Straight Line in Space Complete JEE Notes with examples.

A line in space is the simplest object in three-dimensional geometry and it is also the one JEE returns to every single year. In the nineteen JEE Main 2026 shifts there were twenty-six questions on it, more than one per paper. What makes them worth studying together is that they are not twenty-six different problems. They are four moves, dressed up.

This note builds the line from its definition, then names those four moves and shows each one working on real papers.

What fixes a line in space

In the plane, a point and a slope fix a line. In space there is no single number that plays the part of slope: a direction now needs three numbers, not one. So the data is a point to stand on and a direction to travel in.

A point and a direction pin down exactly one line; without the direction, infinitely many lines pass through the same point

A line in space carries four independent conditions, which is worth knowing before you start any problem. Its direction eats two of them, because a direction in space has three components but only their ratios matter. The point eats the other two, because once the direction is fixed you may slide the point anywhere along the line without changing it. Four conditions, four unknowns: that is why "find the line" problems always hand you exactly four pieces of information, and why a question that seems to give you three is really giving you a family.

Direction cosines and direction ratios

Point the line in one of its two directions and measure the angle it makes with each positive axis. Call those angles $\alpha$, $\beta$, $\gamma$.

The three direction angles measured from the positive x, y and z axes

The direction cosines are $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$. They are not three free numbers. Dropping perpendiculars from a point at distance $r$ along the line onto the three axes gives coordinates $(r\cos\alpha, r\cos\beta, r\cos\gamma)$, and that point is at distance $r$ from the origin, so

$$r^2\cos^2\alpha+r^2\cos^2\beta+r^2\cos^2\gamma=r^2$$ $$l^2+m^2+n^2=1$$

Any three numbers $a,b,c$ proportional to $l,m,n$ are direction ratios. Here is the whole difference between the two, in one table.

The practical rule is short. Write direction ratios while setting the problem up, because they are whatever falls out of the algebra. Convert to direction cosines only at the moment an angle, a projection or a distance-along is asked for, because those need a unit vector.

Example 1 (JEE Main 2026, 23 Jan Shift 1). The direction cosines of two lines satisfy $4l+m-n=0$ and $2mn+10nl+3lm=0$. Find the cosine of the acute angle between the lines.

▸ SHOW SOLUTION

Step 1: Use the linear relation to remove one letter.

From the first equation $n=4l+m$. Substituting into the second:

$$2m(4l+m)+10(4l+m)l+3lm=0$$$$40l^2+21lm+2m^2=0$$

Step 2: Factorise, which splits the two lines.

$$(8l+m)(5l+2m)=0$$

Two factors, two lines. That is the whole point of the quadratic.

Step 3: Read a direction ratio off each branch.

If $8l+m=0$ then $m=-8l$ and $n=4l-8l=-4l$, giving ratios $\langle 1,-8,-4\rangle$.

If $5l+2m=0$ then $m=-\tfrac52 l$ and $n=\tfrac32 l$, giving ratios $\langle 2,-5,3\rangle$.

Step 4: Take the angle, with the absolute value in place.

$$\cos\theta=\frac{\lvert 2+40-12\rvert}{\sqrt{81}\,\sqrt{38}}=\frac{30}{9\sqrt{38}}=\frac{10}{3\sqrt{38}}$$

Answer: (4) $\dfrac{10}{3\sqrt{38}}$

Notice that the question said "direction cosines" but the work was done entirely in direction ratios, and normalising never happened, because the $\cos\theta$ formula divides by the lengths anyway. That is the table above in action.

