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Question
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
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Solution
Let the given point be P`(hati + 3hatj + 4hatk)` and Q be the image of P in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0 as shown in the figure.
Then PQ is the normal to the plane.
Since PQ passes through P and is normal to the given plane
So the equation of PQ is given by `vecr = (hati + 3hatj + 4hatk) + lambda(2hati - hatj + hatk)`
Since Q lies on the line PQ, the position vector of Q can be expressed as `(hati + 3hatj + 4hatk) + lambda(2hati - hatj + hatk)`
i.e., `(1 + 2lambda)hati + (3 - lambda)hatj + (4 + lambda)hatk`
Since R is the mid point of PQ, the position vector of R is
`([(1 + 2lambda)hati + (3 - lambda)hatj + (4 + lambda)hatk] + [hati + 3hatj + 4hatk])/2`
i.e., `(lambda + 1)hati + (3 - lambda/2)hatj + (4 + lambda/2)hatk`
Again, since R lies on the plane `vecr * (2hati - hatj + hatk)` + 3 = 0
We have `{(lambda + 1)hati + (3 - lambda/2)hatj + (4 + lambda/2)hatk} * (2hati - hatj + hatk) + 3` = 0
⇒ λ = –2
Hence, the position vector of Q is `(hati + 3hatj + 4hatk) - 2(2hati - hatj + hatk)`
i.e. `-3hati + 5hatj + 2hatk`.
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