If a=2 hat{i}+hat{k},b=hat{i}+hat{j}+hat{k},a | vedclass

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(D) Given vectors are $a=2 \hat{i}+\hat{k}$,$b=\hat{i}+\hat{j}+\hat{k}$,and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$.Given condition: $r 	imes b = c 	imes

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$-\hat{i}-8 \hat{j}+2 \hat{k}$

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(D) Given vectors are $a=2 \hat{i}+\hat{k}$,$b=\hat{i}+\hat{j}+\hat{k}$,and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$.Given condition: $r 	imes b = c 	imes b$.This implies $(r-c) 	imes b = 0$,which means $(r-c)$ is parallel to $b$.So,$r-c = \lambda b$,or $r = c + \lambda b$ ... $(i)$.Also given $r \cdot a = 0$.Substituting $r$ from $(i)$ into this condition:$(c + \lambda b) \cdot a = 0$$(4 \hat{i}-3 \hat{j}+7 \hat{k} + \lambda(\hat{i}+\hat{j}+\hat{k})) \cdot (2 \hat{i}+\hat{k}) = 0$$((4+\lambda) \hat{i} + (-3+\lambda) \hat{j} + (7+\lambda) \hat{k}) \cdot (2 \hat{i}+\hat{k}) = 0$$2(4+\lambda) + 1(7+\lambda) = 0$$8 + 2\lambda + 7 + \lambda = 0$$3\lambda + 15 = 0 \implies \lambda = -5$.Substituting $\lambda = -5$ into $(i)$:$r = (4 \hat{i}-3 \hat{j}+7 \hat{k}) - 5(\hat{i}+\hat{j}+\hat{k})$$r = (4-5) \hat{i} + (-3-5) \hat{j} + (7-5) \hat{k}$$r = -\hat{i} - 8 \hat{j} + 2 \hat{k}$.

If $a=2 \hat{i}+\hat{k}$,$b=\hat{i}+\hat{j}+\hat{k}$,and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$,then the vector $r$ satisfying $r \times b=c \times b$ and $r \cdot a=0$ is

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