16)$ silently assumes $k \neq \pm 4$ — which is exactly the case the question is about.
Solve $A = 0$ first, then test each root in $B$ . That is what separates the "infinitely many" values of $k$ from the "none" ones.

Interpreting the answer in context

Unique — the information given pins the situation down exactly.
Infinitely many — the information was not enough; one of the conditions repeated another. In context: "the third measurement told us nothing we did not already know."
None — the conditions contradict. In context: "the specification cannot be met; conditions 1 and 3 are incompatible."
Say which is which and why, in the context's own language. That is the interpretation the standard is assessing.

Tips and Tricks

Carry $k$ through the elimination like any other number. The algebra is no harder; the only difference is that you must not divide by anything containing $k$ .

Reduce to a single equation $Az = B$ before classifying. All three cases then read straight off, and it is impossible to muddle them.

Solve $A = 0$ , then substitute each root into $B$ . One root usually gives "infinitely many" and another "no solution" — reporting both as the same case is the standard error.

Check the constants separately when testing for parallel planes. Proportional coefficients alone do not tell you whether the planes coincide or are distinct.

Remember the prism. Every pair of equations can be consistent while the system as a whole has no solution — so "I solved two of them, so there must be a solution" is not valid reasoning.

Verify your boundary values. Substituting $k = 4$ back into the original system and re-solving takes two minutes and confirms the classification is right.

Worked ExampleA system with an unknown constant

For what values of $k$ does the following system have (i) a unique solution, (ii) infinitely many solutions, (iii) no solution? Interpret each case geometrically.

$\begin{cases} x + 2y - 3z = 4 & (1) \\ 3x - y + 5z = 2 & (2) \\ 4x + y + (k^2 - 14)z = k + 2 & (3) \end{cases}$

Step 1 — Eliminate $x$ from the first pair

$(2) - 3 \times (1)$ :

$(3x - y + 5z) - (3x + 6y - 9z) = 2 - 12$

Expand the subtraction in full:

$3x - y + 5z - 3x - 6y + 9z = -10$

$-7y + 14z = -10 \qquad (4)$

Step 2 — Eliminate $x$ from a different pair

$(3) - 4 \times (1)$ :

$\left[4x + y + (k^2 - 14)z\right] - \left[4x + 8y - 12z\right] = (k + 2) - 16$

$-7y + (k^2 - 14 + 12)z = k - 14$

$-7y + (k^2 - 2)z = k - 14 \qquad (5)$

Step 3 — Eliminate $y$

$(5) - (4)$ — the $y$ coefficients are both $-7$ , so subtracting removes $y$ :

$\left[(k^2 - 2) - 14\right]z = (k - 14) - (-10)$

$(k^2 - 16)z = k - 4 \qquad (6)$

$A = k^2 - 16, \qquad B = k - 4$

Do NOT divide by $k^2 - 16$ — that would assume it is non-zero, which is exactly the case in question.

Step 4 — Case (i): unique solution

A unique value of $z$ exists whenever the coefficient is non-zero:

$k^2 - 16 \neq 0$

$k^2 \neq 16 \quad \Rightarrow \quad k \neq 4 \ \text{ and } \ k \neq -4$

$\boxed{\text{Unique solution for all } k \neq \pm 4}$

Once $z$ is determined, (4) gives $y$ and (1) gives $x$ , so the whole solution is unique.

Step 5 — Find where $A = 0$ , and test $B$ at each root

$k^2 - 16 = 0 \quad \Rightarrow \quad k = 4 \ \text{ or } \ k = -4$

At $k = 4$ :

$B = k - 4 = 4 - 4 = 0$

Equation (6) becomes $0 \cdot z = 0$ — TRUE for every $z$ .

At $k = -4$ :

$B = k - 4 = -4 - 4 = -8$

Equation (6) becomes $0 \cdot z = -8$ — FALSE for every $z$ .

Step 6 — Case (ii): infinitely many solutions

$\boxed{k = 4}$

Find the general solution. With $k = 4$ , equation (3) becomes

$4x + y + 2z = 6$

and (6) is $0 = 0$ , so only (1) and (4) remain independent.

