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  • If the lines <img alt="\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx-1%7D%7B1%7D%3D%5Cfrac%7By-2%7D%7B2%7D%3D%5Cfrac%7Bz&plus;3%7D%

If the lines \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1} and  \frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1} intersect at the point P, then the distance of the point P from the plane z=a is :

Option: 1

28


Option: 2

16


Option: 3

10


Option: 4

22


Answers (1)

best_answer

\begin{aligned} & \text { Let } \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}+3}{1}=\lambda \\ & \mathrm{P}(\lambda+1,2 \lambda+2, \lambda-3) \\ & \& \frac{\mathrm{x}-\mathrm{a}}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-3}{1}=\mu \\ & \mathrm{P}(2 \mu+\mathrm{a}, 3 \mu-2, \mu+3) \end{aligned}

\begin{array}{lll} \lambda+1=2 \mu+\mathrm{a} & 2 \lambda+2=3 \mu-2 & \lambda-3=\mu+3 \\ 22+1=2 \times 16+\mathrm{a} & 2(\mu+6)+2=3 \mu-2 & \lambda=\mu+6 \\ \mathrm{a}=23-32 & 2 \mu+12=3 \mu-4 & \\ \mathrm{a}=-9 & \mu=16 & \\ & \therefore \lambda=22 & \end{array}

\therefore \mathrm{P}(23,46,19)                                   Plane is \mathrm{z}=\mathrm{a}
                                                                  \mathrm{z}=-9
The distance of \mathrm{p}$ from $\mathrm{z}=-9$ is $19-(-9)=28

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Gaurav

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