The equation of a line, in the forms you will actually meet

Stand at $A$ with position vector $\vec a$ and walk a multiple of $\vec b$. Every point of the line, and no other point, is reached exactly once.

$$\vec r=\vec a+\lambda\vec b,\qquad \lambda\in\mathbb{R}$$ Each value of the parameter names one point of the line

Writing $\vec a=x_1\hat i+y_1\hat j+z_1\hat k$ and $\vec b=a\hat i+b\hat j+c\hat k$ and comparing components gives the parametric form:

$$x=x_1+a\lambda,\qquad y=y_1+b\lambda$$ $$z=z_1+c\lambda$$

Eliminating $\lambda$ gives the symmetric form, which is the one printed in almost every question:

$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$

Through two points $A$ and $B$, the direction is $\vec b-\vec a$, so

$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

The zero denominator, which is notation and not division

A symmetric form such as

$$\frac{x-1}{2}=\frac{y-3}{0}=\frac{z+4}{5}$$

is not dividing by zero. It is shorthand for the pair of statements $y=3$ and $\dfrac{x-1}{2}=\dfrac{z+4}{5}$. The line lies entirely in the plane $y=3$, and its direction ratios are genuinely $\langle 2,0,5\rangle$. This appears in JEE regularly, and a student who reads it as an error rather than as a plane condition loses the question at the first line.

The three axes are the cleanest examples. The $x$-axis passes through the origin with direction $\langle 1,0,0\rangle$, so it is $y=0$, $z=0$, and the same pattern gives the other two.

Angle between two lines

Position is irrelevant. Slide both direction vectors to a common point and measure between them.

Only the directions matter; where the lines sit does not enter the calculation

$$\cos\theta=\frac{\lvert \vec b_1\cdot\vec b_2\rvert}{\lvert \vec b_1\rvert\,\lvert \vec b_2\rvert}$$

Two consequences follow immediately and are worth stating as separate facts, because questions test them separately. The lines are perpendicular exactly when $a_1a_2+b_1b_2+c_1c_2=0$, and parallel exactly when $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$.

When a question wants a line perpendicular to two given lines, its direction is $\vec b_1\times\vec b_2$. That single observation solves a large fraction of the "let $L$ be perpendicular to both" questions, of which 2026 alone had three.

Example 2 (JEE Main 2026, 6 Apr Shift 1). Let $L$ be perpendicular to both $L_1$ with direction $\langle 3,5,7\rangle$ and $L_2$ with direction $\langle 1,4,7\rangle$. If $\theta$ is the acute angle between $L$ and a line $L_3$ of direction $\langle 2,1,2\rangle$, find $\tan\theta$.

▸ SHOW SOLUTION

Step 1: The cross product gives the direction of $L$.

$$\vec d=\langle 3,5,7\rangle\times\langle 1,4,7\rangle=\langle 7,-14,7\rangle$$

Divide out the common factor and use $\langle 1,-2,1\rangle$. Scaling a direction never changes an angle, so do it early.

Step 2: Take the cosine.

$$\vec d\cdot\vec d_3=2-2+2=2$$$$\lvert\vec d\rvert=\sqrt6,\qquad \lvert\vec d_3\rvert=3$$$$\cos\theta=\frac{2}{3\sqrt6}$$

Step 3: Convert to a tangent.

$$\sin\theta=\sqrt{1-\frac{4}{54}}=\sqrt{\frac{50}{54}}$$$$\tan\theta=\sqrt{\frac{50}{54}}\cdot\frac{3\sqrt6}{2}=\frac{\sqrt{50}}{2}=\frac{5\sqrt2}{2}$$

Answer: (2) $\dfrac{5}{2}\sqrt2$

Foot of the perpendicular: one equation does all the work

This is the single most useful computation in the chapter, and it is one line of algebra.

The foot is the one point of the line whose join to P meets the line at a right angle

Take the general point $F=\vec a+\lambda\vec b$ on the line. The join $\overrightarrow{PF}$ must be perpendicular to the direction, so

$$(\vec a+\lambda\vec b-\vec p)\cdot\vec b=0$$ $$\lambda=\frac{(\vec p-\vec a)\cdot\vec b}{\lvert\vec b\rvert^2}$$

That is a single linear equation in a single unknown. Once $\lambda$ is known, $F$ is known, and the perpendicular distance is just $\lvert PF\rvert$.

Example 3 (JEE Main 2026, 5 Apr Shift 1). Find the square of the distance of $P(5,6,7)$ from the line with point $(2,5,2)$ and direction $\langle 2,3,4\rangle$.

