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Question
Find the image P' of the point P having position vector `hati+ 3hatj+ 4hatk` in the plane `vecr. (2hati - hatj + hatk) + 3 = 0 .` Hence find the length of PP'.
Solution
Given P`(hati + 3hatj + 4hat k)` in the plane `vecr . (2hati - hatj + hat k) + 3 =0.` Then PP' is normal to the plane.
Since PP' passes through the P and is normal to the given plane. So, it is parallel to the normal vector `(2hati- hatj+hatk).`
Therefore, vector equation of line PP' is `vecr (hati +3hatj+ 4hatk)+ λ (2hati -hatj+hatk)`
As P' lies on line PP' so, let the position vector of P' be `(hati+ 3hatj +4hatk) +λ (2hati- hatj +hatk) = (1+2λ)hati +(3-λ)hatj+ (4 +λ)hatk`
Since R is the mid-point of PP'. Therefore, position vector of R is `([(1+2λ)hati+ (3-λ)hatj+(4+λ)hatk]+[hati +3hatj +4hatk])/2 = (λ+1)hati +(3-λ/2)hatj + (4+λ/2)hatk`
Clearly, R lies on the plane `vecr.(2hati-hatj+hatk) +3 =0`
\[\Rightarrow 2\lambda + 2 - 3 + \frac{\lambda}{2} + 4 + \frac{\lambda}{2} + 3 = 0\] \[ \Rightarrow \lambda = - 2\]
Putting
\[\lambda = - 2\] in (i), we obtain the position vector of P' as `(hati+ 3hatj+4hatk)-2 (2hati-hatj+ hatk)= -3hati+5hatj+ 2hatk`
The coordinates of the point corresponding to the position vector P`(hati+ 3hatj+4hatk )`will be (1, 3, 4) and for P' `(-3hati+ 5hatj +2hatk)`will be (−3, 5, 2).
Distance between (1, 3, 4) and (−3, 5, 2) will be
\[d = \sqrt{\left( - 3 - 1 \right)^2 + \left( 5 - 3 \right)^2 + \left( 2 - 4 \right)^2}\]
\[ = \sqrt{\left( - 4 \right)^2 + \left( 2 \right)^2 + \left( - 2 \right)^2}\]
\[ = \sqrt{16 + 4 + 4}\]
\[ = \sqrt{24}\]
\[ = 2\sqrt{6}\]
So, length PP' is \[2\sqrt{6}\].
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