Let a→=i−k , b→=xi+j+(1−x)k and c→=yi+xj+(1+x−y)k. Then a→ , b→ and c→ are non-coplanar for:

a→=i−k and b→=xi+j+(1−x)kc→=yi+xj+(1+x−y)k[a→b→c→]=|10−1x11−xyx1+x−y|=(1+x−y−x+x2)−1(x2−y)=1

Let a→=i−k , b→=xi+j+(1−x)k and c→=yi+xj+(1+x−y)k. Then a→ , b→ and c→ are non-coplanar for:

a→=i−k and b→=xi+j+(1−x)kc→=yi+xj+(1+x−y)k[a→b→c→]=|10−1x11−xyx1+x−y|=(1+x−y−x+x2)−1(x2−y)=1

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Let $\vec{a} = i - k, \vec{b} = x i + j + (1 - x) k a n d \vec{c} = y i + x j + (1 + x - y) k .$ Then $\vec{a}, \vec{b}$ and $\vec{c}$ are non-coplanar for:

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some value of x

some values of y

no values of x and y

for all values of x and y

answer is D.

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$\vec{a} = i - k a n d \vec{b} = x i + j + (1 - x) k$

$\vec{c} = y i + x j + (1 + x - y) k$

$[\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & 0 & - 1 \\ x & 1 & 1 - x \\ y & x & 1 + x - y \end{matrix} | = (1 + x - y - x + x^{2}) - 1 (x^{2} - y) = 1$

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Let a→=i−k , b→=xi+j+(1−x)k and c→=yi+xj+(1+x−y)k. Then a→ , b→ and c→ are non-coplanar for:

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Let $\vec{a} = i - k, \vec{b} = x i + j + (1 - x) k a n d \vec{c} = y i + x j + (1 + x - y) k .$ Then $\vec{a}, \vec{b}$ and $\vec{c}$ are non-coplanar for: