bar{a}, bar{b}, bar{c} are nonzero vectors su | vedclass

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(A) Given $\bar{a} \cdot \bar{b} = 0$,$\bar{a} \cdot \bar{c} = 0$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$,and $\bar{b} \cdot \bar{c} = 1$. Since $\bar

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$13y^2+14y+5$

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(A) Given $\bar{a} \cdot \bar{b} = 0$,$\bar{a} \cdot \bar{c} = 0$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$,and $\bar{b} \cdot \bar{c} = 1$. Since $\bar{d}$ is coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$,we have $\bar{d} = x(\bar{a}+\bar{b}) + y(2\bar{b}-\bar{c}) = x\bar{a} + (x+2y)\bar{b} - y\bar{c}$. Given $\bar{d} \cdot \bar{a} = 1$,we calculate: $(x\bar{a} + (x+2y)\bar{b} - y\bar{c}) \cdot \bar{a} = x|\bar{a}|^2 + (x+2y)(\bar{b} \cdot \bar{a}) - y(\bar{c} \cdot \bar{a}) = x(1) + 0 - 0 = x$. Thus,$x = 1$. Now,$\bar{d} = \bar{a} + (1+2y)\bar{b} - y\bar{c}$. $|\bar{d}|^2 = \bar{d} \cdot \bar{d} = (\bar{a} + (1+2y)\bar{b} - y\bar{c}) \cdot (\bar{a} + (1+2y)\bar{b} - y\bar{c})$. $|\bar{d}|^2 = |\bar{a}|^2 + (1+2y)^2|\bar{b}|^2 + y^2|\bar{c}|^2 + 2(1+2y)(\bar{a} \cdot \bar{b}) - 2y(\bar{a} \cdot \bar{c}) - 2y(1+2y)(\bar{b} \cdot \bar{c})$. Substituting the values: $|\bar{d}|^2 = 1 + (1+4y+4y^2)(4) + y^2(1) + 0 - 0 - 2y(1+2y)(1)$. $|\bar{d}|^2 = 1 + 4 + 16y + 16y^2 + y^2 - 2y - 4y^2$. $|\bar{d}|^2 = 13y^2 + 14y + 5$.

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