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Question Number 124131 by bramlexs22 last updated on 01/Dec/20

Answered by som(math1967) last updated on 01/Dec/20

$$\mathrm{Equation}\mathrm{of}\mathrm{plane}$$$$\left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-1& \mathrm{z}-1\\ 1-1& -1-1& 1-1\\ -7-1& 3-1& -5-1\end{array}\right|=0$$$$\left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-1& \mathrm{z}-1\\ 0& -2& 0\\ -8& 2& -6\end{array}\right|=0$$$$12(\mathrm{x}-1)-(\mathrm{y}-1)\times 0+(\mathrm{z}-1)\times (-16)=0$$$$12\mathrm{x}-16\mathrm{z}+4=0$$$$3\mathrm{x}-4\mathrm{z}+1=0$$$$\mathrm{nowangle}\mathrm{between}3\mathrm{x}-4\mathrm{z}+1=0$$$$\mathrm{andXZ}\mathrm{plane}$$$$\cos \theta =\frac{0}{\sqrt{3^2+0^2+{(-4)}^2}}=0$$$$\theta =\frac{\pi }{2}\therefore \mathrm{it}\mathrm{is}\mathrm{perpendicular}$$$$\mathrm{to}\mathrm{XZ}\mathrm{plane}$$$$$$

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