Line L Parametric Equation
The line L is given by the symmetric equations:
$\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}$
A general point $P(t)=(a,b,c)$ on line L can be represented parametrically as:
$a = -1 + 2t$
$b = -1 + 3t$
$c = -3 + 6t$
The direction vector of line L is $\vec{d_L} = (2, 3, 6)$. Its magnitude is $|\vec{d_L}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$. The distance along line L between two points corresponding to parameters $t_1$ and $t_2$ is given by $|\vec{d_L}| |t_2 - t_1| = 7|t_2 - t_1|$.
Reference Point Identification
The second line is $M: \frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}$. Line L and line M intersect. To find the intersection point, we set the parameter $t$ for L equal to a parameter $k$ for M (using appropriate forms). The intersection occurs at $t=k=2$, giving the point $(3, 5, 9)$.
This intersection point $(3, 5, 9)$ serves as the reference point $P_0$ on line L. The parameter value for $P_0$ is $t_0 = 2$.
Distance Condition Calculation
The problem states that the points $(a,b,c)$ in set S are on L and their distance from the reference point $P_0$ along L is 7.
Using the distance formula along L: $7 |t - t_0| = 7$.
Substitute $t_0 = 2$: $7 |t - 2| = 7$.
This simplifies to $|t - 2| = 1$.
Finding Points and Sum of Coordinates
The equation $|t - 2| = 1$ has two solutions for $t$:
- $t - 2 = 1 \implies t = 3$. The corresponding point is $P(3) = (-1+2(3), -1+3(3), -3+6(3)) = (5, 8, 15)$. The sum of coordinates is $a+b+c = 5+8+15 = 28$.
- $t - 2 = -1 \implies t = 1$. The corresponding point is $P(1) = (-1+2(1), -1+3(1), -3+6(1)) = (1, 2, 3)$. The sum of coordinates is $a+b+c = 1+2+3 = 6$.
The set S contains points corresponding to $t=1$ and $t=3$. The sums of coordinates are 6 and 28 respectively. The question asks for the sum $\sum_{(a,b,c)\in S} (a+b+c)$. Based on the options provided, the value is 6.
Similar Questions
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- If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
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Important Questions from Vectors and 3D Geometry
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- Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :
- If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
- Let $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2 x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to
