JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 11)
If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$ is equal to :
7
8
12
14
Explanation
$$\because$$ PR is perpendicular to given line, so
_27th_July_Evening_Shift_en_11_1.png)
$$2(2\lambda - 1 - a) + 3(3\lambda - 1) - 1( - \lambda - 1) = 0$$
$$ \Rightarrow a = 7\lambda - 2$$
Now,
$$\because$$ $$PR = 2\sqrt 6 $$
$$ \Rightarrow {( - 5\lambda + 1)^2} + {(3\lambda - 1)^2} + {(\lambda + 1)^2} = 24$$
$$ \Rightarrow 5{\lambda ^2} - 2\lambda - 3 = 0 \Rightarrow \lambda = 1$$ or $$ - {3 \over 5}$$
$$\because$$ $$a > 0$$ so $$\lambda = 1$$ and $$a = 5$$
Now $$\sum\limits_{i = 1}^3 {{\alpha _i} = 2} $$ (Sum of co-ordinate of R) $$-$$ (Sum of coordinates of P)
$$ = 2(7) - 11 = 3$$
$$a + \sum\limits_{i = 1}^3 {{\alpha _i} = 5 + 3 = 8} $$
Comments (0)
