3D Geometry for JEE: Lines and Planes, Fully Solved

Three dimensional geometry is coordinate geometry with one more axis, and almost every question in it comes down to one picture: a point, a line through it with a known direction, and a perpendicular dropped from somewhere. Learn to write the general point of a line with one parameter, and most questions become a single equation in that parameter.

This note covers the whole chapter: coordinates, direction cosines, the line in every form JEE uses, the foot, the image, distance along a direction, skew lines, and then the plane, a line meeting a plane, images in a plane, families of planes and three planes together. Every worked example is a JEE Main 2026 or JEE Advanced 2026 question unless it says otherwise, and every answer was re-derived from scratch.

Coordinates in space

Three mutually perpendicular axes through the origin $O$ divide space into eight octants. A point $P(x,y,z)$ is reached by moving $x$ along the $x$-axis, $y$ parallel to the $y$-axis and $z$ parallel to the $z$-axis.

On the $xy$-plane $z=0$, on the $yz$-plane $x=0$, on the $zx$-plane $y=0$. On the $x$-axis $y=z=0$.

The distance between $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is

$$AB^2=(x_2-x_1)^2+(y_2-y_1)^2$$ $$\qquad+(z_2-z_1)^2$$

The point dividing $AB$ internally in the ratio $m:n$ is

$$\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n},\ \frac{mz_2+nz_1}{m+n}\right)$$

and externally, replace $n$ by $-n$. The centroid of a triangle is the average of its three vertices; the centroid of a tetrahedron is the average of its four.

Segment AB with an inner point P dividing it in the ratio m to n and an outer point Q beyond B dividing it externally

Example 1 (IITianForum). Let $A(1,5,3)$, $B(1,1,3)$ and $C(4,1,0)$. Which of the following is (are) correct?

(1) The orthocentre of $\triangle ABC$ is at distance $\sqrt{11}$ from the origin

(2) The circumcentre of $\triangle ABC$ is $\left(\dfrac52,3,\dfrac32\right)$

(3) If $A$, $B$, $C$, $D$ in some order form a parallelogram, $D$ can be $(4,5,0)$

(4) If $A$, $B$, $C$, $D$ in some order form a parallelogram, $D$ can be $(4,-3,0)$

▸ SHOW SOLUTION

Step 1. Find the right angle

$$\overrightarrow{BA}=(0,4,0)$$$$\overrightarrow{BC}=(3,0,-3)$$$$\overrightarrow{BA}\cdot\overrightarrow{BC}=0$$

So $\angle B=90^\circ$.

Step 2. Orthocentre and circumcentre

In a right triangle the orthocentre is the right-angle vertex $B(1,1,3)$:

$$OB=\sqrt{1+1+9}=\sqrt{11}$$

The circumcentre is the midpoint of the hypotenuse $AC$:

$$\left(\frac{1+4}{2},\frac{5+1}{2},\frac{3+0}{2}\right)=\left(\frac52,3,\frac32\right)$$

Step 3. Fourth vertex

The three choices are $A+C-B$, $A+B-C$, $B+C-A$:

$$(4,5,0),\quad(-2,5,6),\quad(4,-3,0)$$

Answer: (1), (2), (3), (4)

Direction cosines and direction ratios

If a line makes angles $\alpha$, $\beta$, $\gamma$ with the positive $x$, $y$, $z$ axes, its direction cosines are

$$l=\cos\alpha,\quad m=\cos\beta,\quad n=\cos\gamma$$ A vector OP from the origin making angles alpha, beta and gamma with the x, y and z axes

They always satisfy

$$l^2+m^2+n^2=1$$

Any three numbers $a,b,c$ proportional to $l,m,n$ are direction ratios. To go back, divide by the length:

$$l=\frac{a}{\sqrt{a^2+b^2+c^2}}$$ $$m=\frac{b}{\sqrt{a^2+b^2+c^2}}$$ $$n=\frac{c}{\sqrt{a^2+b^2+c^2}}$$

The direction ratios of the line through $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ are $x_2-x_1$, $y_2-y_1$, $z_2-z_1$.

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

(1) $\dfrac{10}{\sqrt{38}}$ (2) $\dfrac{20}{3\sqrt{38}}$ (3) $\dfrac{10}{7\sqrt{38}}$ (4) $\dfrac{10}{3\sqrt{38}}$

▸ SHOW SOLUTION

Step 1. Eliminate n

$n=4l+m$. Substitute:

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

Step 2. Factor

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

Step 3. The two directions

$m=-8l$ gives $n=-4l$, direction $(1,-8,-4)$.

$m=-\dfrac52l$ gives $n=\dfrac32l$, direction $(2,-5,3)$.

Step 4. Angle

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

Answer: (4)

Equation of a line

A line is fixed by a point $A$ on it and a direction $\vec d=(a,b,c)$.

A line through the point A with direction vector d, and a general point P on it reached as A plus lambda d

Vector form. $\vec r=\vec a+\lambda\vec d$, where $\vec a$ is the position vector of $A$.

Cartesian form. Through $(x_1,y_1,z_1)$ with direction ratios $a,b,c$:

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

Two-point form. Through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$:

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

The general point. Put each fraction equal to $\lambda$:

$$P=(x_1+a\lambda,\ y_1+b\lambda,\ z_1+c\lambda)$$

This one line is the start of nearly every solution in the chapter. If the direction ratios are the direction cosines, $|\lambda|$ is the actual distance $AP$.

The zero denominator

$\dfrac{x-1}{2}=\dfrac{y+3}{0}=\dfrac{z}{5}$ is not division by zero. It means the direction is $(2,0,5)$, so the $y$-coordinate never changes: every point has $y=-3$. In the general point, write $y=-3+0\cdot\lambda$.

