If P(2, beta, alpha) lies on the plane x+2y-z | vedclass

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(A) Given $P(2, \beta, \alpha)$ lies on $x+2y-z-2=0$,we have $2+2\beta-\alpha-2=0$,which simplifies to $\alpha=2\beta$ $(i)$.Given $Q(\alpha, -1, \bet

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$\left(-\frac{4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right)$

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(A) Given $P(2, \beta, \alpha)$ lies on $x+2y-z-2=0$,we have $2+2\beta-\alpha-2=0$,which simplifies to $\alpha=2\beta$ $(i)$.Given $Q(\alpha, -1, \beta)$ lies on $2x-y+3z+6=0$,we have $2\alpha - (-1) + 3\beta + 6 = 0$,which simplifies to $2\alpha+3\beta+7=0$ $(ii)$.Substituting $(i)$ into $(ii)$: $2(2\beta)+3\beta+7=0 \Rightarrow 7\beta = -7 \Rightarrow \beta = -1$.Then $\alpha = 2(-1) = -2$.Thus,$P = (2, -1, -2)$ and $Q = (-2, -1, -1)$.The vector $\vec{PQ} = (-2-2)\hat{i} + (-1-(-1))\hat{j} + (-1-(-2))\hat{k} = -4\hat{i} + 0\hat{j} + 1\hat{k}$.The magnitude $|\vec{PQ}| = \sqrt{(-4)^2 + 0^2 + 1^2} = \sqrt{16+1} = \sqrt{17}$.The direction cosines are $\left(\frac{-4}{\sqrt{17}}, \frac{0}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right) = \left(-\frac{4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right)$.

If $P(2, \beta, \alpha)$ lies on the plane $x+2y-z-2=0$ and $Q(\alpha, -1, \beta)$ lies on the plane $2x-y+3z+6=0$,then the direction cosines of the line $PQ$ are:

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