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  • Let the image of the point <img alt="\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5Cfrac%7B5%7D%7B3%7D%2C%20%5Cfrac%7B5%7

Let the image of the point \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right) in the plane \mathrm{x}-2 \mathrm{y}+\mathrm{z}-2=0 be  \mathrm{P}. If the distance of the point Q(6,-2$, $\alpha), \alpha>0, from  \mathrm{P} is 13 , then \alpha is equal to___________
 

Option: 1

15


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Image of point }\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right) \\ \end{aligned}

\begin{aligned} & \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{z-\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2} \\ \end{aligned}

                                                                                    =\frac{1}{3}

\begin{aligned} & \begin{array}{l} \therefore x=2, y=1, z=3 \\ 13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \\ \Rightarrow(\alpha-3)^2=144 \Rightarrow \alpha=15(\because \alpha>0) \end{array} \end{aligned}

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manish painkra

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