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Question Number 75317 by vishalbhardwaj last updated on 10/Dec/19

$$\mathrm{Find}\mathrm{the}\mathrm{image}\mathrm{of}\mathrm{the}\mathrm{line}$$$$\frac{x-2}{-1}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-4}{3}\mathrm{in}\mathrm{the}\mathrm{plane}$$$$2x+3y-4z+3=0.$$

Commented by vishalbhardwaj last updated on 10/Dec/19

$$\mathrm{yes}\mathrm{sir},\mathrm{I}\mathrm{corrected}\mathrm{it}$$

Answered by vishalbhardwaj last updated on 10/Dec/19

$$\mathrm{please}\mathrm{solve}\mathrm{this}$$

Answered by mr W last updated on 10/Dec/19

$$line$$$$\frac{x-2}{-1}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-4}{3}=\lambda$$$$\Rightarrow x=-\lambda +2$$$$\Rightarrow y=2\lambda +3$$$$\Rightarrow z=3\lambda +4$$$$using\lambda =1wegetapointPonthe$$$$lineP(1,5,7).$$$$saytheimageofpointPinthe$$$$planeis{P}^{\prime }(u,v,w)$$$$(x,y,z)=(1,5,7)+k(2,3,-4)$$$$2(1+2k)+3(5+3k)-4(7-4k)+3=0$$$$\Rightarrow k=\frac{8}{3}$$$$(u,v,w)=(1,5,7)+\frac{16}{3}(2,3,-4)=(\frac{35}{3},21,\frac{-43}{3})$$$$$$$$intersectionpointQoflineandplane:$$$$2(-\lambda +2)+3(2\lambda +3)-4(3\lambda +4)+3=0$$$$\Rightarrow \lambda =0$$$$\Rightarrow Q(2,3,4)$$$$theimageoflineinplaneisthe$$$$line{P}^{\prime }Q:$$$$\frac{x-2}{\frac{35}{3}-2}=\frac{y-3}{21-3}=\frac{z-4}{-\frac{43}{3}-4}$$$$\Rightarrow \frac{x-2}{\frac{29}{3}}=\frac{y-3}{18}=\frac{z-4}{-\frac{55}{3}}$$

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