CBSE Class 12 Mathematics 2024: question 34(a)

Hints

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Step-by-step solution

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1. Write the line in standard form: $\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}$. It passes through $(4,0,1)$ and has direction ratios $-2,6,-3$.

   ½ mark

2. Let $B$ be the foot of the perpendicular from $A(2,3,-8)$. A general point on the line gives $B(-2\lambda +4,6\lambda ,-3\lambda +1)$.

   ½ mark

3. Direction ratios of $AB$: $(-2\lambda +2,6\lambda -3,-3\lambda +9)$.

   1 mark

4. Since $AB\perp$ the line: $-2(-2\lambda +2)+6(6\lambda -3)-3(-3\lambda +9)=0$.

   This gives $4\lambda -4+36\lambda -18+9\lambda -27=0$, so $49\lambda =49$ and $\lambda =1$.

   1 mark

5. Substituting $\lambda =1$, the foot of the perpendicular is $B(2,6,-2)$.

   1 mark

6. Perpendicular distance:

   $AB=\sqrt{(2-2{)}^2+(6-3{)}^2+(-2+8{)}^2}=\sqrt{0+9+36}=\sqrt{45}=3\sqrt{5}$ units.

   1 mark

Answer: Foot of perpendicular $(2,6,-2)$; perpendicular distance $3\sqrt{5}$ units

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