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The distance between the line \vec{r}=2\vec{i}-2\vec{j}+3\vec{k}+\lambda (\vec{i}-\vec{j}+4\vec{k}) and the plane \vec{r}\cdot (\vec{i}+5\vec{j}+\vec{k})=5  is

  • Option 1)

    \frac{10}{3\sqrt{3}}\;

  • Option 2)

    \; \; \frac{10}{9}\;

  • Option 3)

    \; \frac{10}{3}\;

  • Option 4)

    \; \frac{3}{10}

 

Answers (1)

As we learnt in 

Angle bisector containing origin -

Equation of plane bisecting the angle between the planes 

ax+by+cz+d= 0

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0 containing origin is 

\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}}= \frac{a_{1}x+b_{1}y+c_{1}z+d_{1}}{\sqrt{a^{2}_{1}+b_{1}^{2}+c_{1}^{2}}}

Where     d \: and\: d_{1}  are positive

 

-

 Equation\, \, \, of\, \, \, plane \, \, \,is\, \, \, x+5y+z=5\, \, \, and \, \, \,\frac{x-2}{1}=\frac{y+2}{-1}=\frac{z-3}{4}

Distance of (2,-2,3) from x+5y+z-5=0

is \frac{\left | 2-10+3-5 \right |}{\sqrt{1^{2}+5^{2}+1^{2}}}=\frac{10}{3\sqrt{3}}


Option 1)

\frac{10}{3\sqrt{3}}\;

Correct Option

 

Option 2)

\; \; \frac{10}{9}\;

Incorrect Option

 

Option 3)

\; \frac{10}{3}\;

Incorrect Option

 

Option 4)

\; \frac{3}{10}

Incorrect Option

 

Posted by

Vakul

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