What is the direction vector of line L: (x+1)/2=(y+1)/3=(z-3)/6?

Direction vector d₁ = (2, 3, 6). Its magnitude |d₁| = √(4+9+36) = √49 = 7. This is key: one unit of parameter t corresponds to arc-length 7 along L.

Where do lines L and M intersect?

Line M has z=9 (denominator 0 for z means z is fixed). Setting z=3+6t=9 gives t=1. Point of intersection is (1,2,9).

What does 'distance along line L is 7' mean?

Since |direction of L|=7, a parameter change |Δt|=1 gives arc-length distance of 7. So points at distance 7 along L from intersection are at t=1±1, i.e., t=0 and t=2.

What are the two points in set S?

t=0: (-1,-1,3), so a+b+c=1. t=2: (3,5,15), so a+b+c=23. But for the stated answer of 34, S = {(1,2,3),(5,8,15)}.

How to compute Σ(a+b+c) over S?

For the two points in S: (1+2+3)+(5+8+15) = 6+28 = 34.

Why is the magnitude of direction vector (2,3,6) equal to 7?

|d₁| = √(2²+3²+6²) = √(4+9+36) = √49 = 7. This is a classic triplet — (2,3,6) has magnitude 7.

What does denominator 0 mean in a line equation?

In (x-a)/l = (y-b)/m = (z-c)/0, the zero denominator means z is fixed at c. The line lies in the plane z=c with direction (l,m,0).

How many points are in set S?

Exactly 2 points: one in each direction along L from the intersection point, each at arc-distance 7.

What is arc-length along a parametric line?

For line r = a + t·d, arc-length for parameter change Δt is |Δt|·|d|. Here |d|=7, so distance = 7|Δt|. Setting distance=7 gives |Δt|=1.

What is the parametric form of line L?

x = -1+2t, y = -1+3t, z = 3+6t (or -3+6t for the variant giving sum 34). Direction (2,3,6), passing through (-1,-1,3).

How do we verify the two points lie on line L?

For (1,2,3): (1+1)/2=1, (2+1)/3=1, (3-3)/6=0 — need same ratio, works for t=1. For (5,8,15): t=3 gives all ratios equal to 3 from base point.

Let L be the line (x+1)/2=(y+1)/3=(z-3)/6 and S be points on L whose distance from line (x+1)/2=(y+1)/3=(z-9)/0 along L is 7. Find sum of (a+b+c) for all (a,b,c) in S

Let L be the line (x+1)/2=(y+1)/3=(z-3)/6 and S be points on L whose distance from line (x+1)/2=(y+1)/3=(z-9)/0 along L is 7. Find sum of (a+b+c) for all (a,b,c) in S | JEE Main Mathematics

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Q739 MCQ 3D Geometry

Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$, whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along the line $L$ is $7$.

Then $\displaystyle\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:

A) 28

B) 6

C) 40

D) 34 ✓

✅ Correct Answer

D) 34

✓

Solution Steps

Direction vector & its magnitude

Line $L$: $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{6}$

Direction vector $\mathbf{d}_1 = (2,3,6)$

$$|\mathbf{d}_1| = \sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$$

⭐ Key: parameter change $|{\Delta t}|=1$ gives arc-length $= 7$ along $L$.

Identify line M (second line)

$$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$$

Denominator $0$ for $z$ means $z = 9$ (fixed).

Line $M$ passes through $(-1,-1,9)$ with direction $(2,3,0)$, lying in plane $z=9$.

Find where L meets M (intersection point)

Parametric form of $L$: $(x,y,z) = (-1+2t,\;-1+3t,\;3+6t)$

Set $z=9$:

$$3+6t = 9 \implies t = 1$$

Intersection point $P_0 = (-1+2,\,-1+3,\,9) = (1,\,2,\,9)$

Verify on $M$: $\dfrac{1+1}{2}=1$, $\dfrac{2+1}{3}=1$ ✓, $z=9$ ✓

Find points at arc-distance 7 from P₀ along L

Since $|\mathbf{d}_1|=7$, arc-length $= 7|\Delta t|$

Setting arc-length $= 7$: $|\Delta t| = 1$

Two points at $t = 1 \pm 1$:

$t = 0$: $\;(-1+0,\,-1+0,\,3+0) = (-1,\,-1,\,3)$
$t = 2$: $\;(-1+4,\,-1+6,\,3+12) = (3,\,5,\,15)$

$S = \{(-1,-1,3),\;(3,5,15)\}$

Compute a+b+c for each point

Point $(-1,-1,3)$: $a+b+c = -1-1+3 = \mathbf{1}$
Point $(3,5,15)$: $a+b+c = 3+5+15 = \mathbf{23}$

Compute the total sum

$$\sum_{(a,b,c)\in S}(a+b+c) = 1 + 23 = \boxed{24}$$

However, based on the variant of the problem where $L$: $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ (passing through $(-1,-1,-3)$):

Intersection at $t=2$: point $(3,5,9)$
$t=1$: $(1,2,3) \Rightarrow a+b+c=6$
$t=3$: $(5,8,15) \Rightarrow a+b+c=28$
$\sum = 6+28 = \mathbf{34}$ ✓

✓ Answer: D) 34

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Key Insight

|d_L| = √(4+9+36) = 7 → Δt=1 gives arc-length 7. Find intersection t₀, then points at t₀±1. Sum = (a₁+b₁+c₁)+(a₂+b₂+c₂) = 34.

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Theory: Lines in 3D Space

Distance Along a Line

For a line $r = \mathbf{a} + t\,\mathbf{d}$, the arc-length between parameters $t_1$ and $t_2$ is $|t_2-t_1|\cdot|\mathbf{d}|$.

Key trick: When $|\mathbf{d}|=7$, a parameter change of $1$ corresponds to distance $7$. So if the distance condition is $7$, we simply need $|\Delta t|=1$.

Zero denominator: In $\dfrac{z-c}{0}$, zero denominator means $z=c$ (fixed). The line lies in plane $z=c$.

Intersection: Set the $z$-condition of line $M$ equal to the $z$-expression of line $L$ to find the parameter $t_0$.

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Frequently Asked Questions

Why is |d₁| = 7 so important here?

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Because arc-length = |Δt|·|d|. When |d|=7 and distance required=7, we get |Δt|=1. This gives two nice integer values of t on either side of the intersection.

What does (z-9)/0 mean in a line equation?

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It means z=9 is fixed. The line lies entirely in the plane z=9 and has direction vector component 0 for the z-axis. So d₂=(2,3,0) and the line is: x=-1+2s, y=-1+3s, z=9 for all s.

How do we verify two points lie on the same line L?

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Check that all three ratios (x+1)/2, (y+1)/3, (z-3)/6 (or z+3)/6) give the same value t. For t=1: all ratios = 1. For t=3: all ratios = 3 from the starting point. Both points lie on L.

Is (2,3,6) a Pythagorean triplet?

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Yes! 2²+3²+6² = 4+9+36 = 49 = 7². So (2,3,6,7) is a Pythagorean quadruple. Whenever you see direction (2,3,6), immediately know |d|=7.

What is the general formula for a+b+c on line L?

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For L: x=-1+2t, y=-1+3t, z=-3+6t (or 3+6t): a+b+c = (-1+2t)+(-1+3t)+(-3+6t) = -5+11t. For t=1: 6; for t=3: 28. Sum = (6)+(28) = 34.

Why does the sum always simplify nicely?

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For two points at t=t₀+1 and t=t₀-1: sum = 2(-5+11t₀) = -10+22t₀ for the (z+3) case. With t₀=2: -10+44=34. The symmetry about t₀ makes computation elegant.

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