If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a is equal to :

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If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a is equal to :

The correct answer is

Distance: Point Image in Line

We are given a point $Q(a, 2, 5)$ and a line $L: \frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$. We need to find the distance between $Q$ and the image $P'$ of a point $P(1, 2, 7)$ in the line $L$. This distance is given as 4. Our goal is to find the sum of all possible values of $a$.

Line Parametric Equation

The line $L$ can be represented parametrically:

$x = t$
$y = 1+t$
$z = 2+2t$

The direction vector of the line is $\vec{d} = (1, 1, 2)$.

Foot of Perpendicular Calculation

Let $F(t, 1+t, 2+2t)$ be the foot of the perpendicular from $P(1, 2, 7)$ to the line $L$. The vector $\vec{PF}$ is:

$\vec{PF} = (t-1, (1+t)-2, (2+2t)-7) = (t-1, t-1, 2t-5)$

Since $\vec{PF}$ is perpendicular to the line's direction vector $\vec{d}$, their dot product is zero:

$\vec{PF} \cdot \vec{d} = (t-1)(1) + (t-1)(1) + (2t-5)(2) = 0$

$t-1 + t-1 + 4t-10 = 0$

$6t - 12 = 0 \implies t = 2$

Substituting $t=2$ gives the coordinates of the foot $F$:

$F = (2, 1+2, 2+2(2)) = (2, 3, 6)$

Image Point Determination

The foot $F(2, 3, 6)$ is the midpoint of the segment $PP'$, where $P(1, 2, 7)$ and $P'(x', y', z')$ is the image point.

Using the midpoint formula:

$\frac{1+x'}{2} = 2 \implies 1+x' = 4 \implies x' = 3$
$\frac{2+y'}{2} = 3 \implies 2+y' = 6 \implies y' = 4$
$\frac{7+z'}{2} = 6 \implies 7+z' = 12 \implies z' = 5$

Therefore, the image point is $P'(3, 4, 5)$.

Distance Condition Application

The distance between the point $Q(a, 2, 5)$ and the image point $P'(3, 4, 5)$ is given as 4.

Using the distance formula, $Distance^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$:

$4^2 = (a-3)^2 + (2-4)^2 + (5-5)^2$

$16 = (a-3)^2 + (-2)^2 + 0^2$

$16 = (a-3)^2 + 4$

Sum of 'a' Values Calculation

Rearranging the equation:

$(a-3)^2 = 16 - 4$

$(a-3)^2 = 12$

Taking the square root of both sides:

$a-3 = \pm\sqrt{12} = \pm 2\sqrt{3}$

The possible values for $a$ are:

$a_1 = 3 + 2\sqrt{3}$
$a_2 = 3 - 2\sqrt{3}$

The sum of all possible values of $a$ is:

$a_1 + a_2 = (3 + 2\sqrt{3}) + (3 - 2\sqrt{3}) = 6$

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Important Questions from Vectors and 3D Geometry

Let a vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}, \lambda > 0$, make an obtuse angle with the vector $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle $\theta, \frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta) - \{\gamma\}$, then $\alpha + \beta + \gamma$ is equal to ________.
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