Example 3 (JEE Main 2026, 5 April Shift 2). Points $P$ and $Q$ lie on the line

$$\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$$

and are at distance $6$ from $R(1,2,3)$. If $(\alpha,\beta,\gamma)$ is the centroid of $\triangle PQR$, then $\alpha+\beta+\gamma$ is

(1) $4$ (2) $5$ (3) $6$ (4) $8$

▸ SHOW SOLUTION

Step 1. General point

$$(8\lambda-3,\ 2\lambda+4,\ 2\lambda-1)$$

Step 2. Distance 6 from R

$$(8\lambda-4)^2+(2\lambda+2)^2+(2\lambda-4)^2=36$$

Divide by $4$:

$$(4\lambda-2)^2+(\lambda+1)^2+(\lambda-2)^2=9$$$$18\lambda^2-18\lambda=0\implies\lambda=0,\ 1$$

Step 3. The points and the centroid

$P(-3,4,-1)$, $Q(5,6,1)$, $R(1,2,3)$:

$$G=\left(\frac{3}{3},\frac{12}{3},\frac{3}{3}\right)=(1,4,1)$$$$\alpha+\beta+\gamma=6$$

Answer: (3)

Angle between two lines

The angle between two lines is the angle between their directions $\vec d_1=(a_1,b_1,c_1)$ and $\vec d_2=(a_2,b_2,c_2)$:

$$\cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}$$

• Perpendicular: $a_1a_2+b_1b_2+c_1c_2=0$.
• Parallel: $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$.

A line perpendicular to two lines has direction $\vec d_1\times\vec d_2$:

$$\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}$$

Example 4 (JEE Main 2026, 6 April Shift 1). A line $L$ is perpendicular to both

$$L_1:\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$$ $$L_2:\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$$

If $\theta$ is the acute angle between $L$ and a line $L_3$ with direction ratios $2,1,2$, then $\tan\theta$ is

(1) $\dfrac32\sqrt2$ (2) $\dfrac52\sqrt2$ (3) $\dfrac53\sqrt2$ (4) $\dfrac43\sqrt2$

▸ SHOW SOLUTION

Step 1. Direction of L

$$(3,5,7)\times(1,4,7)=(7,-14,7)$$

So $L$ has direction $(1,-2,1)$.

Step 2. Cosine with L₃

$$\cos\theta=\frac{|2-2+2|}{\sqrt6\cdot3}=\frac{2}{3\sqrt6}$$

Step 3. Tangent

$$\cos^2\theta=\frac{4}{54}$$$$\sin^2\theta=\frac{50}{54}$$$$\tan^2\theta=\frac{50}{4}\implies\tan\theta=\frac{5\sqrt2}{2}$$

Answer: (2)

Example 5 (JEE Main 2026, 24 January Shift 1). A line $L$ through $P(1,1,1)$ is perpendicular to

$$\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$$

and

$$\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$$

$L$ meets the $yz$-plane at $Q$. A line parallel to $L$ through $S(1,0,-1)$ meets the $yz$-plane at $R$. The square of the area of the parallelogram $PQRS$ is ______.

▸ SHOW SOLUTION

Step 1. Direction of L

$$(4,1,1)\times(1,1,0)=(-1,1,3)$$

Step 2. Points Q and R

$L$: $(1-t,\ 1+t,\ 1+3t)$. On $x=0$, $t=1$: $Q=(0,2,4)$.

Through $S$: $(1-u,\ u,\ -1+3u)$. On $x=0$, $u=1$: $R=(0,1,2)$.

Step 3. Area

$$\overrightarrow{PQ}=(-1,1,3)$$$$\overrightarrow{PS}=(0,-1,-2)$$$$\overrightarrow{PQ}\times\overrightarrow{PS}=(1,-2,1)$$

Area $=\sqrt6$, so its square is $6$.

Answer: $6$

Foot of the perpendicular and distance from a line

To drop a perpendicular from $P$ to a line with direction $\vec d$:

1. Write the general point $F(\lambda)$ of the line.
2. Impose $\overrightarrow{PF}\cdot\vec d=0$. It is linear in $\lambda$.
3. Solve, then $PF$ is the distance of $P$ from the line.

A point P off a line, the foot F on the line with PF perpendicular to the line, and the image P-prime on the other side with F the midpoint of P P-prime

There is a shortcut for the distance alone. If $A$ is any point of the line,

$$PF^2=AP^2-\left(\frac{\overrightarrow{AP}\cdot\vec d}{|\vec d|}\right)^2$$

because $AF$ is the projection of $AP$ on the line.

Example 6 (JEE Main 2026, 5 April Shift 1). The square of the distance of $P(5,6,7)$ from the line

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

is

(1) $3$ (2) $5$ (3) $6$ (4) $8$

▸ SHOW SOLUTION

Step 1. General point and the perpendicular condition

$$F=(2\lambda+2,\ 3\lambda+5,\ 4\lambda+2)$$$$\overrightarrow{PF}=(2\lambda-3,\ 3\lambda-1,\ 4\lambda-5)$$$$2(2\lambda-3)+3(3\lambda-1)+4(4\lambda-5)=0$$$$29\lambda-29=0\implies\lambda=1$$

Step 2. Distance

$F=(4,8,6)$:

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

Answer: (3)

Example 7 (JEE Main 2026, 2 April Shift 2). $A$ is the foot of the perpendicular from $P(a,b,0)$ to the line

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

If the midpoint of $PA$ is $\left(0,\dfrac34,-\dfrac14\right)$, then $a^2+b^2+\alpha^2$ is

(1) $1$ (2) $2$ (3) $6$ (4) $9$

▸ SHOW SOLUTION

Step 1. Midpoint conditions

$A=(2r+1,\ r+2,\ 3r+\alpha)$. The midpoint of $PA$ gives

$$a=-1-2r$$$$b=-\frac12-r$$$$\alpha=-\frac12-3r$$

Step 2. Perpendicular condition

$$\overrightarrow{PA}=\left(4r+2,\ 2r+\frac52,\ -\frac12\right)$$$$2(4r+2)+\left(2r+\frac52\right)-\frac32=0$$$$10r+5=0\implies r=-\frac12$$

Step 3. Values

$$a=0,\quad b=0,\quad\alpha=1$$$$a^2+b^2+\alpha^2=1$$

Answer: (1)

Example 8 (JEE Main 2026, 23 January Shift 1). Vertices $B$ and $C$ of $\triangle ABC$ lie on

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

$A=(1,6,3)$, $B=(4,9,\alpha)$ and $BC=10$. The area of $\triangle ABC$ is

(1) $5\sqrt{13}$ (2) $15\sqrt{13}$ (3) $20\sqrt{13}$ (4) $10\sqrt{13}$

▸ SHOW SOLUTION

Step 1. Rewrite the line

$\dfrac{1-y}{-2}=\dfrac{y-1}{2}$, so the line is $(\lambda,\ 2\lambda+1,\ 3\lambda+2)$ with direction $(1,2,3)$.

