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  • If the foot of the perpendicular from the point on the plane <img alt="\mathrm{\mathrm{P}: 2 x+\mathrm{m} y+\ma

If the foot of the perpendicular from the point \mathrm{A}(-1,4,3) on the plane \mathrm{\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4}, is \mathrm{\left(-2, \frac{7}{2}, \frac{3}{2}\right)}, then the distance of the point \mathrm{A} from the plane \mathrm{P}, measured parallel to a line with direction ratios 3,-1,-4, is equal to :
 

Option: 1

1


Option: 2

\sqrt{26}


Option: 3

2 \sqrt{2}


Option: 4

\sqrt{14}


Answers (1)

best_answer

\begin{aligned} &\text{Let B be foot of }\perp\text{ coordinates of }\mathrm{B=\left(-2, \frac{7}{2}, \frac{3}{2}\right)}\\ &\text{Direction ratio of line AB is } \left \langle 2,1,3 \right \rangle \text{So}\\ &\mathrm{m=1, n=3}\\ &\text{So equation of AC is } \mathrm {\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}=\lambda}\\ &\text{So point c is } \mathrm{(3 \lambda-1,-1+9,-4, \lambda+3)}. \text{but c lies on the plane, so}\\ &\mathrm{6 \lambda-2-\lambda-1-12 \lambda+9=4} \\ &\mathrm{\Rightarrow \lambda=1 \Rightarrow C(2,3,-1)} \\ & \mathrm{\Rightarrow \quad A C=\sqrt{26}} \end{aligned}

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