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Three Dimensional Geometry: Mathematics | JEE Main

Q: Why do we divide by 5/42 instead of just 1/42?

Because the problem specifies that must be coprime integers. Dividing by their greatest common divisor 5 yields , which are coprime.

The point of intersection

\mathrm{C}

of the plane

8 x+y+2 z=0

and the line joining the point

\mathrm{A}(-3,-61)

and

\mathrm{B}(2,-4,-3)

divides the line segment

\mathrm{AB}

internally in the ratio

\mathrm{k}

:. If

\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|)

are coprime are the direction ratios of the perpendicular form the point

\mathrm{C}

on the line

\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}

, then

|\mathrm{a}+\mathrm{b}+\mathrm{c}|

is equal to

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Step-by-step solutionView

Correct answer

The coordinates of point C are $(-1, -5, -1)$ and the foot of the perpendicular on the line gives direction ratios proportional to $(11, 4, -5)$ , yielding $|a + b + c| = 10$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Plane: $8x + y + 2z = 0$ , line $AB$ joining $A(-3, -6, 1)$ and $B(2, 4, -3)$ [derived from line equation $(x-2)/5 = (y-4)/10 = (z+3)/(-4)$ ], and line $L: \frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$ .

02approach

First, find the parametric coordinates of the point of intersection $C$ between line $AB$ and the plane $8x + y + 2z = 0$ . Then, define a general point $D$ on the given line $L$ , use the perpendicularity condition $\vec{CD} \cdot \vec{d}_L = 0$ to find the foot of the perpendicular $D$ , get the direction ratios of $CD$ , reduce them to coprime integers $(a, b, c)$ , and calculate $|a+b+c|$ .

03execute

Line $AB$ is given parametrically as $P(\lambda) = (5\lambda + 2, 10\lambda + 4, -4\lambda - 3)$ . Substituting into the plane $8x + y + 2z = 0$ : $8(5\lambda + 2) + (10\lambda + 4) + 2(-4\lambda - 3) = 0$ $40\lambda + 16 + 10\lambda + 4 - 8\lambda - 6 = 0 \implies 42\lambda + 14 = 0 \implies \lambda = -\frac{1}{3}.$ Thus, point $C$ is $\left(5\left(-\frac{1}{3}\right)+2, 10\left(-\frac{1}{3}\right)+4, -4\left(-\frac{1}{3}\right)-3\right) = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right)$ .

04execute

Line $L$ in standard form: $\frac{x - 1}{-1} = \frac{y + 4}{2} = \frac{z + 2}{3} = \mu$ . A general point $D$ on line $L$ is $(1 - \mu, -4 + 2\mu, -2 + 3\mu)$ . The vector $\vec{CD} = \left(1 - \mu - \frac{1}{3}\right)\hat{i} + \left(-4 + 2\mu - \frac{2}{3}\right)\hat{j} + \left(-2 + 3\mu - \left(-\frac{5}{3}\right)\right)\hat{k} = \left(\frac{2}{3} - \mu\right)\hat{i} + \left(2\mu - \frac{14}{3}\right)\hat{j} + \left(3\mu - \frac{1}{3}\right)\hat{k}$ . Since $CD \perp L$ : $(-1)\left(\frac{2}{3} - \mu\right) + 2\left(2\mu - \frac{14}{3}\right) + 3\left(3\mu - \frac{1}{3}\right) = 0$ $\left(\mu - \frac{2}{3}\right) + \left(4\mu - \frac{28}{3}\right) + (9\mu - 1) = 0$ $14\mu - 11 = 0 \implies \mu = \frac{11}{14}.$

05execute

Substitute $\mu = \frac{11}{14}$ into $\vec{CD}$ : $CD_x = \frac{2}{3} - \frac{11}{14} = \frac{28 - 33}{42} = -\frac{5}{42}$ $CD_y = 2\left(\frac{11}{14}\right) - \frac{14}{3} = \frac{11}{7} - \frac{14}{3} = \frac{33 - 98}{21} = -\frac{65}{21} = -\frac{130}{42}$ $CD_z = 3\left(\frac{11}{14}\right) - \frac{1}{3} = \frac{33}{14} - \frac{1}{3} = \frac{99 - 14}{42} = \frac{85}{42}$ Dividing by common factor $\frac{5}{42}$ , the coprime direction ratios are: $(a, b, c) = (-1, -26, 17)$ Check coprimality: $\gcd(|-1|, |-26|, |17|) = \gcd(1, 26, 17) = 1$ . Now, compute $|a + b + c| = |-1 - 26 + 17| = |-10| = 10$ .

✓verify

Check dot product of $(-1, -26, 17)$ with line direction $(-1, 2, 3)$ : $(-1)(-1) + (-26)(2) + (17)(3) = 1 - 52 + 51 = 0$ , confirming orthogonality.