Step 2. Height from A

Foot $D$: $(\lambda-1,\ 2\lambda-5,\ 3\lambda-1)\cdot(1,2,3)=0$:

$$14\lambda-14=0\implies D=(1,3,5)$$$$AD=\sqrt{0+9+4}=\sqrt{13}$$

Step 3. Area

$$\frac12\cdot BC\cdot AD=\frac12\cdot10\cdot\sqrt{13}=5\sqrt{13}$$

Answer: (1)

$\alpha$ never mattered. The height from $A$ is the same for every base on that line.

Example 9 (JEE Advanced 2026, Paper 2). $L$ is the line through $P(1,2,-1)$ and $Q(2,3,1)$. $S$ is the foot of the perpendicular from $R(4,-1,5)$ to $L$. Another line through $R$ meets $L$ at $T$ such that $S$ divides $PT$ internally with $PS:ST=1:2$. Which is (are) TRUE?

(1) The orthocentre of $\triangle PRT$ is $\left(\dfrac{23}{5},-4,\dfrac{31}{5}\right)$

(2) The orthocentre of $\triangle PRT$ is $(4,3,5)$

(3) The area of $\triangle PRT$ is $6\sqrt5$

(4) The area of $\triangle PRT$ is $18\sqrt5$

▸ SHOW SOLUTION

Step 1. Foot S

$\vec d=(1,1,2)$, $\overrightarrow{PR}=(3,-3,6)$.

$$\overrightarrow{PS}=\frac{\overrightarrow{PR}\cdot\vec d}{|\vec d|^2}\,\vec d=\frac{12}{6}\vec d=(2,2,4)$$

$S=(3,4,3)$.

Step 2. Point T

$S$ divides $PT$ as $1:2$, so $S=\dfrac{2P+T}{3}$:

$$T=3S-2P=(7,8,11)$$

Step 3. Area

$RS\perp PT$, so $RS$ is the height:

$$RS=|(-1,5,-2)|=\sqrt{30}$$$$PT=|(6,6,12)|=6\sqrt6$$$$\text{Area}=\frac12\cdot6\sqrt6\cdot\sqrt{30}=18\sqrt5$$

Step 4. Orthocentre

$H$ lies on the altitude $RS$: $H=R+u(-1,5,-2)$. It also lies on the altitude from $P$, so $\overrightarrow{PH}\perp\overrightarrow{RT}=(3,9,6)$:

$$(3-u,\ 5u-3,\ 6-2u)\cdot(1,3,2)=0$$$$6+10u=0\implies u=-\frac35$$$$H=\left(\frac{23}{5},-4,\frac{31}{5}\right)$$

Answer: (1), (4)

Image of a point in a line

The image $P'$ of $P$ in a line is the point with the foot $F$ as the midpoint of $PP'$:

$$P'=2F-P$$

Find the foot first, then double. When the image is given with unknowns, use the two conditions directly: the midpoint of $PP'$ lies on the line, and $\overrightarrow{PP'}$ is perpendicular to the line.

Example 10 (JEE Main 2026, 22 January Shift 1). If the image of $P(1,2,a)$ in the line

$$\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$$

is $Q(5,b,c)$, then $a^2+b^2+c^2$ is

(1) $293$ (2) $264$ (3) $298$ (4) $283$

▸ SHOW SOLUTION

Step 1. Read the line correctly

Point $(6,7,7)$, direction $(3,2,-2)$.

Step 2. Midpoint on the line

$M=\left(3,\ \dfrac{b+2}{2},\ \dfrac{a+c}{2}\right)$. The $x$-coordinate gives parameter $\dfrac{3-6}{3}=-1$:

$$\frac{b+2}{2}=7-2=5\implies b=8$$$$\frac{a+c}{2}=7+2=9\implies a+c=18$$

Step 3. Perpendicular

$$(4,\ 6,\ c-a)\cdot(3,2,-2)=0$$$$24-2(c-a)=0\implies c-a=12$$

So $c=15$, $a=3$.

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

Answer: (3)

Example 11 (JEE Main 2026, 4 April Shift 2). If $\left(2\alpha+1,\ \alpha^2-3\alpha,\ \dfrac{\alpha-1}{2}\right)$ is the image of $(\alpha,\ 2\alpha,\ 1)$ in the line

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

then the possible value(s) of $\alpha$ is (are)

(1) only $3$ (2) only $3$ and $-1$ (3) only $3$, $\dfrac14$ and $-1$ (4) only $3$ and $\dfrac14$

▸ SHOW SOLUTION

Step 1. Midpoint

$$M=\left(\frac{3\alpha+1}{2},\ \frac{\alpha^2-\alpha}{2},\ \frac{\alpha+1}{4}\right)$$

Step 2. M on the line

$$\frac{\alpha-1}{2}=\frac{\alpha^2-\alpha-2}{4}=\frac{\alpha+1}{4}$$

First equals third: $2\alpha-2=\alpha+1$, so $\alpha=3$. The middle one then gives $\dfrac{4}{4}=1$, consistent.

Step 3. Check the perpendicular

$\alpha=3$: from $(3,6,1)$ to $(7,0,1)$ is $(4,-6,0)$, and $(4,-6,0)\cdot(3,2,1)=0$.

Answer: (1)

The middle equation alone also allows $\alpha=-1$, but that value fails the other two. Every coordinate condition must hold, not just one.

Example 12 (JEE Main 2026, 28 January Shift 2). $Q(a,b,c)$ is the image of $P(3,2,1)$ in the line $\dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$. The distance of $Q$ from the line

$$\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$$

is

(1) $6$ (2) $8$ (3) $7$ (4) $5$

▸ SHOW SOLUTION

Step 1. Foot and image

$N=(r+1,\ 2r,\ r+1)$:

$$(r-2)+2(2r-2)+r=0\implies r=1$$

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

Step 2. Distance by the shortcut

$A(9,9,5)$, $\vec d=(3,2,-2)$, $\overrightarrow{AQ}=(-8,-7,-2)$:

$$AQ^2=64+49+4=117$$$$\frac{(\overrightarrow{AQ}\cdot\vec d)^2}{|\vec d|^2}=\frac{(-34)^2}{17}=68$$$$\text{distance}^2=117-68=49$$

Answer: (3) $7$

Example 13 (JEE Main 2026, 6 April Shift 2). $R$ is the image of $P(0,-5,0)$ in the line $\dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{-2}$, and $S$ is the image of $Q\left(0,-\dfrac12,0\right)$ in the line $\dfrac{x-1}{-1}=\dfrac{y+9}{4}=\dfrac{z+1}{1}$. The square of the area of the parallelogram $PQRS$ is ______.