Let $z = t$ . From (4):

$-7y + 14t = -10 \quad \Rightarrow \quad y = \frac{14t + 10}{7} = 2t + \frac{10}{7}$

From (1):

$x = 4 - 2y + 3z = 4 - 2\left(2t + \frac{10}{7}\right) + 3t = 4 - 4t - \frac{20}{7} + 3t$

$x = \frac{28 - 20}{7} - t = \frac{8}{7} - t$

$\boxed{(x, y, z) = \left(\frac{8}{7} - t,\ \ 2t + \frac{10}{7},\ \ t\right), \quad t \in \mathbb{R}}$

Check in the original (3) with $k = 4$ :

$4\left(\frac{8}{7} - t\right) + \left(2t + \frac{10}{7}\right) + 2t = \frac{32}{7} - 4t + 2t + \frac{10}{7} + 2t = \frac{42}{7} = 6 \ \checkmark$

Every $t$ cancels ✓ and $k + 2 = 6$ ✓

Check in (2): $3\left(\frac{8}{7} - t\right) - \left(2t + \frac{10}{7}\right) + 5t = \frac{24 - 10}{7} - 3t - 2t + 5t = \frac{14}{7} = 2$ ✓

Step 7 — Case (iii): no solution

$\boxed{k = -4}$

Naming the inconsistency. With $k = -4$ , equation (3) becomes

$4x + y + 2z = -2$

while the first two equations together force $4x + y + 2z = 6$ (from the $k = 4$ working, where the same left-hand side arose). The two demands are contradictory:

$6 \neq -2$

Notice that the left-hand side of (3) is IDENTICAL for $k = 4$ and $k = -4$ , since $k^2 - 14 = 2$ in both cases — only the constant $k + 2$ differs. That is precisely why one value gives a repeated plane and the other a parallel one.

Step 8 — Interpret geometrically

$k$	Configuration	Solutions
$k \neq \pm 4$	The three planes meet at a single point	Unique
$k = 4$	The third plane contains the line in which the first two meet — a sheaf	Infinitely many (a line)
$k = -4$	The third plane is parallel to that line but offset from it — a triangular prism	None

Step 9 — Verify a boundary value independently

Take $k = 4$ and a specific $t$ , say $t = 0$ : the solution should be $\left(\frac{8}{7},\ \frac{10}{7},\ 0\right)$ .

Equation (1): $\frac{8}{7} + \frac{20}{7} - 0 = \frac{28}{7} = 4$ ✓ Equation (2): $\frac{24}{7} - \frac{10}{7} + 0 = \frac{14}{7} = 2$ ✓ Equation (3): $\frac{32}{7} + \frac{10}{7} + 0 = \frac{42}{7} = 6 = k + 2$ ✓

And take $k = 5$ , which should give a unique solution. From (6):

$(25 - 16)z = 5 - 4 \quad \Rightarrow \quad 9z = 1, \quad z = \frac{1}{9}$

A definite value ✓ — so the classification holds at a general value as well as at the boundaries.

Three planes, and systems containing an unknown constant

Covers AS91587

Every equation is a plane

$ax + by + cz = d$ is the equation of a plane in three-dimensional space.
A solution of the system is a point lying on all three planes at once.
The nature of the solution is exactly the geometry of how the three planes are arranged, which is why the geometric picture is worth knowing rather than just the algebraic signatures.

The configurations

Configuration	Solutions	How to recognise it
Meet at a point	Unique	three independent equations
Meet along a line ("sheaf")	Infinitely many, 1 parameter	one equation is a combination of the other two
All three coincide	Infinitely many, 2 parameters	two equations are multiples of the third
Triangular prism	None	each pair meets in a line, but the three lines are parallel and distinct
Two parallel, one crossing	None	two equations have proportional coefficients but non-proportional constants
All three parallel	None	all three have proportional coefficients, different constants

The prism is the subtle case. Every pair of equations is perfectly consistent — you can solve any two of them — but no point satisfies all three. It is the geometric picture behind "the pieces all agree, but not together".

Reading the coefficients

Proportional coefficient rows mean parallel planes:
- $2x - y + z = 4$ and $4x - 2y + 2z = 9$ have coefficients in the ratio $1:2$ but constants $4$ and $9$ , which is not $1:2$ — parallel and distinct, no solution.
- If the constant had been $8$ , the whole rows would be proportional — the same plane.
Check the constant separately from the coefficients. The coefficients decide whether the planes are parallel; the constant decides whether they coincide.

