CBSE Class 12 Mathematics 2025: question 34(a)

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Step-by-step solution

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1. Write the line as $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. A general point on it is $M(\lambda ,2\lambda +1,3\lambda +2)$. Take $M$ as the foot of the perpendicular from $A(1,6,3)$.

   1 mark

2. Direction ratios of $AM$ are $(\lambda -1,2\lambda -5,3\lambda -1)$. Since $AM\perp$ the line, whose d.r.'s are $(1,2,3)$:
   $1(\lambda -1)+2(2\lambda -5)+3(3\lambda -1)=0$

   1 mark

3. $14\lambda -14=0\Rightarrow \lambda =1$, so the foot of the perpendicular is $M(1,3,5)$.

   ½ mark

4. Let the image be $A^{\prime }(\alpha ,\beta ,\gamma )$. $M$ is the mid-point of $A{A}^{\prime }$:
   $\left(\frac{1+\alpha }{2},\frac{6+\beta }{2},\frac{3+\gamma }{2}\right)=(1,3,5)$

   ½ mark

5. $\alpha =1,\beta =0,\gamma =7$, so the image is $A^{\prime }(1,0,7)$.

   1 mark

6. D.r.'s of $A{A}^{\prime }$: $(1-1,0-6,7-3)=(0,-6,4)$, i.e. proportional to $(0,-3,2)$. The line through $A(1,6,3)$ is
   $\frac{x-1}{0}=\frac{y-6}{-3}=\frac{z-3}{2}$

   1 mark

Answer: $A^{\prime }(1,0,7)$; line $A{A}^{\prime }$: $\frac{x-1}{0}=\frac{y-6}{-3}=\frac{z-3}{2}$

Revise the chapter

• Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
• Topic: Foot of Perpendicular and Image of a Point