If overline{p}=2 hat{i}+hat{k},overline{q}=ha | vedclass

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(C) Given $\overline{m} 	imes \overline{q} = \overline{r} 	imes \overline{q}$,we can write $(\overline{m} - \overline{r}) 	imes \overline{q} = 0$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $\overline{m} 	imes \overline{q} = \overline{r} 	imes \overline{q}$,we can write $(\overline{m} - \overline{r}) 	imes \overline{q} = 0$. This implies that $(\overline{m} - \overline{r})$ is parallel to $\overline{q}$. So,$\overline{m} - \overline{r} = t \overline{q}$ for some scalar $t$,which gives $\overline{m} = \overline{r} + t \overline{q}$. Substituting the given vectors: $\overline{m} = (4 \hat{i} - 3 \hat{j} + 7 \hat{k}) + t(\hat{i} + \hat{j} + \hat{k}) = (4+t) \hat{i} + (-3+t) \hat{j} + (7+t) \hat{k}$. We are given $\overline{m} \cdot \overline{p} = 0$,where $\overline{p} = 2 \hat{i} + \hat{k}$. So,$((4+t) \hat{i} + (-3+t) \hat{j} + (7+t) \hat{k}) \cdot (2 \hat{i} + \hat{k}) = 0$. $2(4+t) + 0(-3+t) + 1(7+t) = 0$. $8 + 2t + 7 + t = 0 \implies 3t + 15 = 0 \implies t = -5$. Substituting $t = -5$ into the expression for $\overline{m}$: $\overline{m} = (4-5) \hat{i} + (-3-5) \hat{j} + (7-5) \hat{k} = -\hat{i} - 8 \hat{j} + 2 \hat{k}$.

If $\overline{p}=2 \hat{i}+\hat{k}$,$\overline{q}=\hat{i}+\hat{j}+\hat{k}$,$\overline{r}=4 \hat{i}-3 \hat{j}+7 \hat{k}$ and a vector $\overline{m}$ is such that $\overline{m} \times \overline{q}=\overline{r} \times \overline{q}$ and $\overline{m} \cdot \overline{p}=0$,then $\overline{m} = \dots$

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