Systems containing an unknown constant

A standard Excellence question: for what values of $k$ does the system have a unique solution, infinitely many, or none?
The strategy is always the same:
- Eliminate normally, carrying $k$ along as an ordinary symbol.
- Reduce to a single equation of the form $(\text{expression in } k)\,z = (\text{expression in } k)$ .
- Then read off the three cases:

Final equation $Az = B$	Condition	Outcome
$A \neq 0$	whatever $B$ is	Unique solution
$A = 0$ and $B = 0$		Infinitely many
$A = 0$ and $B \neq 0$		No solution

Do not divide by an expression containing $k$ . Dividing by $(k^2 - 16)$ silently assumes $k \neq \pm 4$ — which is exactly the case the question is about.
Solve $A = 0$ first, then test each root in $B$ . That is what separates the "infinitely many" values of $k$ from the "none" ones.

Interpreting the answer in context

Unique — the information given pins the situation down exactly.
Infinitely many — the information was not enough; one of the conditions repeated another. In context: "the third measurement told us nothing we did not already know."
None — the conditions contradict. In context: "the specification cannot be met; conditions 1 and 3 are incompatible."
Say which is which and why, in the context's own language. That is the interpretation the standard is assessing.

Tips and Tricks

Carry $k$ through the elimination like any other number. The algebra is no harder; the only difference is that you must not divide by anything containing $k$ .

Reduce to a single equation $Az = B$ before classifying. All three cases then read straight off, and it is impossible to muddle them.

Solve $A = 0$ , then substitute each root into $B$ . One root usually gives "infinitely many" and another "no solution" — reporting both as the same case is the standard error.

Check the constants separately when testing for parallel planes. Proportional coefficients alone do not tell you whether the planes coincide or are distinct.

Remember the prism. Every pair of equations can be consistent while the system as a whole has no solution — so "I solved two of them, so there must be a solution" is not valid reasoning.

Verify your boundary values. Substituting $k = 4$ back into the original system and re-solving takes two minutes and confirms the classification is right.

Worked ExampleA system with an unknown constant

For what values of $k$ does the following system have (i) a unique solution, (ii) infinitely many solutions, (iii) no solution? Interpret each case geometrically.

$\begin{cases} x + 2y - 3z = 4 & (1) \\ 3x - y + 5z = 2 & (2) \\ 4x + y + (k^2 - 14)z = k + 2 & (3) \end{cases}$

Step 1 — Eliminate $x$ from the first pair

$(2) - 3 \times (1)$ :

$(3x - y + 5z) - (3x + 6y - 9z) = 2 - 12$

Expand the subtraction in full:

$3x - y + 5z - 3x - 6y + 9z = -10$

$-7y + 14z = -10 \qquad (4)$

Step 2 — Eliminate $x$ from a different pair

$(3) - 4 \times (1)$ :

$\left[4x + y + (k^2 - 14)z\right] - \left[4x + 8y - 12z\right] = (k + 2) - 16$

$-7y + (k^2 - 14 + 12)z = k - 14$

$-7y + (k^2 - 2)z = k - 14 \qquad (5)$

Step 3 — Eliminate $y$

$(5) - (4)$ — the $y$ coefficients are both $-7$ , so subtracting removes $y$ :

$\left[(k^2 - 2) - 14\right]z = (k - 14) - (-10)$

$(k^2 - 16)z = k - 4 \qquad (6)$

$A = k^2 - 16, \qquad B = k - 4$

Do NOT divide by $k^2 - 16$ — that would assume it is non-zero, which is exactly the case in question.

Step 4 — Case (i): unique solution

A unique value of $z$ exists whenever the coefficient is non-zero:

$k^2 - 16 \neq 0$

$k^2 \neq 16 \quad \Rightarrow \quad k \neq 4 \ \text{ and } \ k \neq -4$

$\boxed{\text{Unique solution for all } k \neq \pm 4}$

Once $z$ is determined, (4) gives $y$ and (1) gives $x$ , so the whole solution is unique.

Step 5 — Find where $A = 0$ , and test $B$ at each root

$k^2 - 16 = 0 \quad \Rightarrow \quad k = 4 \ \text{ or } \ k = -4$

At $k = 4$ :

$B = k - 4 = 4 - 4 = 0$

Equation (6) becomes $0 \cdot z = 0$ — TRUE for every $z$ .

