generic · 12th TN - English Medium · MATHEMATICS-VOLUME 1 · Page 281question

6.9 Image of a Point in a Plane

Chapter 6: Chapter 6 · MATHEMATICS-VOLUME 1 · EN medium

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Let A be the given point whose position vector is u  . Let r n be the equation of the plane. Let v  be the position vector of the mirror image A ′ of A in the plane. Then ' AA is perpendicular to the plane. So it is parallel to n  . Then AA ′ =  or v u  ⇒ v u ... ( ) Let M be the middle point of AA ′ . Then the position vector of M is u v . But M lies on the plane. So, u v  ⋅ ... ( ) Sustituting ( ) in ( ), we get u u ⋅   ⇒ u Therefore, the position vector of A ′ is [ )] u n v u Note The mid point of M of AA ′ is the foot of the perpendicular from the point A to the plane r n . So the position vector of the foot M of the perpendicular is given by .

📖 Class 12 Mathematics English Volume 1 2025 Edition www.tntextbooks.in · Page 281

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Let A be the given point whose position vector is u  . Let r n be the equation of the plane. Let v  be the position vector of the mirror image A ′ of A in the plane. Then ' AA is perpendicular to the plane.

So it is parallel to n  . Then AA ′ =  or v u  ⇒ v u ... ( ) Let M be the middle point of AA ′ . Then the position vector of M is u v .

But M lies on the plane. So, u v  ⋅ ... ( ) Sustituting ( ) in ( ), we get u u ⋅   ⇒ u Therefore, the position vector of A ′ is [ )] u n v u Note The mid point of M of AA ′ is the foot of the perpendicular from the point A to the plane r n . So the position vector of the foot M of the perpendicular is given by .

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