Let vec{AB} = 2 hat{i} + 4 hat{j} - 5 hat{k} | vedclass

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(C) In a parallelogram $ABCD$,the diagonal $\vec{AC} = \vec{AB} + \vec{AD}$.Given $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{

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$3$

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(C) In a parallelogram $ABCD$,the diagonal $\vec{AC} = \vec{AB} + \vec{AD}$.Given $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$.Therefore,$\vec{AC} = (2+1) \hat{i} + (4+2) \hat{j} + (-5+\lambda) \hat{k} = 3 \hat{i} + 6 \hat{j} + (\lambda - 5) \hat{k}$.The projection of $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on $\vec{AC}$ is given by $\frac{|\vec{v} \cdot \vec{AC}|}{|\vec{AC}|} = 1$.$|\vec{v} \cdot \vec{AC}| = |(1)(3) + (1)(6) + (1)(\lambda - 5)| = |3 + 6 + \lambda - 5| = |\lambda + 4|$.$|\vec{AC}| = \sqrt{3^2 + 6^2 + (\lambda - 5)^2} = \sqrt{9 + 36 + \lambda^2 - 10\lambda + 25} = \sqrt{\lambda^2 - 10\lambda + 70}$.So,$\frac{|\lambda + 4|}{\sqrt{\lambda^2 - 10\lambda + 70}} = 1 \Rightarrow (\lambda + 4)^2 = \lambda^2 - 10\lambda + 70$.$\lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 70 \Rightarrow 18\lambda = 54 \Rightarrow \lambda = 3$.The quadratic equation becomes $3^2 x^2 - 6(3)x + 5 = 0$,which is $9x^2 - 18x + 5 = 0$.Solving for $x$: $x = \frac{18 \pm \sqrt{324 - 180}}{18} = \frac{18 \pm \sqrt{144}}{18} = \frac{18 \pm 12}{18}$.$x = \frac{30}{18} = \frac{5}{3}$ and $x = \frac{6}{18} = \frac{1}{3}$.Since $\alpha > \beta$,we have $\alpha = \frac{5}{3}$ and $\beta = \frac{1}{3}$.Then $2\alpha - \beta = 2(\frac{5}{3}) - \frac{1}{3} = \frac{10-1}{3} = \frac{9}{3} = 3$.

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