▸ SHOW SOLUTION

Step 1. Image of P

Foot $(1+2k,\ k,\ -1-2k)$ with $(1+2k,\ k+5,\ -1-2k)\cdot(2,1,-2)=0$:

$$9k+9=0\implies k=-1$$

Foot $F=(-1,-1,1)$, so $R=(-2,3,2)$.

Step 2. Image of Q

Foot $(1-\lambda,\ -9+4\lambda,\ -1+\lambda)$ with the perpendicular condition:

$$18\lambda-36=0\implies\lambda=2$$

The foot is again $F=(-1,-1,1)$, so $S=\left(-2,-\dfrac32,2\right)$.

Step 3. Area from the diagonals

$PR$ and $QS$ have the same midpoint $F$, so they are the diagonals.

$$\overrightarrow{PR}=(-2,8,2)$$$$\overrightarrow{QS}=(-2,-1,2)$$$$\overrightarrow{PR}\times\overrightarrow{QS}=(18,0,18)$$$$\text{Area}=\frac12\cdot18\sqrt2=9\sqrt2$$

Answer: $162$

Distance along a given line

"The distance of $P$ from the line $L$ along the line $M$" is not the perpendicular distance. It is the length $PQ$, where $Q$ is the point at which the line through $P$ parallel to $M$ meets $L$.

A point P, a line L, the perpendicular from P to L shown faint, and a slanting segment from P in a given direction meeting L at Q

1. Write the line through $P$ with the given direction: $P+t\vec d$.
2. Make that point satisfy $L$.
3. The answer is $|t|\,|\vec d|$.

Example 14 (JEE Main 2026, 4 April Shift 1). The square of the distance of $(-2,-8,6)$ from the line

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

along the line $\dfrac{x+5}{1}=\dfrac{y+5}{-1}=\dfrac{z}{2}$ is

(1) $3$ (2) $6$ (3) $8$ (4) $12$

▸ SHOW SOLUTION

Step 1. Line through the point along (1, −1, 2)

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

Step 2. B on the first line

$$\frac{t-3}{1}=\frac{-9-t}{2}=\frac{6+2t}{-1}$$

First equals third: $t-3=-6-2t$, so $t=-1$. The middle gives $-4$, consistent.

Step 3. Distance

$$PB^2=t^2(1+1+4)=6$$

Answer: (2)

Example 15 (JEE Main 2026, 28 January Shift 1). The distances of $(1,2,a)$ from the line $\dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$ along the lines

$$L_1:\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$$ $$L_2:\frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$$

are equal. Then $a+b+c$ is

(1) $7$ (2) $5$ (3) $6$ (4) $4$

▸ SHOW SOLUTION

Step 1. Along L₁

$A=(3\lambda+1,\ 4\lambda+2,\ b\lambda+a)$ must satisfy the line:

$$3\lambda=\frac{4\lambda+2}{2}\implies\lambda=1$$

Then $3=b+a-1$, so $a+b=4$, and $A=(4,6,4)$.

Step 2. Along L₂

$B=(\mu+1,\ 4\mu+2,\ c\mu+a)$:

$$\mu=\frac{4\mu+2}{2}\implies\mu=-1$$

Then $-1=a-c-1$, so $c=a$, and $B=(0,-2,0)$.

Step 3. Equal distances

$$9+16+(a-4)^2=1+16+a^2$$$$41-8a=17\implies a=3$$

$b=1$, $c=3$, so $a+b+c=7$.

Answer: (1)

Example 16 (JEE Main 2026, 28 January Shift 2). The distance of $P(43,\alpha,\beta)$, $\beta<0$, from the line $\vec r=4\hat i-\hat k+\mu(2\hat i+3\hat k)$ along a line with direction ratios $3,-1,0$ is $13\sqrt{10}$. Then $\alpha^2+\beta^2$ is ______.

▸ SHOW SOLUTION

Step 1. Meet the given line

$P+t(3,-1,0)=(43+3t,\ \alpha-t,\ \beta)$ must equal $(4+2\mu,\ 0,\ -1+3\mu)$:

$$\alpha=t$$$$2\mu=39+3t$$$$\beta=3\mu-1$$

So $\beta=\dfrac{115+9t}{2}$.

Step 2. Use the distance

$$|t|\sqrt{10}=13\sqrt{10}\implies t=\pm13$$

$t=13$ gives $\beta=116>0$, rejected. $t=-13$ gives $\alpha=-13$, $\beta=-1$.

$$\alpha^2+\beta^2=169+1=170$$

Answer: $170$

The sign of $\alpha$ is fixed by $\beta<0$: $\alpha=-13$, not $13$. The final answer $170$ is the same either way, which is why the slip is easy to miss.

Example 17 (JEE Main 2026, 22 January Shift 2). $L$ is the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$, and $S$ is the set of points on $L$ whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along $L$ is $7$. The sum of $a+b+c$ over all $(a,b,c)\in S$ is

(1) $34$ (2) $28$ (3) $40$ (4) $6$

▸ SHOW SOLUTION

Step 1. Where L meets the second line

The zero denominator says the second line lies in $z=9$. On $L$, $(2t-1,\ 3t-1,\ 6t-3)$, so $6t-3=9$ gives $t=2$:

$$M=(3,5,9)$$

Step 2. Points at distance 7 along L

$|\vec d|=\sqrt{4+9+36}=7$, so a step of $1$ in $t$ is a distance of $7$: $t=1$ or $t=3$.

$$(1,2,3)$$$$(5,8,15)$$

Step 3. Add

$$6+28=34$$

Answer: (1)

As printed the second line read $\dfrac{z+9}{0}$, which gives no option; the intended line is $z=9$, used above.

Skew lines and the shortest distance

Two lines in space can intersect, be parallel, or be skew: neither parallel nor meeting. Between skew lines there is exactly one segment perpendicular to both, and its length is the shortest distance.

Two skew lines, the rear one broken where the front one passes over it, with the common perpendicular segment joining them

For $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$:

$$\text{SD}=\frac{\left|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\right|}{|\vec b_1\times\vec b_2|}$$

The numerator is the volume of the box on $\vec a_2-\vec a_1$, $\vec b_1$, $\vec b_2$, the denominator is its base, so SD is its height.

