24. Vektorja $$ \vec{e}=(-2,4,3-x) $$ in $$ \vec{f}=(2+x, 3,-6) $$ sta diagonali paralelograma. Določite $$ x $$ tako, da bo ta paralelogram romb, zapišite vektorja na stra- nicah in izračunajte dolžino stranice.

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We are given the diagonal vectors  
$$
\vec{e}=(-2,4,3-x) \quad \text{and} \quad \vec{f}=(2+x,3,-6)
$$
of a parallelogram that is a rhombus. In any rhombus the diagonals are perpendicular. Therefore, we require
$$
\vec{e}\cdot\vec{f}=0.
$$

1. Calculate the dot product:
   $$
   \vec{e}\cdot\vec{f}= (-2)(2+x)+4\cdot 3+(3-x)(-6).
   $$
   Compute each term:
   $$
   (-2)(2+x)=-4-2x,\quad 4\cdot 3=12,\quad (3-x)(-6)=-18+6x.
   $$
   Thus,
   $$
   \vec{e}\cdot\vec{f} = (-4-2x)+12+(-18+6x) = (-10) + 4x.
   $$
2. Set the dot product equal to zero:
   $$
   4x-10=0 \quad\Longrightarrow\quad 4x=10\quad\Longrightarrow\quad x=\frac{10}{4}=\frac{5}{2}.
   $$

3. With $$ x=\frac{5}{2} $$, determine the diagonals:
   $$
   \vec{e} = \left(-2,\, 4,\, 3-\frac{5}{2}\right) = \left(-2,\, 4,\, \frac{6-5}{2}\right)=\left(-2,\, 4,\, \frac{1}{2}\right),
   $$
   $$
   \vec{f} = \left(2+\frac{5}{2},\, 3,\, -6\right)=\left(\frac{4+5}{2},\, 3,\,-6\right)=\left(\frac{9}{2},\, 3,\,-6\right).
   $$

4. The sides of the parallelogram are given by
   $$
   \vec{a}=\frac{1}{2}(\vec{e}+\vec{f}) \quad \text{and} \quad \vec{b}=\frac{1}{2}(\vec{e}-\vec{f}).
   $$

- Compute $$\vec{a}$$:
     $$
     \vec{e}+\vec{f} = \left(-2+\frac{9}{2},\, 4+3,\, \frac{1}{2}+(-6)\right).
     $$
     For each coordinate:
     $$
     -2+\frac{9}{2} = \frac{-4+9}{2}=\frac{5}{2},\quad 4+3=7,\quad \frac{1}{2}-6=\frac{1-12}{2}=-\frac{11}{2}.
     $$
     Therefore,
     $$
     \vec{a}=\frac{1}{2}\left(\frac{5}{2},\, 7,\, -\frac{11}{2}\right)=\left(\frac{5}{4},\, \frac{7}{2},\, -\frac{11}{4}\right).
     $$

- Compute $$\vec{b}$$:
     $$
     \vec{e}-\vec{f} = \left(-2-\frac{9}{2},\, 4-3,\, \frac{1}{2}-(-6)\right).
     $$
     For each coordinate:
     $$
     -2-\frac{9}{2} = \frac{-4-9}{2}=-\frac{13}{2},\quad 4-3=1,\quad \frac{1}{2}+6=\frac{1+12}{2}=\frac{13}{2}.
     $$
     Therefore,
     $$
     \vec{b}=\frac{1}{2}\left(-\frac{13}{2},\, 1,\, \frac{13}{2}\right)=\left(-\frac{13}{4},\, \frac{1}{2},\, \frac{13}{4}\right).
     $$

5. The side lengths are given by the magnitude of either $$\vec{a}$$ or $$\vec{b}$$. We use $$\vec{a}$$:
   $$
   |\vec{a}|=\sqrt{\left(\frac{5}{4}\right)^2+\left(\frac{7}{2}\right)^2+\left(-\frac{11}{4}\right)^2}.
   $$
   Compute each squared term:
   $$
   \left(\frac{5}{4}\right)^2=\frac{25}{16},\quad \left(\frac{7}{2}\right)^2=\frac{49}{4}=\frac{196}{16},\quad \left(\frac{11}{4}\right)^2=\frac{121}{16}.
   $$
   Sum:
   $$
   |\vec{a}|=\sqrt{\frac{25+196+121}{16}}=\sqrt{\frac{342}{16}}=\frac{\sqrt{342}}{4}.
   $$
   Note that $$342=9\cdot38$$, so
   $$
   \sqrt{342}=3\sqrt{38},
   $$
   and
   $$
   |\vec{a}|=\frac{3\sqrt{38}}{4}.
   $$

Thus, the value of $$ x $$ is $$\frac{5}{2}$$, the side vectors are
$$
\vec{a}=\left(\frac{5}{4},\, \frac{7}{2},\, -\frac{11}{4}\right) \quad \text{and} \quad \vec{b}=\left(-\frac{13}{4},\, \frac{1}{2},\, \frac{13}{4}\right),
$$
and the length of a side is
$$
\frac{3\sqrt{38}}{4}.
$$
$$ x = \frac{5}{2} $$. The side vectors are $$ \vec{a} = \left(\frac{5}{4}, \frac{7}{2}, -\frac{11}{4}\right) $$ and $$ \vec{b} = \left(-\frac{13}{4}, \frac{1}{2}, \frac{13}{4}\right) $$. The length of each side is $$ \frac{3\sqrt{38}}{4} $$.

24. Vektorja \( \vec{e}=(-2,4,3-x) \) in \( \vec{f}=(2+x, 3,-6) \) sta diagonali paralelograma. Določite \( x \) tako, da bo ta paralelogram romb, zapišite vektorja na stra- nicah in izračunajte dolžino stranice.

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