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 The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular

to both the lines \frac{x-1}{1}= \frac{y+2}{-2}= \frac{z-4}{3}  and \frac{x-2}{2}= \frac{y+1}{-1}= \frac{z+7}{-1},is:
 

  • Option 1)

    \frac{10}{\sqrt{83}}

  • Option 2)

    \frac{5}{\sqrt{83}}

  • Option 3)

    \frac{10}{\sqrt{74}}

  • Option 4)

    \frac{20}{\sqrt{74}}

 

Answers (1)

best_answer

As we learnt in 

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

 Normal vector of Plane= \begin{bmatrix} \hat{i} & \hat{j}& \hat{k}\\ 1 & -2 & 3\\ 2 & -1 & -1 \end{bmatrix}

=5\hat{i}+7\hat{j}+3\hat{k}

Equation of plane is of the form

5x+7y+3y=C

With Point (1,-1,-1) we get

5-7-3=C=-5

So plane is 5x+7y+3y+5=0

distance from (1,3,-7) is

d=\left | \frac{5(1)+7(3)+3(-7)+5)}{\sqrt{5^{2}+7^{2}+3^{2}}} \right |=\frac{10}{\sqrt{83}}


Option 1)

\frac{10}{\sqrt{83}}

This option is correct

Option 2)

\frac{5}{\sqrt{83}}

This option is incorrect

Option 3)

\frac{10}{\sqrt{74}}

This option is incorrect

Option 4)

\frac{20}{\sqrt{74}}

This option is incorrect

Posted by

divya.saini

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