For parallel lines with common direction $\vec b$:

$$d=\frac{\left|(\vec a_2-\vec a_1)\times\vec b\right|}{|\vec b|}$$

Example 18 (JEE Main 2026, 4 April Shift 2). The shortest distance between

$$\vec r=\frac13\hat i+2\hat j+\frac83\hat k+\lambda(2\hat i-5\hat j+6\hat k)$$ $$\vec r=-\frac23\hat i-\frac13\hat k+\mu(\hat j-\hat k)$$

is

(1) $\sqrt5$ (2) $3$ (3) $2\sqrt3$ (4) $\sqrt{15}$

▸ SHOW SOLUTION

Step 1. Cross product of directions

$$(2,-5,6)\times(0,1,-1)=(-1,2,2)$$

with length $3$.

Step 2. Joining vector

$$\vec a_1-\vec a_2=(1,2,3)$$

Step 3. SD

$$\frac{|-1+4+6|}{3}=3$$

Answer: (2)

Example 19 (JEE Main 2026, 24 January Shift 2). The sum of all values of $\alpha$ for which the shortest distance between

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

and

$$\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha}$$

is $\sqrt2$, is

(1) $8$ (2) $-6$ (3) $6$ (4) $-8$

▸ SHOW SOLUTION

Step 1. Cross product

With $\vec d_1=(\alpha,-1,-\alpha)$ and $\vec d_2=(\alpha,2,2\alpha)$:

$$\vec d_1\times\vec d_2=(0,\ -3\alpha^2,\ 3\alpha)$$

with length $3|\alpha|\sqrt{\alpha^2+1}$.

Step 2. SD

Joining vector $(1,-1,-3)$:

$$\text{SD}=\frac{|3\alpha^2-9\alpha|}{3|\alpha|\sqrt{\alpha^2+1}}=\frac{|\alpha-3|}{\sqrt{\alpha^2+1}}$$

Step 3. Solve

$$(\alpha-3)^2=2(\alpha^2+1)$$$$\alpha^2+6\alpha-7=0$$

Sum of roots $=-6$.

Answer: (2)

Example 20 (JEE Main 2026, 22 January Shift 1). $P(\alpha,\beta,\gamma)$ is the point on $\dfrac{x-1}{2}=\dfrac{y+1}{-3}=z$ at distance $4\sqrt{14}$ from $(1,-1,0)$, nearer to the origin. The shortest distance between

$$\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$$

and

$$\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$$

is

(1) $7\sqrt{\dfrac54}$ (2) $4\sqrt{\dfrac75}$ (3) $4\sqrt{\dfrac57}$ (4) $2\sqrt{\dfrac74}$

▸ SHOW SOLUTION

Step 1. The point P

$P=(2\lambda+1,\ -3\lambda-1,\ \lambda)$ with $14\lambda^2=224$, so $\lambda=\pm4$.

$$(9,-13,4):\ OP^2=266$$$$(-7,11,-4):\ OP^2=186$$

So $P=(-7,11,-4)$.

Step 2. Cross product

$$(1,2,3)\times(2,1,1)=(-1,5,-3)$$

with length $\sqrt{35}$.

Step 3. SD

Joining vector $(2,-1,7)$:

$$\text{SD}=\frac{|-2-5-21|}{\sqrt{35}}=\frac{28}{\sqrt{35}}=4\sqrt{\frac75}$$

Answer: (2)

Example 21 (JEE Main 2026, 8 April Shift 2). The foot of the perpendicular from $(\lambda,2,3)$ to $\dfrac{x-4}{1}=\dfrac{y-9}{2}=\dfrac{z-5}{1}$ is $(1,\mu,2)$. The distance between the lines

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$$

and

$$\frac{x-\lambda}{2}=\frac{y-\mu}{3}=\frac{z+5}{6}$$

is

(1) $\dfrac{12}{7}$ (2) $\dfrac{\sqrt{145}}{7}$ (3) $\dfrac{\sqrt{146}}{7}$ (4) $\dfrac{\sqrt{143}}{7}$

▸ SHOW SOLUTION

Step 1. Find μ and λ

$(1,\mu,2)$ is on the line at parameter $-3$, so $\mu=9-6=3$.

Perpendicular: $(1-\lambda)+2(1)+(-1)=0$, so $\lambda=2$.

Step 2. Parallel lines

Both have direction $(2,3,6)$. Joining vector $(1,1,-1)$:

$$(1,1,-1)\times(2,3,6)=(9,-8,1)$$$$d=\frac{\sqrt{81+64+1}}{7}=\frac{\sqrt{146}}{7}$$

Answer: (3)

Lines that meet, and transversals

Two lines meet when one parameter on each makes the general points equal. That gives three equations in two unknowns: solve two, and the third must check. If it fails, the lines are skew.

A transversal is a line meeting two given lines. If its direction is given, take a general point $P$ on the first line and $Q$ on the second, and make $\overrightarrow{PQ}$ parallel to the given direction.

Example 22 (JEE Main 2026, 2 April Shift 1). The point of intersection of

$$\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}$$

and

$$\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}$$

lies on the $xy$-plane. Then $a+b$ is

(1) $2$ (2) $5$ (3) $7$ (4) $9$

▸ SHOW SOLUTION

Step 1. General points

$$(3r-1,\ 5r-a,\ 7r-b-1)$$$$(s+2,\ 4s+b,\ 7s+2a)$$

Step 2. Four equations

$z=0$ on both: $b=7r-1$, $a=-\dfrac72s$. Equal $x$: $s=3r-3$.

Step 3. Equal y

$$5r-a=4s+b$$$$5r+\frac{21}{2}r-\frac{21}{2}=12r-12+7r-1$$$$7r=5\implies r=\frac57$$

Step 4. Values

$b=4$, $s=-\dfrac67$, $a=3$:

$$a+b=7$$

Answer: (3)

Example 23 (JEE Main 2026, 4 April Shift 1). A line with direction ratios $1,-1,2$ meets

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

and

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

at $P$ and $Q$. If $PQ=\alpha$, then $225\alpha^2$ is

(1) $1024$ (2) $1014$ (3) $1104$ (4) $1204$

▸ SHOW SOLUTION

Step 1. General points

$$P=(2\lambda,\ 3\lambda,\ 3\lambda-1)$$$$Q=(-\mu-1,\ \mu+2,\ 4\mu)$$

The components of $\overrightarrow{QP}$ are

$$2\lambda+\mu+1$$$$3\lambda-\mu-2$$$$3\lambda-4\mu-1$$

Step 2. Parallel to (1, −1, 2)

