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Question 57 of 59

Q.Find the image of the point (1,6,3) with respect to the line x/1 = (y-1)/2 = (z-2)/3, and also find the equation of the line passing through the point and its image. (3+2)

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 5mImportance★★★★★
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Find the foot of the perpendicular from PP to the line first — it is the midpoint of PP and its image.

Write the line x1=y−12=z−23=t\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=t in parametric form: a general point is Q=(t, 1+2t, 2+3t)Q=(t,\,1+2t,\,2+3t), with direction vector d⃗=(1,2,3)\vec d=(1,2,3).

Step 1 — find the foot of perpendicular NN from P(1,6,3)P(1,6,3): the vector PQ⃗\vec{PQ} must be perpendicular to d⃗\vec d:

PQ⃗=(t−1, 2t−5, 3t−1)\vec{PQ} = (t-1,\ 2t-5,\ 3t-1)

PQ⃗⋅d⃗=(t−1)(1)+(2t−5)(2)+(3t−1)(3)=14t−14=0  ⇒  t=1\vec{PQ}\cdot\vec d = (t-1)(1)+(2t-5)(2)+(3t-1)(3) = 14t-14=0 \;\Rightarrow\; t=1

So N=(1,3,5)N=(1,3,5).

Step 2 — find the image P′P': since NN is the midpoint of PP and its image P′P': P′=2N−P=(2−1, 6−6, 10−3)=(1,0,7)P' = 2N-P = (2-1,\,6-6,\,10-3) = (1,0,7).

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