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Vector equation of plane passing through point with position vector \hat{i}+\hat{j}+\hat{k} and normal to  a given vector 2\hat{i}-\hat{j}+\hat{k} is

  • Option 1)

    (\vec{r}-\hat{i}-\hat{j}-\hat{k})\cdot(2\hat{i}-\hat{j}+\hat{k})=0(\vec{r}-\hat{i}-\hat{j}-\hat{k})\cdot(2\hat{i}-\hat{j}+\hat{k})=0

  • Option 2)

    (\vec{r}-\hat{i}+\hat{j}+\hat{k})\cdot(2\hat{i}+\hat{j}-\hat{k})=0

  • Option 3)

    (\vec{r}-\hat{i}-\hat{j}+\hat{k})\cdot(2\hat{i}-\hat{j}-\hat{k})=0

  • Option 4)

    (\vec{r}+\hat{i}-\hat{j}+\hat{k})\cdot(2\hat{i}+\hat{j}+\hat{k})=0

 

Answers (1)

best_answer

As we have learned 

Vector equation of plane passing through a given point and normal to a given vector -

\left ( r-\vec{r_{0}} \right )\cdot \vec{n}= 0

- wherein

Let a plane be passing through P_{0}\left ( \vec{r_{0}} \right )and normal to \vec{n},P\left ( \vec{r} \right )be any point on plane.

Now\vec{P_{0}P} will be perpendicular to \vec{r}

\vec{P_{0}P}= \left ( \vec{r}-\vec{r_{0}} \right )

\left ( \vec{r}-\vec{r_{0}} \right )\cdot \vec{n}= 0

 

 Vector equation is (\vec{r}-\vec{r_0})\cdot\vec{n}=0

option A


Option 1)

(\vec{r}-\hat{i}-\hat{j}-\hat{k})\cdot(2\hat{i}-\hat{j}+\hat{k})=0(\vec{r}-\hat{i}-\hat{j}-\hat{k})\cdot(2\hat{i}-\hat{j}+\hat{k})=0

Option 2)

(\vec{r}-\hat{i}+\hat{j}+\hat{k})\cdot(2\hat{i}+\hat{j}-\hat{k})=0

Option 3)

(\vec{r}-\hat{i}-\hat{j}+\hat{k})\cdot(2\hat{i}-\hat{j}-\hat{k})=0

Option 4)

(\vec{r}+\hat{i}-\hat{j}+\hat{k})\cdot(2\hat{i}+\hat{j}+\hat{k})=0

Posted by

Himanshu

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