If a = 2i + k,b = i + j + k and c = 4i - 3j + | vedclass

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(D) Given $d 	imes b = c 	imes b$,this implies $(d - c) 	imes b = 0$. This means $(d - c)$ is parallel to $b$,so $d - c = \lambda b$ for some scala

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$-i - 8j + 2k$

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(D) Given $d 	imes b = c 	imes b$,this implies $(d - c) 	imes b = 0$. This means $(d - c)$ is parallel to $b$,so $d - c = \lambda b$ for some scalar $\lambda$. Thus,$d = c + \lambda b = (4i - 3j + 7k) + \lambda(i + j + k) = (4 + \lambda)i + (-3 + \lambda)j + (7 + \lambda)k$. We are given $d \cdot a = 0$,where $a = 2i + k$. Substituting $d$ and $a$: $((4 + \lambda)i + (-3 + \lambda)j + (7 + \lambda)k) \cdot (2i + 0j + k) = 0$. $2(4 + \lambda) + 0(-3 + \lambda) + 1(7 + \lambda) = 0$. $8 + 2\lambda + 7 + \lambda = 0$. $15 + 3\lambda = 0 \implies \lambda = -5$. Substituting $\lambda = -5$ back into the expression for $d$: $d = (4 - 5)i + (-3 - 5)j + (7 - 5)k = -i - 8j + 2k$.

If $a = 2i + k$,$b = i + j + k$ and $c = 4i - 3j + 7k$. If $d \times b = c \times b$ and $d \cdot a = 0$,then $d$ is equal to:

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