Determining condition of coplanarity

Question

Determine the value of $\lambda$ such that the vectors $$5\vec{a}+6\vec{b}+7\vec{c},7\vec{a}+\lambda\vec{b}+9\vec{c},3\vec{a}+20\vec{b}+5\vec{c}$$ are coplanar given that $\vec{a},\vec{b},\vec{c}$ are non coplanar.

For the vectors to be coplanar,there must exist scalars $x_1,x_2,x_3$ not all $0$ such that their linear combination is $0$. Then after rearranging,the condition we get is $$(5x_1+7x_2+3x_3)\vec{a}+(6x_1+\lambda x_2+20x_3)\vec{b}+(7x_1+9x_2+5x_3)\vec{c}=0$$ Since $\vec{a},\vec{b},\vec{c}$ are non coplanar,hence each of the coefficients must be $0$.

But here,we are getting $3$ equations for $4$ variables(the fourth one being $\lambda$). So,how can $\lambda$ be determined? Can I assign any one of $x_1,x_2,x_3$ any value I like,for example can I make $x_1$ or $x_3$ $0$ or $1$ or any other number? I am really confused.

Answer 1

There's another approach which is imposing $$\mathbf u\cdot(\mathbf v\wedge\mathbf w)=0,$$ where $$\left\{\begin{align} &\mathbf u=5\mathbf a+6\mathbf b+7\mathbf c\\ &\mathbf v=7\mathbf a+\lambda\mathbf b+9\mathbf c\\ &\mathbf w=3\mathbf a+20\mathbf b+5\mathbf c \end{align}\right.$$ Let us now substitute and proceed: $$\begin{align} (5\mathbf a+6\mathbf b+7\mathbf c)\cdot((7\mathbf a+\lambda\mathbf b+9\mathbf c)\wedge(3\mathbf a+20\mathbf b+5\mathbf c)) &\overset{(1)}{=}(5\mathbf a+6\mathbf b+7\mathbf c)\cdot((140-3\lambda)\mathbf a\wedge\mathbf b+8\mathbf a\wedge\mathbf c+(5\lambda-180)\mathbf b\wedge\mathbf c)\\ &\overset{(2)}{=}7(140-3\lambda)\mathbf c\cdot(\mathbf a\wedge\mathbf b)+48\mathbf b\cdot(\mathbf a\wedge\mathbf c)+5(5\lambda-180)\mathbf a\cdot(\mathbf b\wedge\mathbf c)\\ &\overset{(3)}{=}[7(140-3\lambda)-48+5(5\lambda-180)]\underbrace{\mathbf a\cdot(\mathbf b\wedge\mathbf c)}_{\neq 0\, (4)}\overset{!}{=}0\\ &\iff 980-21\lambda-48+25\lambda-900=0\\ &\implies 32+4\lambda=0\implies \lambda=-8 \end{align}$$ Where at $(1)$ we used the fact that $\mathbf v_i\wedge\mathbf v_j=0$ if $i=j$, and $\mathbf v_i\wedge\mathbf v_j=-\mathbf v_j\wedge\mathbf v_i$; at $(2)$ that $\mathbf v_i\cdot(\mathbf v_j\wedge \mathbf v_k)=0$ if $i=j$ or $j=k$; at $(3)$ that every time you interchange places of two near vectors $\mathbf v_i\cdot(\mathbf v_j\wedge \mathbf v_k)$ changes its sign, and that at $(4)$ it is told that $\mathbf a$, $\mathbf b$, $\mathbf c$ aren't coplanar so $\mathbf a\cdot(\mathbf b\wedge\mathbf c)\neq 0$.

Answer 2

Even if you don't know about matrix associated with your system, You can try to translate them into manipulations on your system :

$\begin{bmatrix}5 &7 &3 \\6 &\lambda & 20 \\7 & 9 & 5\end{bmatrix}\overset{C_1'=7C_1\color{red}.C'_2=5C_2}\longleftrightarrow \begin{bmatrix}35 &35 &3 \\42 &5\lambda & 20 \\49 & 45 & 5\end{bmatrix}\overset{C_1''=\frac37C_1'\color{red}.C''_2=C'_2-C'_1}\longleftrightarrow \begin{bmatrix}15 &0 &3 \\18 &5\lambda-42 & 20 \\21 & -4 & 5\end{bmatrix}\overset{C_3'=5C_3} \longleftrightarrow \begin{bmatrix}15 &0 &15 \\18 &5\lambda-42 & 100 \\21 & -4 & 25\end{bmatrix}\overset{C_3''=C''_1-C'_3} \longleftrightarrow \begin{bmatrix}15 &0 &0 \\18 &5\lambda-42 & -82 \\21 & -4 & -4\end{bmatrix}$ $\overset{R_2\leftrightarrow R_3} \longleftrightarrow \begin{bmatrix}15 &0 &0 \\21 & -4 & -4 \\18 &5\lambda-42 & -82 \end{bmatrix}\longleftrightarrow \begin{bmatrix}15 &0 &0 \\21 & -4 & 0\\18 &5\lambda-42 & 5\lambda+42 \end{bmatrix}$

Then for $\boxed{\lambda=-8}$, $$5\vec{a}+6\vec{b}+7\vec{c},7\vec{a}+\lambda\vec{b}+9\vec{c},3\vec{a}+20\vec{b}+5\vec{c}$$ are coplanar.

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Note : The dot "$\color{red}.$" above the arrows means "and"

Answer 3

If they are coplanar then

$\det\begin{bmatrix}5 &7 &3 \\6 &\lambda & 20 \\7 & 9 & 5\end{bmatrix}=0$,

where the given matrix represents linear transformation in the basis with vectors $\bf{a,b,c}$.

Answer 4

The idea is any two nonparallel vectors can span out a plane in $\mathbb{R}^3$. Let $$\textbf{u}:=5⃗+6⃗+7⃗,\textbf{v}:=7⃗+⃗+9⃗,\textbf{w}:=3⃗+20⃗+5⃗$$ Then the coplanar condition implies v lies on the plane having u $\times$ w as the normal vector. This means you should solve $$\textbf{v}\cdot(\textbf{u}\times\textbf{w})=0$$ After computation you should get out something like $k(\textbf{a}\cdot(\textbf{b}\times\textbf{c}))=0$ for some constant $k$, but by the noncoplanar condition of a,b and c, this means $k=0$, so you can solve for lambda.

Answer 5

You can indeed view this problem as presenting 3 equations with 4 variables. However the assumption that you need as many equations as variables is false for several reasons.

Example 1: Find $x, y \in \mathbb{R}$ such that $x^2 + y^2 = 0$

Here we have one equation with two variables, but it has still a unique solution: $x = y = 0$. The setting where the "as many equations as variables" rule of thumb shines is when the equations are linear.

Example 2: Find $x, y \in \mathbb{R}$ not all $0$ such that $x+y = 0$ and $x-y = 0$

Here we have two equations and two variables, but solving the equation system leads to $x=y=0$ which is not a valid solution. The question is not to solve the system, having a unique solution to the equation system can prevent having a solution to the question.

Example 3: Find $\lambda \in \mathbb{R}$ and $x, y \in \mathbb{R}$ not all $0$ such that $x + y = 0$ and $x + \lambda y = 0$

Here we have two equation with 3 variables, but the variables don't all have the same status. If we choose one value for $\lambda$, then we have a linear system for x and y. We actually want to choose $\lambda$ so that the linear system won't have a unique solution, as it will be all $0$.

The third example is very close to your problem, if you can solve it you will be have a good understanding of it. I suggest you then look for the criteria for a unique solution if you are still stuck, and you should treat $\lambda$ as a constant of the linear system while calculating the criteria.