At $k = -4$ :

$B = k - 4 = -4 - 4 = -8$

Equation (6) becomes $0 \cdot z = -8$ — FALSE for every $z$ .

Step 6 — Case (ii): infinitely many solutions

$\boxed{k = 4}$

Find the general solution. With $k = 4$ , equation (3) becomes

$4x + y + 2z = 6$

and (6) is $0 = 0$ , so only (1) and (4) remain independent.

Let $z = t$ . From (4):

$-7y + 14t = -10 \quad \Rightarrow \quad y = \frac{14t + 10}{7} = 2t + \frac{10}{7}$

From (1):

$x = 4 - 2y + 3z = 4 - 2\left(2t + \frac{10}{7}\right) + 3t = 4 - 4t - \frac{20}{7} + 3t$

$x = \frac{28 - 20}{7} - t = \frac{8}{7} - t$

$\boxed{(x, y, z) = \left(\frac{8}{7} - t,\ \ 2t + \frac{10}{7},\ \ t\right), \quad t \in \mathbb{R}}$

Check in the original (3) with $k = 4$ :

$4\left(\frac{8}{7} - t\right) + \left(2t + \frac{10}{7}\right) + 2t = \frac{32}{7} - 4t + 2t + \frac{10}{7} + 2t = \frac{42}{7} = 6 \ \checkmark$

Every $t$ cancels ✓ and $k + 2 = 6$ ✓

Check in (2): $3\left(\frac{8}{7} - t\right) - \left(2t + \frac{10}{7}\right) + 5t = \frac{24 - 10}{7} - 3t - 2t + 5t = \frac{14}{7} = 2$ ✓

Step 7 — Case (iii): no solution

$\boxed{k = -4}$

Naming the inconsistency. With $k = -4$ , equation (3) becomes

$4x + y + 2z = -2$

while the first two equations together force $4x + y + 2z = 6$ (from the $k = 4$ working, where the same left-hand side arose). The two demands are contradictory:

$6 \neq -2$

Step 8 — Interpret geometrically

$k$	Configuration	Solutions
$k \neq \pm 4$	The three planes meet at a single point	Unique
$k = 4$	The third plane contains the line in which the first two meet — a sheaf	Infinitely many (a line)
$k = -4$	The third plane is parallel to that line but offset from it — a triangular prism	None

Step 9 — Verify a boundary value independently

Take $k = 4$ and a specific $t$ , say $t = 0$ : the solution should be $\left(\frac{8}{7},\ \frac{10}{7},\ 0\right)$ .

And take $k = 5$ , which should give a unique solution. From (6):

$(25 - 16)z = 5 - 4 \quad \Rightarrow \quad 9z = 1, \quad z = \frac{1}{9}$

A definite value ✓ — so the classification holds at a general value as well as at the boundaries.

Three planes, and systems containing an unknown constant

Every equation is a plane

The configurations

Reading the coefficients

Systems containing an unknown constant

Interpreting the answer in context

Step 1 — Eliminate xxx from the first pair

Step 2 — Eliminate xxx from a different pair

Step 3 — Eliminate yyy

Step 4 — Case (i): unique solution

Step 5 — Find where A=0A = 0A=0, and test BBB at each root

Step 6 — Case (ii): infinitely many solutions

Step 7 — Case (iii): no solution

Step 8 — Interpret geometrically

Step 9 — Verify a boundary value independently

Three planes, and systems containing an unknown constant

Every equation is a plane

The configurations

Reading the coefficients

Systems containing an unknown constant

Interpreting the answer in context

Step 1 — Eliminate xxx from the first pair

Step 2 — Eliminate xxx from a different pair

Step 3 — Eliminate yyy

Step 4 — Case (i): unique solution

Step 5 — Find where A=0A = 0A=0, and test BBB at each root

Step 6 — Case (ii): infinitely many solutions

Step 7 — Case (iii): no solution

Step 8 — Interpret geometrically

Step 9 — Verify a boundary value independently

Step 1 — Eliminate $x$ from the first pair

Step 2 — Eliminate $x$ from a different pair

Step 3 — Eliminate $y$

Step 5 — Find where $A = 0$ , and test $B$ at each root

Step 1 — Eliminate $x$ from the first pair

Step 2 — Eliminate $x$ from a different pair

Step 3 — Eliminate $y$

Step 5 — Find where $A = 0$ , and test $B$ at each root