Find vec x from vec x·vec a=1 , vec x·vec b=2 , vec x·vec…
Find $\vec x$ from $\vec x\cdot\vec a=1$, $\vec x\cdot\vec b=2$, $\vec x\cdot\vec c=3$, where $\vec a=\{2,-4,3\}$, $\vec b=\{3,-1,5\}$, $\vec c=\{1,-2,4\}$.
Explanation of Method
The vector x from three scalar-product conditions.
Step 1: System
\[ \begin{aligned} 2x_1-4x_2+3x_3&=1 &&(\mathrm I)\\ 3x_1-x_2+5x_3&=2 &&(\mathrm{II})\\ x_1-2x_2+4x_3&=3 &&(\mathrm{III}) \end{aligned} \]
The determinant is non-zero, so the solution is unique: \[\det\begin{pmatrix}2&-4&3\\3&-1&5\\1&-2&4\end{pmatrix}=2(-4+10)+4(12-5)+3(-6+1)=25\neq0.\]
Step 2: Elimination
From (III), $x_1=3+2x_2-4x_3$. Then: \[(\mathrm I):\ 6-5x_3=1\Rightarrow x_3=1;\qquad (\mathrm{II}):\ 9+5x_2-7x_3=2\Rightarrow x_2=0;\qquad x_1=3-4=-1.\]
\[\vec x=\{-1,\ 0,\ 1\}=-\vec i+\vec k\]
Step 3: Check
| Condition | Computation | Value |
|---|---|---|
| $\vec x\cdot\vec a$ | $-2+0+3$ | $1$ $\checkmark$ |
| $\vec x\cdot\vec b$ | $-3+0+5$ | $2$ $\checkmark$ |
| $\vec x\cdot\vec c$ | $-1+0+4$ | $3$ $\checkmark$ |
Because $\vec a,\vec b,\vec c$ are not coplanar (determinant $25$), the three projections determine $\vec x$ uniquely.
Result
\[\boxed{\,\vec x=\{-1,\ 0,\ 1\}=-\vec i+\vec k\,}\]