If the image of the point P(1, 2, a) in the line $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{7-z}{2}$ is Q(5, b, c), then $a^2 + b^2 + c^2$ is equal to
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Solution Steps
1
Understand the line equation
Line: $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{7-z}{2}$
Rewrite: $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{z-7}{-2}$
Point on line: $(6, 7, 7)$
Direction ratios: $(3, 2, -2)$
2
General point on the line
Any point M on line: $(6+3λ, 7+2λ, 7-2λ)$ for parameter $λ$
3
Use midpoint condition
Given Q(5, b, c) is image of P(1, 2, a)
Let M be foot of perpendicular from P to line
M is midpoint of PQ
$M = \left(\dfrac{1+5}{2}, \dfrac{2+b}{2}, \dfrac{a+c}{2}\right) = (3, \dfrac{2+b}{2}, \dfrac{a+c}{2})$
4
M lies on the line
Since M is on line: $(3, \dfrac{2+b}{2}, \dfrac{a+c}{2}) = (6+3λ, 7+2λ, 7-2λ)$
From x-coordinate: $3 = 6 + 3λ$ → $λ = -1$
5
Find M coordinates
$M = (6+3(-1), 7+2(-1), 7-2(-1))$
$M = (3, 5, 9)$
6
Use M as midpoint of PQ
$M = \left(\dfrac{1+5}{2}, \dfrac{2+b}{2}, \dfrac{a+c}{2}\right) = (3, 5, 9)$
From y: $\dfrac{2+b}{2} = 5$ → $2+b = 10$ → $b = 8$
Wait, but Q is given as (5, b, c). Let me recalculate using Q = 2M – P
7
Correct approach: Find a and c
We have M(3, 5, 9) and P(1, 2, a)
Q = 2M – P = $(2(3)-1, 2(5)-2, 2(9)-a) = (5, 8, 18-a)$
Given Q = (5, b, c), so $b = 8$ and $c = 18 – a$
Also, PM must be perpendicular to line direction (3, 2, -2)
$\vec{PM} = M – P = (3-1, 5-2, 9-a) = (2, 3, 9-a)$
$\vec{PM} \cdot (3, 2, -2) = 0$
$2(3) + 3(2) + (9-a)(-2) = 0$
$6 + 6 – 18 + 2a = 0$
$2a = 6$ → $a = 3$
So $c = 18 – 3 = 15$
8
Calculate a² + b² + c²
$a = 3, b = 8, c = 15$
Wait, let me verify: Given answer is 298
$3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$ ✓
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Final Answer
a² + b² + c² = 298
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Related Theory
📌 Reflection of Point in a Line (3D)
Image or reflection of point P in line L is point Q such that L is perpendicular bisector of segment PQ. This means: (1) Line PQ is perpendicular to line L, (2) Midpoint M of PQ lies on line L. To find image: First, find foot of perpendicular M from P to L. Let line L pass through point A with direction vector d. Any point on L is A + λd for some λ. Point M on L is such that vector PM is perpendicular to d, i.e., PM · d = 0. Substitute M = A + λd and solve: (M – P) · d = 0, giving ((A + λd) – P) · d = 0, which gives (A – P) · d + λ(d · d) = 0, so λ = -[(A – P) · d]/(d · d). Once M is found, image Q = 2M – P (since M is midpoint). This formula Q = 2M – P is key. In 2D, similar process with 2 coordinates. Common JEE error: forgetting perpendicularity condition or midpoint condition. Both are essential for reflection.
📌 Equation of Line in 3D Space
A line in 3D can be represented in multiple forms. Symmetric form (most common in JEE): (x-x₁)/a = (y-y₁)/b = (z-z₁)/c where (x₁, y₁, z₁) is point on line and (a, b, c) are direction ratios (not necessarily unit). Vector form: r = a + λb where a is position vector of point on line, b is direction vector, λ is parameter. Any point on line: (x₁+λa, y₁+λb, z₁+λc). Parametric form: x = x₁ + λa, y = y₁ + λb, z = z₁ + λc. Two-point form: line through (x₁,y₁,z₁) and (x₂,y₂,z₂) has direction ratios (x₂-x₁, y₂-y₁, z₂-z₁). Cartesian form (intersection of two planes): a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0. Converting between forms is important skill. Direction ratios are proportional to actual direction; direction cosines l, m, n satisfy l² + m² + n² = 1. For JEE, master symmetric form conversions and finding points on line using parameter λ.
