First slide

Planes in 3D

Question

Consider three planes 2x+py+6z=8, x + 2y + qz = 5 and x + y + 3z = 4

Moderate

Question

Three planes intersect at a point if

A

$p = 2, q \neq 3$
B

$p \neq 2, q \neq 3$
C

$p \neq 2, q = 3$
D

$p = 2, q = 3$

Solution

The given system of equations is
$\begin{array}{l} 2 x + py + 6 z = 8 \\ x + 2 y + qz = 5 \\ x + y + 3 z = 4 \\ Δ = |\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}| = (2 - p) (3 - q) \end{array}$
$By Cramer's rule, if Δ \neq 0, i.e., p \neq 2 and q \neq 3, the$ system has a unique solution.
$If p = 2 or q = 3, Δ = 0, then if Δ_{x} = Δ_{v} = Δ_{z} = 0, the$ system has infinite solutions and if any one of
$Δ_{x}, Δ_{y} and Δ_{z} \neq 0$ the system has no solution
now,
$\begin{array}{l} Δ_{x} = |\begin{array}{lll} 8 & p & 6 \\ 5 & 2 & q \\ 4 & 1 & 3 \end{array}| \\ = 30 - 8 q - 15 p + 4 pq = (4 q - 15) \cdot (p - 2) \\ = |\begin{array}{lll} 2 & 8 & 6 \\ 1 & 5 & q \\ 1 & 4 & 3 \end{array}| \\ = - 8 q + 8 q = 0 \\ Δ_{y} = |\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}| \\ = p - 2 \end{array}$
Thus, if $p = 2, Δ_{x} = Δ_{y} = Δ_{z} = 0$ for all $q \in R$ the system has infinite solutions.
if $p \neq 2, q = 3 and Δ_{z} \neq 0$ then the system has no solution.
Hence the system has (i) no solution if $p \neq 2$ and q = 3,(ii) a unique solution if $p \neq 2$ and $q \neq 2$ and (iii) infinite solutions if p = 2 and $q \in R$

Question

three planes do not have any common point of intersection if

A

$p = 2, q \neq 3$
B

$p \neq 2, q \neq 3$
C

$p \neq 2, q = 3$
D

$p = 2, q = 3$

Solution

The given system of equations is
$\begin{array}{l} 2 x + py + 6 z = 8 \\ x + 2 y + qz = 5 \\ x + y + 3 z = 4 \\ Δ = |\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}| = (2 - p) (3 - q) \end{array}$
$By Cramer's rule, if Δ \neq 0, i.e., p \neq 2 and q \neq 3, the$ system has a unique solution.
$If p = 2 or q = 3, Δ = 0, then if Δ_{x} = Δ_{v} = Δ_{z} = 0, the$ system has infinite solutions and if any one of
$Δ_{x}, Δ_{y} and Δ_{z} \neq 0$ the system has no solution
now,
$\begin{array}{l} Δ_{x} = |\begin{array}{lll} 8 & p & 6 \\ 5 & 2 & q \\ 4 & 1 & 3 \end{array}| \\ = 30 - 8 q - 15 p + 4 pq = (4 q - 15) \cdot (p - 2) \\ = |\begin{array}{lll} 2 & 8 & 6 \\ 1 & 5 & q \\ 1 & 4 & 3 \end{array}| \\ = - 8 q + 8 q = 0 \\ Δ_{y} = |\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}| \\ = p - 2 \end{array}$
Thus, if $p = 2, Δ_{x} = Δ_{y} = Δ_{z} = 0$ for all $q \in R$ the system has infinite solutions.
if $p \neq 2, q = 3 and Δ_{z} \neq 0$ then the system has no solution.
Hence the system has (i) no solution if $p \neq 2$ and q = 3,(ii) a unique solution if $p \neq 2$ and $q \neq 2$ and (iii) infinite solutions if p = 2 and $q \in R$

Question

null

A

$p = 2, q \in 3$
B

$p \in 2, q \in 3$
C

$p \neq 2, q = 3$
D

$p = 2, q = 3$

Solution

The given system of equations is
$\begin{array}{l} 2 x + py + 6 z = 8 \\ x + 2 y + qz = 5 \\ x + y + 3 z = 4 \\ Δ = |\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}| = (2 - p) (3 - q) \end{array}$
$By Cramer's rule, if Δ \neq 0, i.e., p \neq 2 and q \neq 3, the$ system has a unique solution.
$If p = 2 or q = 3, Δ = 0, then if Δ_{x} = Δ_{v} = Δ_{z} = 0, the$ system has infinite solutions and if any one of
$Δ_{x}, Δ_{y} and Δ_{z} \neq 0$ the system has no solution
now,
$\begin{array}{l} Δ_{x} = |\begin{array}{lll} 8 & p & 6 \\ 5 & 2 & q \\ 4 & 1 & 3 \end{array}| \\ = 30 - 8 q - 15 p + 4 pq = (4 q - 15) \cdot (p - 2) \\ = |\begin{array}{lll} 2 & 8 & 6 \\ 1 & 5 & q \\ 1 & 4 & 3 \end{array}| \\ = - 8 q + 8 q = 0 \\ Δ_{y} = |\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}| \\ = p - 2 \end{array}$
Thus, if $p = 2, Δ_{x} = Δ_{y} = Δ_{z} = 0$ for all $q \in R$ the system has infinite solutions.
if $p \neq 2, q = 3 and Δ_{z} \neq 0$ then the system has no solution.
Hence the system has (i) no solution if $p \neq 2$ and q = 3,(ii) a unique solution if $p \neq 2$ and $q \neq 2$ and (iii) infinite solutions if p = 2 and $q \in R$

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