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The perpendicular distance from the point (3, 1, 1) on the plane passing through the point (1, 2, 3) and containing the line,
\vec{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}+\hat{j}+4\hat{k}),is
  • Option 1)

    \frac{3}{\sqrt{11}}

  • Option 2)

    \frac{1}{\sqrt{11}}

  • Option 3)

    \frac{4}{\sqrt{41}}

  • Option 4)

    0

 

Answers (1)

best_answer

As we learnt in 

Unit vector perpendicular to the plane of vector a and b -

\frac{\vec{a}\times \vec{b}}{\left | \vec{a}\times \vec{b} \right |}

- wherein

Here \vec{a} and \vec{b} are two vectors.

 

Plane passing through (1,2,3)

Containing line

\underset{r}{\rightarrow}=\hat{i}+\hat{j}+\lambda \left ( 2\hat{i}+\hat{j}+4\hat{k} \right )

Normal vector of plane =\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 2& 1 &4 \\ 0& 1 & 3 \end{vmatrix}

=-\hat{i}-6\hat{j}+2\hat{k}

Equation of plane

x+6y-2z=1+12-6

x+6y-2z=7

Distance = \frac{\left | 3+6-2-7 \right |}{\sqrt{1^{2}+6^{2}+2^{2}}}=0 

 


Option 1)

\frac{3}{\sqrt{11}}

This option is incorrect.

Option 2)

\frac{1}{\sqrt{11}}

This option is incorrect.

Option 3)

\frac{4}{\sqrt{41}}

This option is incorrect.

Option 4)

0

This option is correct.

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Aadil

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