📌 Perpendicularity and Dot Product in 3D
Two vectors are perpendicular if and only if their dot product is zero. For vectors a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃), dot product a · b = a₁b₁ + a₂b₂ + a₃b₃. If a · b = 0, then a ⊥ b. This extends to direction ratios: if line 1 has direction ratios (a₁, b₁, c₁) and line 2 has (a₂, b₂, c₂), they are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0. Angle θ between vectors: cos θ = (a · b)/(|a||b|). For perpendicular lines/vectors, θ = 90°, so cos θ = 0, giving a · b = 0. In reflection problems, vector joining point to its foot must be perpendicular to line direction. Common applications: (1) finding perpendicular distance from point to line/plane, (2) checking if lines are perpendicular, (3) finding angle between lines/planes, (4) projection calculations. Remember: perpendicularity is algebraic condition (dot product zero) with geometric meaning (90° angle). Master both interpretations for JEE success.
📌 Midpoint Formula and Applications
Midpoint M of two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in 3D is M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2). This averages coordinates component-wise. If M is known and P is known, find Q using Q = 2M – P, which rearranges midpoint formula. Component-wise: x₂ = 2x_M – x₁, similarly for y and z. In 2D, drops z-coordinate. Applications in JEE: (1) finding image/reflection (M is on mirror line/plane), (2) finding vertex of triangle given midpoints, (3) centroid problems (centroid is average of vertices), (4) section formula (midpoint is special case m:n = 1:1). Section formula generalizes: point dividing PQ in ratio m:n is ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n)). For external division, use m:(-n). Midpoint is fundamental in coordinate geometry – appears in almost every reflection, symmetry, or centroid problem. Practice converting between “M is midpoint of PQ” and algebraic equations.
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FAQs
Q
What is image of point in a line?
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Image Q of point P in line L is such that L is perpendicular bisector of PQ. This means: (1) PQ ⊥ L, (2) midpoint of PQ lies on L. Find foot M of perpendicular from P to L, then Q = 2M – P.
Q
How to find foot of perpendicular to line in 3D?
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Let line be r = a + λd (point a, direction d). Foot M = a + λd where λ is found from (M – P) · d = 0. Expand: (a + λd – P) · d = 0 → λ = -(a – P) · d / (d · d). Substitute back to get M.
Q
Why use Q = 2M – P formula?
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Since M is midpoint of P and Q: M = (P + Q)/2. Solving for Q: 2M = P + Q → Q = 2M – P. This directly gives image once foot of perpendicular M is known. Very useful shortcut!
Q
How to handle (7-z) in line equation?
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Rewrite (7-z)/2 as (z-7)/(-2) to get standard form. Now direction ratio for z is -2 instead of 2. Point on line remains (6, 7, 7). General point: (6+3λ, 7+2λ, 7-2λ).
Q
What are direction ratios vs direction cosines?
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Direction ratios (a, b, c) are proportional to direction – any scalar multiple represents same direction. Direction cosines (l, m, n) are normalized: l² + m² + n² = 1. Convert: l = a/√(a²+b²+c²), similarly m, n.
Q
Can we verify answer by checking perpendicularity?
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Yes! With a=3, b=8, c=15: P(1,2,3), Q(5,8,15), M(3,5,9). Check: PM = (2,3,6), direction d = (3,2,-2). PM · d = 6 + 6 – 12 = 0 ✓. Also M lies on line ✓.
Q
What if perpendicularity condition gives different λ?
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Then there’s calculation error. Perpendicularity and “M on line” must give same λ. In our case, both correctly gave λ = -1. If different, recheck dot product calculation and line equation.
Q
How to calculate a² + b² + c² quickly?
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Square each: 3² = 9, 8² = 64, 15² = 225. Add: 9 + 64 = 73, then 73 + 225 = 298. Know perfect squares up to 20² for speed. Alternatively: 3² + 8² = 73 (memorize), then add 15² = 225.
Q
What is distance from P to line?
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Distance = |PM| where M is foot. PM = (2, 3, 6), so |PM| = √(4 + 9 + 36) = √49 = 7 units. This is perpendicular distance from P(1,2,3) to given line.
Q
Can we solve using symmetric properties?
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Yes! Since Q is reflection of P in line, distance from P to line = distance from Q to line. Also PQ is perpendicular to line. These symmetric properties can verify answer or provide alternate solution paths.
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