▸ SHOW SOLUTION

Step 1: General point and join.

$$Q=(2\lambda+2,\ 3\lambda+5,\ 4\lambda+2)$$$$\overrightarrow{PQ}=(2\lambda-3,\ 3\lambda-1,\ 4\lambda-5)$$

Step 2: One dot product.

$$2(2\lambda-3)+3(3\lambda-1)+4(4\lambda-5)=0$$$$29\lambda-29=0$$

So $\lambda=1$ and $Q=(4,8,6)$.

Step 3: Measure.

$$PQ^2=1+4+1=6$$

Answer: (3) $6$

Example 4 (JEE Main 2026, 21 Jan Shift 1). Let $(\alpha,\beta,\gamma)$ be the foot of the perpendicular from $(5,4,2)$ on $\vec r=(-\hat i+3\hat j+\hat k)+\lambda(2\hat i+3\hat j-\hat k)$. Find the length of the projection of $\alpha\hat i+\beta\hat j+\gamma\hat k$ on $6\hat i+2\hat j+3\hat k$.

▸ SHOW SOLUTION

Step 1: Foot, by the same one equation.

With $P=(2\lambda-1,\ 3\lambda+3,\ -\lambda+1)$ and $A=(5,4,2)$:

$$2(2\lambda-6)+3(3\lambda-1)-(-\lambda-1)=0$$$$14\lambda-14=0$$

So $\lambda=1$ and $(\alpha,\beta,\gamma)=(1,6,0)$.

Step 2: Project, which is where the direction cosines finally earn their keep.

$$\text{projection}=\frac{\lvert \vec u\cdot\vec w\rvert}{\lvert\vec w\rvert}=\frac{18}{7}$$

Answer: (3) $\dfrac{18}{7}$

The pattern here is worth naming, because JEE uses it constantly. The foot is rarely the answer. It is the intermediate object the real question is built on.

Example 5 (JEE Main 2025, 22 Jan Shift 2). Find the perpendicular distance of the line through $(1,-2,-3)$ with direction $\langle 2,-1,2\rangle$ from the point $P(2,-10,1)$.

▸ SHOW SOLUTION

Step 1: General point and join.

$$F=(2\mu+1,\ -\mu-2,\ 2\mu-3)$$$$\overrightarrow{PF}=(2\mu-1,\ -\mu+8,\ 2\mu-4)$$

Step 2: Dot with the direction.

$$2(2\mu-1)-(-\mu+8)+2(2\mu-4)=0$$$$9\mu-18=0$$

So $\mu=2$ and $F=(5,-4,1)$.

Step 3: Measure.

$$PF=\sqrt{9+36+0}=\sqrt{45}=3\sqrt5$$

Answer: (2) $3\sqrt5$

Three papers apart, three different wrappings, one identical calculation. That is the argument for learning the move rather than the question.

Image of a point in a line: find the foot, then double it

Reflecting $P$ in the line means finding the point $Q$ for which the line is the perpendicular bisector of $PQ$. The foot $F$ is therefore the midpoint of $PQ$.

The foot is the midpoint of P and its image, so the image costs one extra line of work

$$\vec q=2\vec f-\vec p$$

There is nothing else to it. Every image question in this chapter is a foot question with one subtraction on the end, and recognising that turns a question students find long into a short one.

Example 6 (JEE Main 2026, 28 Jan Shift 2). Let $Q(a,b,c)$ be the image of $P(3,2,1)$ in the line through $(1,0,1)$ with direction $\langle 1,2,1\rangle$. Find the distance of $Q$ from the line through $(9,9,5)$ with direction $\langle 3,2,-2\rangle$.

▸ SHOW SOLUTION

Step 1: Foot on the first line.

$$N=(r+1,\ 2r,\ r+1)$$$$\overrightarrow{PN}=(r-2,\ 2r-2,\ r)$$$$(r-2)+2(2r-2)+r=6r-6=0$$

So $r=1$ and $N=(2,2,2)$.

