If $\vec a=\hat i-\hat k,\ \vec b=x\hat i+\hat j+(1-x)\hat k,\ \vec c=y\hat i+x\hat j+(1+x-y)\hat k$, show that $[\vec a,\vec b,\vec c]$ depends on neither $x$ nor $y$.

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> [!TLDR]
>
> Expanding the determinant along the row with the most zeros ($\vec a$ has a zero in the middle) reduces it to two $2	imes2$ minors that, remarkably, combine to the constant $1$ regardless of $x,y$.

**Step 1. Write the determinant** with rows $\vec a=(1,0,-1),\ \vec b=(x,1,1-x),\ \vec c=(y,x,1+x-y)$:
$$[\vec a,\vec b,\vec c]=\begin{vmatrix}1&0&-1\ x&1&1-x\ y&x&1+x-y\end{vmatrix}.$$

**Step 2. Expand along row 1** (middle entry is $0$, so only two cofactors are needed):
$$=1\cdot\begin{vmatrix}1&1-x\x&1+x-y\end{vmatrix}-0+(-1)\begin{vmatrix}x&1\y&x\end{vmatrix}.$$

**Step 3. Evaluate the first minor.**
$$\begin{vmatrix}1&1-x\x&1+x-y\end{vmatrix}=1(1+x-y)-(1-x)(x)=1+x-y-x+x^2=1-y+x^2.$$

**Step 4. Evaluate the second minor.**
$$\begin{vmatrix}x&1\y&x\end{vmatrix}=x^2-y.$$ …

Q.If $\vec a=\hat i-\hat k,\ \vec b=x\hat i+\hat j+(1-x)\hat k,\ \vec c=y\hat i+x\hat j+(1+x-y)\hat k$ , show that $[\vec a,\vec b,\vec c]$ depends on neither $x$ nor $y$ .

Concept understanding — Scalar Triple Product and Coplanarity

Scalar Triple Product and Coplanarity

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