The image of the point (1, 6, 3) with respect to the line \( \displaystyle\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (1, 0, 7)
A general point of the line is M(λ, 1 + 2λ, 2 + 3λ). For M to be the foot of the perpendicular from P(1, 6, 3), PM must be perpendicular to the direction (1, 2, 3):
\( \displaystyle \overrightarrow{PM}=(\lambda-1,\ 2\lambda-5,\ 3\lambda-1) \)
\( \displaystyle (\lambda-1)\cdot 1+(2\lambda-5)\cdot 2+(3\lambda-1)\cdot 3=0 \)
\( \displaystyle 14\lambda-14=0\ \Rightarrow\ \lambda=1 \)
So the foot is M(1, 3, 5). M is the mid-point of P and its image Q(x, y, z):
\( \displaystyle \frac{1+x}{2}=1\ \Rightarrow\ x=1 \)
\( \displaystyle \frac{6+y}{2}=3\ \Rightarrow\ y=0 \)
\( \displaystyle \frac{3+z}{2}=5\ \Rightarrow\ z=7 \)
So the image is Q(1, 0, 7).
Shortcut: For each option, the mid-point with P must lie on the line and PQ must be perpendicular to (1, 2, 3). For (1, 0, 7): mid-point (1, 3, 5) satisfies \( \frac{1}{1}=\frac{3-1}{2}=\frac{5-2}{3} \), and PQ = (0, −6, 4) gives 0 − 12 + 12 = 0.
NCERT Exemplar:
Three Dimensional Geometry