First and second components are opposite:

$$5\lambda=1\implies\lambda=\frac15$$

Third is twice the first:

$$\lambda+6\mu=-3\implies\mu=-\frac{8}{15}$$

Step 3. Length

$$\overrightarrow{QP}=\frac{13}{15}(1,-1,2)$$$$225\alpha^2=169\cdot6=1014$$

Answer: (2)

Example 24 (JEE Main 2026, 21 January Shift 2). $L_1$ is parallel to $-3\hat i+2\hat j+4\hat k$ through $(2,6,7)$, and $L_2$ is parallel to $2\hat i+\hat j+3\hat k$ through $(4,3,5)$. A line $L_3$ parallel to $-3\hat i+5\hat j+16\hat k$ meets $L_1$ at $C$ and $L_2$ at $D$. Then $CD^2$ is

(1) $171$ (2) $290$ (3) $312$ (4) $89$

▸ SHOW SOLUTION

Step 1. General points

$$C=(2-3p,\ 6+2p,\ 7+4p)$$$$D=(4+2q,\ 3+q,\ 5+3q)$$

Step 2. CD parallel to (−3, 5, 16)

$$\frac{2q+3p+2}{-3}=\frac{q-2p-3}{5}$$$$\frac{q-2p-3}{5}=\frac{3q-4p-2}{16}$$$$13q+9p+1=0$$$$q-12p-38=0$$$$p=-3$$$$q=2$$

Step 3. Length

$C=(11,0,-5)$, $D=(8,5,11)$:

$$CD^2=9+25+256=290$$

Answer: (2)

Here $\overrightarrow{CD}$ came out exactly equal to the given direction, a quick check that the arithmetic is right.

The plane

A plane is fixed by a point $A$ on it and a normal $\vec n=(a,b,c)$ perpendicular to every line in it.

A plane drawn as a parallelogram, a point A on it, the normal vector n standing perpendicular at A, and a general point P on the plane

Through three points as a determinant:

$$\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$$

Angle between planes is the angle between their normals:

$$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1|\,|\vec n_2|}$$ Two planes meeting along a line, with their normals n1 and n2 and the angle theta between the planes marked

Distance of a point $(x_1,y_1,z_1)$ from $ax+by+cz+d=0$:

$$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$$

Distance between parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$:

$$\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$$

Example 25 (IITianForum). Find the plane through $A(1,1,0)$, $B(1,2,1)$, $C(-2,2,-1)$, its intercepts on the axes, and its distance from the origin.

▸ SHOW SOLUTION

Step 1. Normal

$$\overrightarrow{AB}=(0,1,1)$$$$\overrightarrow{AC}=(-3,1,-1)$$$$\overrightarrow{AB}\times\overrightarrow{AC}=(-2,-3,3)$$

Step 2. Equation

Normal $(2,3,-3)$ through $A$:

$$2x+3y-3z=5$$

Check $B$: $2+6-3=5$. Check $C$: $-4+6+3=5$.

Step 3. Intercept form

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

Step 4. Distance from O

$$\frac{5}{\sqrt{4+9+9}}=\frac{5}{\sqrt{22}}$$

Answer: $2x+3y-3z=5$; intercepts $\dfrac52$, $\dfrac53$, $-\dfrac53$; distance $\dfrac{5}{\sqrt{22}}$.

Example 26 (JEE Advanced 2026, Paper 1). $P$ is the plane containing the line $\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{1}$ and perpendicular to the plane $x+2y+3z=4$. $P_1$ is the plane through $(4,2,2)$ parallel to $P$. Which is (are) TRUE?

(1) $P$ is $7x-5y+z=-10$

(2) The distance between $P$ and $P_1$ is $30$

(3) The distance of $P$ from the origin is $2\sqrt3$

(4) The acute angle between $P$ and $2x+2y+z=3$ is $\cos^{-1}\left(\dfrac{1}{3\sqrt3}\right)$

▸ SHOW SOLUTION

Step 1. Normal of P

It is perpendicular to the line direction $(2,3,1)$ and to the normal $(1,2,3)$:

$$(2,3,1)\times(1,2,3)=(7,-5,1)$$

Step 2. Equation of P

Through $(1,3,-2)$:

$$7x-5y+z=7-15-2=-10$$

So (1) is true.

Step 3. P₁ and the distances

$P_1$: $7x-5y+z=28-10+2=20$. $|\vec n|=\sqrt{75}=5\sqrt3$.

$$\text{between }P,P_1:\ \frac{30}{5\sqrt3}=2\sqrt3$$$$\text{from }O:\ \frac{10}{5\sqrt3}=\frac{2}{\sqrt3}$$

So (2) and (3) are false.

Step 4. Angle

$$\cos\theta=\frac{|14-10+1|}{5\sqrt3\cdot3}=\frac{1}{3\sqrt3}$$

Answer: (1), (4)

A line and a plane

For a line with direction $\vec d$ and a plane with normal $\vec n$:

Angle. The line makes angle $\varphi$ with the plane, and $90^\circ-\varphi$ with the normal:

$$\sin\varphi=\frac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}$$ A line meeting a plane at M, the normal at M, the projection of the line on the plane, and the angle phi between the line and its projection

Intersection. Put the general point of the line into the plane. It is one linear equation in the parameter.

Projection of a line on a plane. It passes through the meeting point $M$, and its direction is $\vec d$ with the normal part removed:

$$\vec d\,'=\vec d-\frac{\vec d\cdot\vec n}{|\vec n|^2}\,\vec n$$

Example 27 (IITianForum). Find where the line $\dfrac{x-1}{2}=\dfrac{y-2}{-1}=\dfrac{z-3}{2}$ meets the plane $2x+y-2z+9=0$, and the sine of the angle between them.

▸ SHOW SOLUTION

Step 1. Substitute the general point

$(1+2t,\ 2-t,\ 3+2t)$:

$$2+4t+2-t-6-4t+9=0$$$$7-t=0\implies t=7$$

The point is $(15,-5,17)$.

Step 2. Angle

$$\sin\varphi=\frac{|4-1-4|}{3\cdot3}=\frac19$$

Answer: $(15,-5,17)$ and $\sin\varphi=\dfrac19$.

Example 28 (IITianForum). The planes $x+y+z=2$ and $x-y+z=0$ meet in a line $L$. The distance of $L$ from the plane $2x+y+2z=9$ is

(1) $1$ (2) $2$ (3) $3$ (4) $\sqrt3$

▸ SHOW SOLUTION

Step 1. Direction of L

It is perpendicular to both normals:

$$(1,1,1)\times(1,-1,1)=(2,0,-2)$$

Step 2. L is parallel to the third plane

$$(2,0,-2)\cdot(2,1,2)=0$$

So every point of $L$ is at the same distance.

