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  • Let <img alt="\mathrm{\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx-2%7D%7B3%7D%3D%5Cfrac%7By&plus;1%7D%7B-2%7D

Let \mathrm{\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}} lie on the plane \mathrm{p x-q y+z=5} for some \mathrm{p, q \in \mathbb{R}} . The shortest distance of the plane from the origin is:

Option: 1

\mathrm{ \sqrt{\frac{3}{109}} \\ }


Option: 2

\mathrm{\sqrt{\frac{5}{142}} \\ }


Option: 3

\mathrm{\frac{5}{\sqrt{71}} }


Option: 4

\mathrm{\frac{1}{\sqrt{142}}}


Answers (1)

best_answer

For line to lie on the plane, point \mathrm{A(2,-1,-3)} lying on line should lie on plane as well

\mathrm{\therefore \quad 2 p+q-3=5 \Rightarrow 2 p+q=8}    .........(i)

Also vector parallel to line \mathrm{(3i-2j-k)} should be perpendicular to normal vector \vec{n} of plane \mathrm{\left ( pi-qj+k \right )}

\mathrm{\therefore (3 i-2 j-k) \cdot(p i-q j+k)=0} \\

\mathrm{\Rightarrow 3 p+2 q-1=0} \\

\mathrm{\Rightarrow 3 p+2 q=1 \quad \ldots . \text { (ii) }}

From (i) and (ii) 

\mathrm{p=15, q=-22}

\mathrm{\therefore \text { plane is } 15 p+22 y+z-5=0}

Distance from origin \mathrm{(0,0,0)}

\mathrm{=\frac{|-5|}{\sqrt{15^{2}+22^{2}+1^{2}}}} \\

\mathrm{=\frac{5}{\sqrt{710}}} \\

\mathrm{=\sqrt{\frac{5}{142}}}

hence answer is option 2

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