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Exercise 6.2 · Q8

Q.If a⃗=i^−k^, b⃗=xi^+j^+(1−x)k^, c⃗=yi^+xj^+(1+x−y)k^\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 [a⃗,b⃗,c⃗][\vec a,\vec b,\vec c] depends on neither xx nor yy.

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Expanding the determinant along the row with the most zeros (a⃗\vec a has a zero in the middle) reduces it to two 2×22\times2 minors that, remarkably, combine to the constant 11 regardless of x,yx,y.

Step 1. Write the determinant with rows a⃗=(1,0,−1), b⃗=(x,1,1−x), c⃗=(y,x,1+x−y)\vec a=(1,0,-1),\ \vec b=(x,1,1-x),\ \vec c=(y,x,1+x-y):

[a⃗,b⃗,c⃗]=∣10−1x11−xyx1+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 00, so only two cofactors are needed):

=1⋅∣11−xx1+x−y∣−0+(−1)∣x1yx∣.=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.

∣11−xx1+x−y∣=1(1+x−y)−(1−x)(x)=1+x−y−x+x2=1−y+x2.\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.

∣x1yx∣=x2−y.\begin{vmatrix}x&1\\y&x\end{vmatrix}=x^2-y. …

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