Hints that build this answer step by step

If point C divides the segment joining $A(-3, -6, 1)$ and $B(2, -4, -3)$ internally in the ratio $k : 1$ , what are the coordinates of C in terms of $k$ ?
$\left(\frac{2k - 3}{k + 1}, \frac{-4k - 6}{k + 1}, \frac{-3k + 1}{k + 1}\right)$
Substituting the coordinates of C into the plane equation $8x + y + 2z = 0$ , what is the value of $k$ and the resulting coordinates of C?
$k = 2$ , so $C = (-1, -5, -1)$
Let $P$ be a point on the line $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3} = \lambda$ . If $CP$ is perpendicular to the line, what are the coprime direction ratios $(a, b, c)$ of vector $\vec{CP}$ ?
$(11, 4, -5)$ or $(-11, -4, 5)$
Using the coprime direction ratios $(a, b, c) = (11, 4, -5)$ , what is the value of $|a + b + c|$ ?
$10$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

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Students also ask

Why do we divide by 5/42 instead of just 1/42?

Because the problem specifies that $|a|, |b|, |c|$ must be coprime integers. Dividing $-5, -130, 85$ by their greatest common divisor 5 yields $-1, -26, 17$ , which are coprime.

Answer

The coordinates of point C are $(-1, -5, -1)$ and the foot of the perpendicular on the line gives direction ratios proportional to $(11, 4, -5)$ , yielding $|a + b + c| = 10$ .

Why each option works or fails

Step-by-step solution

given: Plane: $8x + y + 2z = 0$ , line $AB$ joining $\mathrm{B}(2,-4,-3)$ and $\mathrm{AB}$ [derived from line equation $(x-2)/5 = (y-4)/10 = (z+3)/(-4)$ ], and line $L: \frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$ .
approach: First, find the parametric coordinates of the point of intersection $C$ between line $AB$ and the plane $8x + y + 2z = 0$ . Then, define a general point $D$ on the given line $L$ , use the perpendicularity condition $\vec{CD} \cdot \vec{d}_L = 0$ to find the foot of the perpendicular $D$ , get the direction ratios of $CD$ , reduce them to coprime integers $(a, b, c)$ , and calculate $|a+b+c|$ .
execute: Line $AB$ is given parametrically as $P(\lambda) = (5\lambda + 2, 10\lambda + 4, -4\lambda - 3)$ . Substituting into the plane $8x + y + 2z = 0$ : $8(5\lambda + 2) + (10\lambda + 4) + 2(-4\lambda - 3) = 0$ $40\lambda + 16 + 10\lambda + 4 - 8\lambda - 6 = 0 \implies 42\lambda + 14 = 0 \implies \lambda = -\frac{1}{3}.$ Thus, point $C$ is $\left(5\left(-\frac{1}{3}\right)+2, 10\left(-\frac{1}{3}\right)+4, -4\left(-\frac{1}{3}\right)-3\right) = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right)$ .
execute: Line $L$ in standard form: $\frac{x - 1}{-1} = \frac{y + 4}{2} = \frac{z + 2}{3} = \mu$ . A general point $D$ on line $L$ is $(1 - \mu, -4 + 2\mu, -2 + 3\mu)$ . The vector $\vec{CD} = \left(1 - \mu - \frac{1}{3}\right)\hat{i} + \left(-4 + 2\mu - \frac{2}{3}\right)\hat{j} + \left(-2 + 3\mu - \left(-\frac{5}{3}\right)\right)\hat{k} = \left(\frac{2}{3} - \mu\right)\hat{i} + \left(2\mu - \frac{14}{3}\right)\hat{j} + \left(3\mu - \frac{1}{3}\right)\hat{k}$ . Since $CD \perp L$ : $(-1)\left(\frac{2}{3} - \mu\right) + 2\left(2\mu - \frac{14}{3}\right) + 3\left(3\mu - \frac{1}{3}\right) = 0$ $\left(\mu - \frac{2}{3}\right) + \left(4\mu - \frac{28}{3}\right) + (9\mu - 1) = 0$ $14\mu - 11 = 0 \implies \mu = \frac{11}{14}.$
execute: Substitute $\mu = \frac{11}{14}$ into $\vec{CD}$ : $CD_x = \frac{2}{3} - \frac{11}{14} = \frac{28 - 33}{42} = -\frac{5}{42}$ $CD_y = 2\left(\frac{11}{14}\right) - \frac{14}{3} = \frac{11}{7} - \frac{14}{3} = \frac{33 - 98}{21} = -\frac{65}{21} = -\frac{130}{42}$ $CD_z = 3\left(\frac{11}{14}\right) - \frac{1}{3} = \frac{33}{14} - \frac{1}{3} = \frac{99 - 14}{42} = \frac{85}{42}$ Dividing by common factor $\frac{5}{42}$ , the coprime direction ratios are: $(a, b, c) = (-1, -26, 17)$ Check coprimality: $\gcd(|-1|, |-26|, |17|) = \gcd(1, 26, 17) = 1$ . Now, compute $|a + b + c| = |-1 - 26 + 17| = |-10| = 10$ .
verify: Check dot product of $(-1, -26, 17)$ with line direction $(-1, 2, 3)$ : $(-1)(-1) + (-26)(2) + (17)(3) = 1 - 52 + 51 = 0$ , confirming orthogonality.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The coordinates of point C are $(-1, -5, -1)$ and the foot of the perpendicular on the line gives direction ratios proportional to $(11, 4, -5)$ , yielding $|a + b + c| = 10$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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Three Dimensional Geometry: Mathematics | JEE Main

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