Step 3. One point of L

$y=1$ from subtracting the planes; take $x=1$, $z=0$: $(1,1,0)$.

$$\frac{|2+1+0-9|}{3}=2$$

Answer: (2)

Example 29 (IITianForum). Find the projection of the line $\dfrac{x-2}{1}=\dfrac{y}{2}=\dfrac{z-1}{0}$ on the plane $x+y+z=6$.

▸ SHOW SOLUTION

Step 1. Meeting point

$(2+t,\ 2t,\ 1)$ in the plane: $3+3t=6$, so $t=1$.

$$M=(3,2,1)$$

Step 2. Remove the normal part

$\vec d=(1,2,0)$, $\vec n=(1,1,1)$, $\vec d\cdot\vec n=3$, $|\vec n|^2=3$:

$$\vec d\,'=(1,2,0)-(1,1,1)=(0,1,-1)$$

Step 3. The projection

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

Answer: the line through $(3,2,1)$ with direction $(0,1,-1)$.

Check: $(0,1,-1)\cdot(1,1,1)=0$, so the projection lies in the plane.

Foot and image of a point in a plane

From $P(x_1,y_1,z_1)$ to $ax+by+cz+d=0$, move along the normal:

$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=k$$ $$\text{foot: }k=-\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$

For the image, double it:

$$\text{image: }k=-2\,\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$ A point P above a plane, its foot F on the plane and its image P-prime the same distance below, all on one normal line

Image of a line in a plane. If the line meets the plane at $M$, its image passes through $M$ and through the image of any other point of the line.

Reflection of a ray. A ray with direction $\vec d$ reflected by a plane mirror with normal $\vec n$ leaves along

$$\vec d\,'=\vec d-2\,\frac{\vec d\cdot\vec n}{|\vec n|^2}\,\vec n$$

Example 30 (IITianForum). A plane cuts the axes at $2$, $4$ and $4$. If $(\alpha,\beta,\gamma)$ is the image of the origin in it, find $3(\alpha+\beta+\gamma)$.

▸ SHOW SOLUTION

Step 1. The plane

$$\frac x2+\frac y4+\frac z4=1\implies2x+y+z=4$$

Step 2. Image of O

$$k=-2\cdot\frac{0-4}{6}=\frac43$$$$(\alpha,\beta,\gamma)=\frac43(2,1,1)=\left(\frac83,\frac43,\frac43\right)$$$$3(\alpha+\beta+\gamma)=8+4+4=16$$

Answer: $16$

Example 31 (IITianForum). A ray from $P(2,1,3)$ travelling parallel to $(1,2,-6)$ strikes the mirror $x+y+z=0$ at $Q$ and reflects along $QR$. Find $Q$, the direction of $QR$, and the plane $PQR$.

▸ SHOW SOLUTION

Step 1. Point Q

$(2+t,\ 1+2t,\ 3-6t)$ in the plane:

$$6-3t=0\implies t=2$$$$Q=(4,5,-9)$$

Step 2. Reflected direction

$\vec d\cdot\vec n=1+2-6=-3$, $|\vec n|^2=3$:

$$\vec d\,'=(1,2,-6)+2(1,1,1)=(3,4,-4)$$

Step 3. Plane PQR

It contains both directions:

$$(1,2,-6)\times(3,4,-4)=(16,-14,-2)$$

Normal $(8,-7,-1)$ through $P$:

$$8x-7y-z=6$$

Check $Q$: $32-35+9=6$.

Answer: $Q(4,5,-9)$, $QR$ along $(3,4,-4)$, plane $8x-7y-z=6$.

The plane of incidence always contains the normal too: $(8,-7,-1)\cdot(1,1,1)=0$.

Example 32 (IITianForum). Find the image of the line $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{1}$ in the plane $x+y+z=2$, and the plane containing the image that is perpendicular to the mirror.

▸ SHOW SOLUTION

Step 1. Where the line meets the mirror

$(1+2t,\ 2+t,\ 3+t)$: $6+4t=2$, so $t=-1$.

$$M=(-1,1,2)$$

Step 2. Image of A(1, 2, 3)

$$k=-2\cdot\frac{6-2}{3}=-\frac83$$$$A'=\left(-\frac53,-\frac23,\frac13\right)$$

Step 3. Image line

$$\overrightarrow{MA'}=-\frac13(2,5,5)$$$$\frac{x+1}{2}=\frac{y-1}{5}=\frac{z-2}{5}$$

Step 4. The perpendicular plane

Normal $(2,5,5)\times(1,1,1)=(0,3,-3)$, through $M$:

$$y-z+1=0$$

Answer: image $\dfrac{x+1}{2}=\dfrac{y-1}{5}=\dfrac{z-2}{5}$; plane $y-z+1=0$.

The original line lies in the same plane: $A(1,2,3)$ gives $2-3+1=0$. A line and its image always share the plane perpendicular to the mirror.

Planes through a line, and coplanar lines

Every plane through the line of intersection of $P_1=0$ and $P_2=0$ is

$$P_1+kP_2=0$$

for some $k$ (and $P_2=0$ itself). One extra condition fixes $k$.

Several planes fanning out around a common line like the pages of an open book

A plane containing a line through $A$ with direction $\vec d$ has a normal perpendicular to $\vec d$. With a second line, or a point $B$, the normal is $\vec d_1\times\vec d_2$ or $\vec d\times\overrightarrow{AB}$.

Bisector planes. The planes bisecting the angles between $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$ are

Write $P_1=a_1x+b_1y+c_1z+d_1$ and $P_2=a_2x+b_2y+c_2z+d_2$. The bisectors are

$$\frac{P_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm\frac{P_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$$

With $d_1$, $d_2$ of the same sign, the $+$ sign gives the bisector of the angle containing the origin. That angle is acute when $\vec n_1\cdot\vec n_2<0$ and obtuse when $\vec n_1\cdot\vec n_2>0$.

Coplanar lines. Two lines lie in one plane exactly when

$$(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0$$

and that plane has normal $\vec d_1\times\vec d_2$ (for non-parallel lines).

