Given vec a=\0,2λ,λ\ , vec b=\2,2,1\ , vec c=\-1,-2,-1\…

Given $\vec a=\{0,2\lambda,\lambda\}$, $\vec b=\{2,2,1\}$, $\vec c=\{-1,-2,-1\}$: $1^\circ$ Find $\vec d$ from $\vec a\times\vec b=\vec c\times\vec d$ and $\vec a\times\vec c=\vec b\times\vec d$. $2^\circ$ Show that $\vec a-\vec d$ and $\vec b-\vec c$ are collinear. $3^\circ$ Prove that $\vec a\times\vec b$, $\vec a\times\vec c$ and $\vec d$ are coplanar. $4^\circ$ Find $\lambda$ from $(\vec a-\vec b)\cdot\vec c=\vec a\cdot\vec c+\lambda$.

Step 1: Sum and difference of the two conditions

Sum:

\[ \vec a\times(\vec b+\vec c)=(\vec b+\vec c)\times\vec d=-\vec d\times(\vec b+\vec c), \] so $(\vec a+\vec d)\times(\vec b+\vec c)=\vec 0\ \Longrightarrow\ \vec a+\vec d\parallel\vec b+\vec c$.

Difference:

\[ \vec a\times(\vec b-\vec c)=(\vec c-\vec b)\times\vec d=\vec d\times(\vec b-\vec c), \] so $(\vec a-\vec d)\times(\vec b-\vec c)=\vec 0\ \Longrightarrow\ \vec a-\vec d\parallel\vec b-\vec c$.

These two statements are together equivalent to the two given conditions.

Step 2: $1^\circ$ Solving for $\vec d$

\[ \vec b+\vec c=\{1,0,0\},\qquad \vec b-\vec c=\{3,4,2\}. \]

Write $\vec a+\vec d=s\{1,0,0\}$ and $\vec a-\vec d=t\{3,4,2\}$. Adding: \[2\vec a=\{s+3t,\ 4t,\ 2t\}=\{0,\ 4\lambda,\ 2\lambda\}\ \Longrightarrow\ t=\lambda,\quad s=-3\lambda.\]

\[ \vec d=\vec a-\lambda(\vec b-\vec c)=\{0,2\lambda,\lambda\}-\{3\lambda,4\lambda,2\lambda\}. \]

\[\vec d=-\lambda\{3,\ 2,\ 1\}=\{-3\lambda,\ -2\lambda,\ -\lambda\}\]

Check (for every $\lambda$):

\[ \vec a\times\vec b=\{0,2\lambda,-4\lambda\}=\vec c\times\vec d,\qquad \vec a\times\vec c=\{0,-\lambda,2\lambda\}=\vec b\times\vec d.\ \checkmark \]

Step 3: $2^\circ$ Collinearity

\[ \vec a-\vec d=\{3\lambda,4\lambda,2\lambda\}=\lambda\,(\vec b-\vec c). \]

\[\vec a-\vec d=\lambda(\vec b-\vec c)\ \Longrightarrow\ \text{collinear}\]

Step 4: $3^\circ$ Coplanarity

First, $\vec a,\vec b,\vec c$ are themselves coplanar: \[[\vec a,\vec b,\vec c]=\begin{vmatrix}0&2\lambda&\lambda\\2&2&1\\-1&-2&-1\end{vmatrix}=0-2\lambda(-2+1)+\lambda(-4+2)=0.\]

Both $\vec a\times\vec b$ and $\vec a\times\vec c$ are normal to that common plane, so they are collinear. Indeed, $\vec a\times\vec c=-\tfrac12\,\vec a\times\vec b$. Two collinear vectors and any third vector always lie in one plane. Formally, with $(\vec u\times\vec v)\times(\vec u\times\vec w)=[\vec u,\vec v,\vec w]\,\vec u$:

\[ [\vec a\times\vec b,\ \vec a\times\vec c,\ \vec d]=[\vec a,\vec b,\vec c]\,(\vec a\cdot\vec d)=0. \]

\[\vec a\times\vec b,\ \vec a\times\vec c,\ \vec d\ \text{ are coplanar}\]

Step 5: $4^\circ$ $\lambda$

\[ (\vec a-\vec b)\cdot\vec c=\vec a\cdot\vec c-\vec b\cdot\vec c=\vec a\cdot\vec c+\lambda\ \Longrightarrow\ \lambda=-\vec b\cdot\vec c=-(-2-4-1). \]

\[\lambda=7,\qquad \vec a=\{0,14,7\},\qquad \vec d=\{-21,-14,-7\}\]