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(A) Since $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{c} = x\overrightarrow{a} +

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) Since $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{c} = x\overrightarrow{a} + y\overrightarrow{b}$.Given $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,we have $\overrightarrow{b} \cdot (x\overrightarrow{a} + y\overrightarrow{b}) = 0$,which implies $x(\overrightarrow{a} \cdot \overrightarrow{b}) + y|\overrightarrow{b}|^{2} = 0$.Calculating $\overrightarrow{a} \cdot \overrightarrow{b} = (-1)(2) + (1)(0) + (1)(1) = -1$ and $|\overrightarrow{b}|^{2} = 2^{2} + 0^{2} + 1^{2} = 5$.So,$-x + 5y = 0 \Rightarrow x = 5y$.Thus,$\overrightarrow{c} = y(5\overrightarrow{a} + \overrightarrow{b}) = y(5(-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k})) = y(-3\hat{i} + 5\hat{j} + 6\hat{k})$.Given $\overrightarrow{a} \cdot \overrightarrow{c} = 7$,we have $y((-1)(-3) + (1)(5) + (1)(6)) = 7 \Rightarrow y(3 + 5 + 6) = 7 \Rightarrow 14y = 7 \Rightarrow y = \frac{1}{2}$.So,$\overrightarrow{c} = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.Then $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = (-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k}) + (-\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}) = (2 - 1 - \frac{3}{2})\hat{i} + (1 + \frac{5}{2})\hat{j} + (1 + 1 + 3)\hat{k} = -\frac{1}{2}\hat{i} + \frac{7}{2}\hat{j} + 5\hat{k}$.Finally,$2|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^{2} = 2((\frac{-1}{2})^{2} + (\frac{7}{2})^{2} + 5^{2}) = 2(\frac{1}{4} + \frac{49}{4} + 25) = 2(\frac{50}{4} + 25) = 2(12.5 + 25) = 2(37.5) = 75$.

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