Example 33 (IITianForum). $L$ is the line through $A(1,2,1)$ with direction $(1,0,-1)$. $P_1$ contains $L$ and has normal parallel to $(1,2,1)$. $P_2$ contains $L$ and the point $B(0,1,0)$. Find $P_1$, $P_2$ and the cosine of the acute angle between them.

▸ SHOW SOLUTION

Step 1. P₁

$(1,2,1)\cdot(1,0,-1)=0$, so such a plane exists. Through $A$:

$$x+2y+z=6$$

Step 2. P₂

$$\overrightarrow{AB}=(-1,-1,-1)$$$$\vec d\times\overrightarrow{AB}=(-1,2,-1)$$

Through $A$:

$$x-2y+z+2=0$$

Check $B$: $0-2+0+2=0$.

Step 3. Angle

$$\cos\theta=\frac{|1-4+1|}{\sqrt6\cdot\sqrt6}=\frac13$$

Answer: $x+2y+z=6$, $x-2y+z+2=0$, $\cos\theta=\dfrac13$.

Example 34 (IITianForum). Among the planes containing the line $x+2y=5,\ z=2$, find the one farthest from the origin.

▸ SHOW SOLUTION

Step 1. Foot of O on the line

The line is $(5-2t,\ t,\ 2)$ with direction $(-2,1,0)$:

$$(5-2t)(-2)+t=0\implies t=2$$$$F=(1,2,2)$$$$OF=3$$

Step 2. Why OF is the answer

Every plane through the line contains $F$, so its distance from $O$ is at most $OF$. The distance equals $OF$ exactly when $OF$ is the normal.

Step 3. The plane

Normal $(1,2,2)$ through $F$:

$$x+2y+2z=9$$

Answer: $x+2y+2z=9$, at distance $3$.

As a member of the family $(x+2y-5)+k(z-2)=0$ this is $k=2$, and the distance $\dfrac{|5+2k|}{\sqrt{5+k^2}}$ peaks there.

Example 35 (IITianForum). Find the plane bisecting the acute angle between $2x-y+2z=3$ and $x+2y+2z=6$.

▸ SHOW SOLUTION

Step 1. Both normals have length 3

$$\frac{2x-y+2z-3}{3}=\pm\frac{x+2y+2z-6}{3}$$

Step 2. The two bisectors

$$+:\ x-3y+3=0$$$$-:\ 3x+y+4z-9=0$$

Step 3. Which is acute

$d_1=-3$ and $d_2=-6$ have the same sign, and $\vec n_1\cdot\vec n_2=2-2+4=4>0$, so the $+$ sign bisects the obtuse angle. The acute bisector is

$$3x+y+4z=9$$

Check: its angle with the first plane has cosine $\dfrac{6-1+8}{3\sqrt{26}}=\dfrac{\sqrt{26}}{6}$, which is $\cos\dfrac\theta2$ for $\cos\theta=\dfrac49$.

Answer: $3x+y+4z=9$

Example 36 (IITianForum). Find $k$ so that

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

and

$$\frac{x-2}{1}=\frac{y-k}{2}=\frac{z-6}{3}$$

are coplanar, and find the plane containing them.

▸ SHOW SOLUTION

Step 1. Cross product of directions

$$(2,3,4)\times(1,2,3)=(1,-2,1)$$

Step 2. Coplanarity

$$(1,\ k-2,\ 3)\cdot(1,-2,1)=0$$$$8-2k=0\implies k=4$$

Step 3. The plane

Normal $(1,-2,1)$ through $(1,2,3)$:

$$x-2y+z=0$$

Check $(2,4,6)$: $2-8+6=0$.

Answer: $k=4$; plane $x-2y+z=0$.

Example 37 (IITianForum). Find the shortest distance between the $z$-axis and the line $x+y+z=1,\ x+2y+3z=6$.

▸ SHOW SOLUTION

Step 1. The line as point and direction

$$(1,1,1)\times(1,2,3)=(1,-2,1)$$

Put $z=0$: $x+y=1$, $x+2y=6$, so $(-4,5,0)$.

Step 2. SD with the z-axis

The $z$-axis: point $O$, direction $(0,0,1)$.

$$(1,-2,1)\times(0,0,1)=(-2,-1,0)$$$$\text{SD}=\frac{|(-4,5,0)\cdot(-2,-1,0)|}{\sqrt5}=\frac{3}{\sqrt5}$$

Answer: $\dfrac{3}{\sqrt5}$

Three planes

Three planes with normals $\vec n_1$, $\vec n_2$, $\vec n_3$ can meet in one point, in a common line, or not at all.

Three panels: three planes meeting at a single point, three planes sharing one common line, and three planes forming a triangular prism

The determinant test is the Determinants chapter again: the three equations have a unique solution exactly when the coefficient determinant is non-zero.

Example 38 (IITianForum). Consider $x+y+z=3$, $x-y+2z=1$ and $3x+y+4z=k$. Which is (are) correct?

(1) For $k=7$ the three planes have a common line

(2) For $k=9$ they form a triangular prism

(3) For some $k$ they meet in exactly one point

(4) For $k=7$ the common line is parallel to $(3,-1,-2)$

▸ SHOW SOLUTION

Step 1. Determinant of the normals

$$\begin{vmatrix}1&1&1\\1&-1&2\\3&1&4\end{vmatrix}=0$$

So (3) is false for every $k$.

Step 2. The combination

$$(3,1,4)=2(1,1,1)+(1,-1,2)$$

The third plane is $2P_1+P_2$ exactly when $k=2\cdot3+1=7$. Then all three share a line; for $k\ne7$ no point lies on all three, and since no two normals are parallel, they form a prism.

Step 3. Direction of the common line

$$(1,1,1)\times(1,-1,2)=(3,-1,-2)$$

Answer: (1), (2), (4)

Which result to use

A decision chart: from the object in the question, arrows lead to the general point of a line, the foot condition, the cross product, the SD formula, the plane through three points, the point to plane distance, or the family of planes

What JEE asks now

In the IITianForum syllabus the JEE Main unit covers everything up to and including lines that meet (with the coplanarity determinant) and the reflection of a line in a plane as a basic idea. The plane sections belong to the JEE Advanced unit.

In JEE Main 2026 the line questions fell into five types, in this order of frequency: foot and image in a line, shortest distance between lines, distance along a direction, lines that meet, and the line perpendicular to two lines. Every one of these types has a worked example above.

Quick recall

Practise these methods on real exam questions in the doMath Three Dimensional Geometry chapter, and read the Vectors note alongside this one: the dot product, the cross product and the scalar triple product do all the work here.