Step 2: Double it.

$$Q=2N-P=(1,2,3)$$

Step 3: Now a fresh perpendicular-distance calculation.

With $A=(9,9,5)$ and $\overrightarrow{AQ}=(-8,-7,-2)$:

$$\lvert AQ\rvert^2=64+49+4=117$$$$\overrightarrow{AQ}\cdot\vec d=-24-14+4=-34$$$$\lvert \vec d\rvert=\sqrt{17}$$

The projection of $\overrightarrow{AQ}$ along the line has square $\dfrac{34^2}{17}=68$, so by Pythagoras

$$QM=\sqrt{117-68}=\sqrt{49}=7$$

Answer: (3) $7$

Step 3 is worth a second look. Rather than solving for the foot again, it used $\lvert AQ\rvert^2$ minus the square of the along-line component. On a numeric question that is two lines shorter than finding the foot, and it is the version to reach for when the foot itself is not asked for.

Example 7 (JEE Main 2026, 22 Jan Shift 1). If the image of $P(1,2,a)$ in the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{7-z}{2}$ is $Q(5,b,c)$, find $a^2+b^2+c^2$.

▸ SHOW SOLUTION

Step 1: Repair the direction before anything else.

The third fraction is written $\dfrac{7-z}{2}$, which equals $\dfrac{z-7}{-2}$. The direction ratios are $\langle 3,2,-2\rangle$ and the point is $(6,7,7)$. Skip this and every later sign is wrong.

Step 2: Midpoint on the line.

$$M=\left(3,\ \frac{b+2}{2},\ \frac{a+c}{2}\right)$$

The $x$-coordinate gives $t=\dfrac{3-6}{3}=-1$ at once, and then

$$\frac{b+2}{2}-7=-2\ \Rightarrow\ b=8$$$$\frac{a+c}{2}-7=2\ \Rightarrow\ a+c=18$$

Step 3: Perpendicularity, the second condition.

$$\overrightarrow{PQ}=(4,\ 6,\ c-a)$$$$12+12-2(c-a)=0\ \Rightarrow\ c-a=12$$

Step 4: Solve the pair.

From $a+c=18$ and $c-a=12$: $c=15$, $a=3$.

$$a^2+b^2+c^2=9+64+225=298$$

Answer: (3) $298$

Example 8 (JEE Main 2024, 31 Jan Shift 2). Let $(\alpha,\beta,\gamma)$ be the mirror image of $(2,3,5)$ in the line $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$. Find $2\alpha+3\beta+4\gamma$.

▸ SHOW SOLUTION

Step 1: Foot.

$$F=(2t+1,\ 3t+2,\ 4t+3)$$$$\overrightarrow{PF}=(2t-1,\ 3t-1,\ 4t-2)$$$$2(2t-1)+3(3t-1)+4(4t-2)=0$$$$29t-13=0$$

So $t=\dfrac{13}{29}$.

Step 2: Double it and combine.

Doubling gives the image, and substituting into $2\alpha+3\beta+4\gamma$ collapses to

$$2\alpha+3\beta+4\gamma=33$$

Answer: (4) $33$

The word that changes everything: along

Read these two questions side by side.

Find the distance of $P$ from the line $L$.

Find the distance of $P$ from the line $L$ measured along the line $M$.

They are not the same question and they do not share a method. The first is a perpendicular distance. The second says: leave $P$ travelling in the direction of $M$, and stop when you hit $L$. Since that path is not perpendicular, it is longer.

Perpendicular distance versus distance measured along a stated direction

The method for the second is mechanical once you see it. Write the line through $P$ with the given direction, write the target line, and force them to meet. Two parameters, three coordinate equations, and the system is over-determined in your favour: solve two, check the third.

Example 9 (JEE Main 2026, 4 Apr Shift 1). Find the square of the distance of $(-2,-8,6)$ from the line $\dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z}{-1}$ along the line $\dfrac{x+5}{1}=\dfrac{y+5}{-1}=\dfrac{z}{2}$.

