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If P_{1}:\vec{r}\cdot \vec{n}_{1}-d_{1}=0,P_{2}:\vec{r}\cdot \vec{n}_{2}-d_{2}=0,P_{3}:\vec{r}\cdot \vec{n}_{3}-d_{3}=0 are three planes and \vec{n}_{1},\vec{n}_{2} and \vec{n}_{3} are three non-coplanar vectors then, the three lines P_{1}=0,P_{2}=0; P_{2}=0,P_{3}=0; and P_{3}=0,P_{1}=0 are

Option: 1

Parallel lines


Option: 2

coplanar lines


Option: 3

coincident lines


Option: 4

concurrent lines


Answers (1)

best_answer

 

 

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

 

 

P_{1}=P_{2}=0,P_{2}=P_{3}=0 and P_{3}=P_{1}=0 are lines of intersection of the three planes P_{1},P_{2} and P_{3}. As \vec{n}_{1},\vec{n}_{2} and \vec{n}_{3} are non-coplanar , planes P_{1},P_{2} and P_{3} will intersect at unique point. So the given lines will pass through a fixed point.

Posted by

Devendra Khairwa

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