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Let the line L pass through the point (0, 1, 2), intersect the line \mathrm{\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}} and be parallel to the plane \mathrm{2x + y -3z = 4}. Then the distance of the point \mathrm{ P(1, -9, 2)} from the line L is :

Option: 1

9


Option: 2

\sqrt{54}


Option: 3

\sqrt{69}


Option: 4

\sqrt{74}


Answers (1)

best_answer

\begin{aligned} & \overrightarrow{\mathrm{PQ}}=(2 \lambda+1,3 \lambda+1,4 \lambda+1) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{n}}=0 \quad \Rightarrow(2 \lambda+1) \cdot(2)+(3 \lambda+1)(1)+(4 \lambda+1)(-3)=0 \\ & \Rightarrow-5 \lambda=0 \\ & \Rightarrow \lambda=0 \\ & Q=(1,2,3) \\ & \mathrm{eq}^{\mathrm{n}} \text { of line } \\ & \end{aligned}

\qquad \frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu \\
\quad \text { distance of line from }(1,-9,2) \\
\quad\left(P^{\prime} \mathrm{Q}\right) \cdot(1,1,1)=0 \\
\Rightarrow[\mu-1, \mu+10, \mu] \cdot[1,1,1]=0
\Rightarrow \mu-1+\mu+10+\mu=0 \\
\mu=-3 \\
Q^{\prime}=(-3,-2,1) \\
P^{\prime} Q^{\prime}=\sqrt{16+49+9}=\sqrt{74}

Posted by

Rakesh

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