Let a = 2i + j + k,b = i + 2j - k,and a unit | vedclass

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(A) Since $c$ is coplanar with $a$ and $b$,we can write $c = xa + yb$. Substituting the vectors: $c = x(2i + j + k) + y(i + 2j - k) = (2x + y)i + (x +

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$\frac{1}{\sqrt{2}} (-j + k)$

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(A) Since $c$ is coplanar with $a$ and $b$,we can write $c = xa + yb$. Substituting the vectors: $c = x(2i + j + k) + y(i + 2j - k) = (2x + y)i + (x + 2y)j + (x - y)k$. Given that $c$ is perpendicular to $a$,$a \cdot c = 0$. $2(2x + y) + 1(x + 2y) + 1(x - y) = 0$. $4x + 2y + x + 2y + x - y = 0 \implies 6x + 3y = 0 \implies y = -2x$. Substituting $y = -2x$ into the expression for $c$: $c = (2x - 2x)i + (x - 4x)j + (x + 2x)k = -3xj + 3xk = 3x(-j + k)$. Since $c$ is a unit vector,$|c| = 1$. $|3x(-j + k)| = 1 \implies |3x| \sqrt{(-1)^2 + 1^2} = 1 \implies |3x| \sqrt{2} = 1$. $|x| = \frac{1}{3\sqrt{2}}$. Thus,$c = 3(\pm \frac{1}{3\sqrt{2}})(-j + k) = \pm \frac{1}{\sqrt{2}}(-j + k)$. Comparing with the options,the correct answer is $\frac{1}{\sqrt{2}}(-j + k)$.

Let $a = 2i + j + k$,$b = i + 2j - k$,and a unit vector $c$ be coplanar. If $c$ is perpendicular to $a$,then $c = \dots$

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