CBSE Class 12 Mathematics 2025: question 34(b)

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Step-by-step solution

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1. Write the line as $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda$. Any point on it is $P(\lambda -5,4\lambda -3,-9\lambda +6)$, and $Q$ is $(2,4,-1)$.

   1 mark

2. Since $PQ=7$:
   $\sqrt{(\lambda -7{)}^2+(4\lambda -7{)}^2+(-9\lambda +7{)}^2}=7$

   1 mark

3. Squaring gives $98{\lambda }^2-196\lambda +147=49$.
   So $98({\lambda }^2-2\lambda +1)=0\Rightarrow (\lambda -1{)}^2=0\Rightarrow \lambda =1$.

   1 mark

4. Hence the required point is $P(-4,1,-3)$.

   1 mark

5. D.r.'s of $PQ$: $(2-(-4),4-1,-1-(-3))=(6,3,2)$. The line through $P$ is
   $\frac{x+4}{6}=\frac{y-1}{3}=\frac{z+3}{2}$
   (equivalently, through $Q$: $\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$).

   1 mark

Answer: $P(-4,1,-3)$; line $PQ$: $\frac{x+4}{6}=\frac{y-1}{3}=\frac{z+3}{2}$

Revise the chapter

• Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)