If the image of the point (4, 4, 3) in the line x-12 = y-21 = z-13 is (a,,), then a + + is equal to:

Given: The point is P(4,4,3) and the line is x-12 = y-21 = z-13 Find: The image of P in the given line, and then the value of a+ +. Write the line in parametric form: x = 1 + 2t, y = 2 + t, z = 1 + 3t So a point on the line is A(1,2,1) and its direction vector is d = 2,1,3 Now, AP = 4-1, 4-2, 3-1 = 3,2,2 The projection of AP on d is AP d d d d Compute the dot products: AP d = 3 2 + 2 1 + 2 3 = 14 d d = 2^2 + 1^2 + 3^2 = 14 Hence the projection is 1414 2,1,3 = 2,1,3 Therefore the foot of the perpendicular from P to the line is A + 2,1,3 = (1,2,1) + (2,1,3) = (3,3,4) If Q(a,,) is the image of P in the line, then this foot is the midpoint of PQ. So, Q = 2(3,3,4) - (4,4,3) = (2,2,5) Thus, a + + = 2 + 2 + 5 = 9 Therefore, the correct option is A. The solution marks option D on the page, but the extracted working concludes the image is (2,2,5) and the required sum is 9, so the working implies option A.

If the image of the point (4, 4, 3) in the line x-12 = y-21 = z-1… — Solution | JEE Mathematics

MCQMediumJEE 2025Equation of Line in 3D

If the image of the point $(4, 4, 3)$ in the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$ is $(a, \beta, \gamma)$ , then $a + \beta + \gamma$ is equal to:

A
$9$
B
$7$
C
$8$
D
$12$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The point is $P(4,4,3)$ and the line is

\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}

Find: The image of $P$ in the given line, and then the value of $a+\beta+\gamma$ .

Write the line in parametric form:

x = 1 + 2t, \quad y = 2 + t, \quad z = 1 + 3t

So a point on the line is $A(1,2,1)$ and its direction vector is

\vec d = \langle 2,1,3 \rangle

Now,

\vec{AP} = \langle 4-1, 4-2, 3-1 \rangle = \langle 3,2,2 \rangle

The projection of $\vec{AP}$ on $\vec d$ is

\frac{\vec{AP}\cdot \vec d}{\vec d\cdot \vec d}\vec d

Compute the dot products:

\vec{AP}\cdot \vec d = 3\cdot 2 + 2\cdot 1 + 2\cdot 3 = 14

\vec d\cdot \vec d = 2^2 + 1^2 + 3^2 = 14

Hence the projection is

\frac{14}{14}\langle 2,1,3 \rangle = \langle 2,1,3 \rangle

Therefore the foot of the perpendicular from $P$ to the line is

A + \langle 2,1,3 \rangle = (1,2,1) + (2,1,3) = (3,3,4)

If $Q(a,\beta,\gamma)$ is the image of $P$ in the line, then this foot is the midpoint of $PQ$ . So,

Q = 2(3,3,4) - (4,4,3) = (2,2,5)

Thus,

a + \beta + \gamma = 2 + 2 + 5 = 9

Therefore, the correct option is A.

The solution marks option D on the page, but the extracted working concludes the image is $(2,2,5)$ and the required sum is $9$ , so the working implies option A.

Common mistakes

Treating the given line as a plane or using a 2D reflection formula is incorrect because the question is about reflection of a point in a line in 3D. First find the foot of the perpendicular on the line using vector projection, then reflect across that point.
Using the direction vector incorrectly is a common error. For the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$ , the direction vector is $\langle 2,1,3 \rangle$ , not $\langle 1,2,3 \rangle$ or any normalized variant unless handled consistently.
After finding the foot $(3,3,4)$ , some students mistakenly take it as the image point. This is wrong because the foot is the midpoint of the original point and its reflection. Use $Q = 2M - P$ .

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