If the length of the perpendicular drawn from the point on the line is units and is the image of the point P on this line, then is equal to

Question

If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to

Smart Achievers · Accepted Answer

(8)

$Given line \frac{X + 1}{2} = \frac{Y - 3}{3} = \frac{Z - 1}{- 1} = λ$

$X = 2 λ - 1, Y = 3 λ + 3, Z = - λ + 1$

$a + \sum_{i = 1}^{3} α_{i}$

$\Rightarrow 4 λ - 2 - 2 a + 9 λ - 3 + λ + 1 = 0$

$14 λ - 4 - 2 a = 0$ $\Rightarrow 7 λ - 2 - a = 0$

$\Rightarrow {(5 λ - 1)}^{2} + {(3 λ - 1)}^{2} + {(λ - 1)}^{2} = 24$

$35 λ^{2} - 14 λ - 21 = 0$ $\Rightarrow (λ - 1) (35 λ + 21) = 0$

$For, λ = 1 \Rightarrow a = 5$

$Let (α_{1}, α_{2}, α_{3}) be reflection point P$

$α_{1} + 5, α_{2} + 4 = 12, α_{3} + 2 = 0$

$α_{1} = - 3, α_{2} = 8, α_{3} = - 2$

$α_{1} + α_{2} + α_{3} = 8$

Solution

Q.
If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to

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Solution

Q. If the length of the perpendicular drawn from the point (a,4,2), a>0 on the line x+12=y-33=z-1-1 is 26 units and Q(α1,α2,α3) is the image of the point P on this line, then a+∑i=13αi is equal to

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Q.
If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to