▸ SHOW SOLUTION

Step 1: Travel from the point in the stated direction.

Only the direction $\langle 1,-1,2\rangle$ of the second line is used; its position is irrelevant.

$$B=(\lambda-2,\ -\lambda-8,\ 2\lambda+6)$$

Step 2: Land on the target line.

$$\lambda-2=\mu+1$$$$-\lambda-8=2\mu+1$$$$2\lambda+6=-\mu$$

From the first, $\mu=\lambda-3$. Substituting into the third gives $3\lambda=-3$, so $\lambda=-1$ and $\mu=-4$.

Step 3: Check the spare equation, then measure.

The second equation reads $-7=-7$, which confirms the point. So $B=(-3,-7,4)$ and

$$AB^2=1+1+4=6$$

Answer: (2) $6$

Example 10 (JEE Main 2025, 3 Apr Shift 2). Find the distance of the point $(7,10,11)$ from the line $\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}$ along the line $\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}$.

▸ SHOW SOLUTION

Step 1: Note the zero denominator.

The target line has direction $\langle 1,0,3\rangle$ and lies in the plane $y=4$. That single fact does most of the work.

Step 2: Travel and land.

$$B=(2\lambda+7,\ 3\lambda+10,\ 6\lambda+11)$$

Landing on $y=4$ gives $3\lambda+10=4$, so $\lambda=-2$ immediately, and $B=(3,4,-1)$. Checking, $\dfrac{3-4}{1}=-1=\dfrac{-1-2}{3}$, so $B$ really is on the line.

Step 3: Measure.

$$PB=\sqrt{16+36+144}=\sqrt{196}=14$$

Answer: (1) $14$

The zero denominator turned a two-parameter system into a one-line calculation. That is what the earlier section on notation was for.

Skew lines and the shortest distance

Two lines in space fall into exactly three cases, and the first question to ask is always which one you are in.

Skew lines neither meet nor run parallel, and one join is perpendicular to both

For skew lines,

$$d=\frac{\lvert (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\rvert}{\lvert \vec b_1\times\vec b_2\rvert}$$

The formula reads better than it looks. The vector $\vec b_1\times\vec b_2$ points along the one direction perpendicular to both lines, and the numerator projects the gap between the lines onto it. Everything else cancels.

Example 11 (JEE Main 2026, 6 Apr Shift 2). Find the shortest distance between $\dfrac{x-4}{1}=\dfrac{y-3}{2}=\dfrac{z-2}{-3}$ and $\dfrac{x+2}{2}=\dfrac{y-6}{4}=\dfrac{z-5}{-5}$.

▸ SHOW SOLUTION

Step 1: Check the case first.

$$\vec d_1\times\vec d_2=\langle 1,2,-3\rangle\times\langle 2,4,-5\rangle$$$$\vec d_1\times\vec d_2=\langle 2,-1,0\rangle$$

Non-zero, so the lines are not parallel, despite $\langle 1,2,-3\rangle$ and $\langle 2,4,-5\rangle$ looking suspiciously close. That near-miss is deliberate.

Step 2: Apply the formula.

$$\overrightarrow{AB}=(-6,3,3),\qquad \lvert \vec d_1\times\vec d_2\rvert=\sqrt5$$$$d=\frac{\lvert -12-3+0\rvert}{\sqrt5}=\frac{15}{\sqrt5}=3\sqrt5$$

Answer: (3) $3\sqrt5$

Example 12 (JEE Main 2026, 24 Jan Shift 2). Find the sum of all values of $\alpha$ for which the shortest distance between $\dfrac{x+1}{\alpha}=\dfrac{y-2}{-1}=\dfrac{z-4}{-\alpha}$ and $\dfrac{x}{\alpha}=\dfrac{y-1}{2}=\dfrac{z-1}{2\alpha}$ is $\sqrt2$.

