CBSE Class 12 Mathematics 2023: question 34(b)

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Step-by-step solution

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1. Direction ratios of $AB$: $3-1,5-2,9-3=2,3,6$. The equation of line $AB$ is $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{6}$, or in vector form $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$.

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2. Any point on this line is $P(2\lambda +1,3\lambda +2,6\lambda +3)$ for some $\lambda$.

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3. Given $PB=14$ with $B(3,5,9)$: $(2\lambda -2{)}^2+(3\lambda -3{)}^2+(6\lambda -6{)}^2=14^2$, so $49(\lambda -1{)}^2=196$, which gives ${\lambda }^2-2\lambda -3=0$.

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4. $(\lambda -3)(\lambda +1)=0$, so $\lambda =3$ or $\lambda =-1$.

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5. $\lambda =3$ gives $(7,11,21)$ and $\lambda =-1$ gives $(-1,-1,-3)$. These are the required points.

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Answer: Line: $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{6}$; points $(7,11,21)$ and $(-1,-1,-3)$

Revise the chapter

• Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
• Topic: Equation of a Line in Space