ParabolaMedium

Question

Let the point A be the foot of perpendicular drawn from the point $P(a,b,0)$ on the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}$. If the midpoint of the line segment PA is $\left( 0,\frac{3}{4},\frac{- 1}{4} \right)$, then the value of $a^{2} + b^{2} + \alpha^{2}$ is equal to :

Options

A.1
B.2
C.6
D.9

Solution

Sol. [Key Concept: Foot on the line plus the midpoint condition give $a,b,\alpha$; then $a^{2} + b^{2} + \alpha^{2}$.]

$${\frac{2r + 1 + a}{2} = 0 \Rightarrow 2r + a = - 1 }{\frac{r + 2 + b}{2} = \frac{3}{4} \Rightarrow r + b = - \frac{1}{2} }$$

$${\frac{3r + \alpha}{2} = \frac{- 1}{4} \Rightarrow 3r + \alpha = \frac{- 1}{2} }{2 \cdot a + 1 \cdot (\text{ }b - 3/4) + 3 \cdot \frac{1}{4} = 0 \Rightarrow a + b = 0 }{\Rightarrow 2( - 1 - 2r) + ( - r - 1/2) = 0 }{\Rightarrow - 3/2 = 3r = 1 \Rightarrow r = - 1/2,a = 0,\text{ }b = 0,\alpha = 1 }{\therefore a^{2} + b^{2} + \alpha^{2} = 1 }$$

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