▸ SHOW SOLUTION

Step 1: Cross product, keeping $\alpha$ symbolic.

$$\vec d_1\times\vec d_2=\langle 0,\ -3\alpha^2,\ 3\alpha\rangle$$$$\lvert \vec d_1\times\vec d_2\rvert=3\lvert\alpha\rvert\sqrt{\alpha^2+1}$$

Step 2: Build the distance and watch it simplify.

With $\vec r=(1,-1,-3)$, the numerator is $\lvert 3\alpha^2-9\alpha\rvert$, so

$$d=\frac{\lvert 3\alpha^2-9\alpha\rvert}{3\lvert\alpha\rvert\sqrt{\alpha^2+1}}=\frac{\lvert\alpha-3\rvert}{\sqrt{\alpha^2+1}}$$

The $\lvert\alpha\rvert$ cancels, which is the whole reason this question is doable in two minutes.

Step 3: Set it equal and solve.

$$\frac{(\alpha-3)^2}{\alpha^2+1}=2\ \Longrightarrow\ \alpha^2+6\alpha-7=0$$$$(\alpha+7)(\alpha-1)=0$$

The roots are $-7$ and $1$, summing to $-6$.

Answer: (2) $-6$

Look back at the condition count from the first section. Two unknowns' worth of freedom, one condition, so a family and a "sum of all values" ending. The shape of the answer was predictable before any algebra happened.

Example 13 (JEE Main 2025, 2 Apr Shift 2). $L_1$ is parallel to $-3\hat i+2\hat j+4\hat k$ through $(7,6,2)$; $L_2$ is parallel to $2\hat i+\hat j+3\hat k$ through $(5,3,4)$. Find the shortest distance.

▸ SHOW SOLUTION

Step 1: Cross product.

$$\vec b_1\times\vec b_2=\langle -3,2,4\rangle\times\langle 2,1,3\rangle$$$$\vec b_1\times\vec b_2=\langle 2,17,-7\rangle$$$$\lvert \vec b_1\times\vec b_2\rvert=\sqrt{4+289+49}=3\sqrt{38}$$

Step 2: Project the gap.

$$\vec a_2-\vec a_1=(-2,-3,2)$$$$(-2)(2)+(-3)(17)+(2)(-7)=-69$$$$d=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}$$

Answer: (1) $\dfrac{23}{\sqrt{38}}$

When two lines meet: intersection and coplanarity

If two lines do meet, the fastest route is almost never the determinant. Write both general points, set them equal, solve two of the three coordinate equations, and use the third as a check. The determinant condition is for questions that ask whether they meet, or for what value of a parameter they meet.

Example 14 (JEE Main 2026, 2 Apr Shift 1). If the point of intersection of $\dfrac{x+1}{3}=\dfrac{y+a}{5}=\dfrac{z+b+1}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-b}{4}=\dfrac{z-2a}{7}$ lies on the $xy$-plane, find $a+b$.

▸ SHOW SOLUTION

Step 1: Two general points.

$$P=(3r_1-1,\ 5r_1-a,\ 7r_1-b-1)$$$$Q=(r_2+2,\ 4r_2+b,\ 7r_2+2a)$$

Step 2: Use the extra condition early.

Lying on the $xy$-plane means $z=0$, and that applies to both expressions for the same point:

$$7r_1-b-1=0,\qquad 7r_2+2a=0$$

Two equations for free, before touching the harder ones.

Step 3: Equate the remaining coordinates.

Matching $x$ gives $r_2=3r_1-3$. Matching $y$, and substituting, gives $7r_1=5$, so $r_1=\dfrac57$.

Step 4: Back-substitute.

$$b=7\cdot\tfrac57-1=4$$$$r_2=-\tfrac67,\qquad a=3$$$$a+b=7$$

Answer: (3) $7$

Step 2 is the lesson. The condition "lies on the $xy$-plane" was not an afterthought to apply at the end; used first, it cut the work roughly in half.

Which move, when

A short decision chart covering almost every straight-line question in space

Read the question for one of four signals and the method follows.

Quick recall

Work these methods on the past papers until the recognition is automatic, then move on to the plane, where every one of them reappears with an extra equation attached.