Q.Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$.

Concept & Intuition

Two lines in 3D that are not parallel and do not intersect are called skew lines. The shortest distance between them is the length of the common perpendicular segment. This length equals the absolute value of the scalar projection of the vector connecting any point on one line to any point on the other line onto the direction perpendicular to both lines — i.e., onto the cross product of their direction vectors.

Step-by-step solution

1. Identify points and direction vectors

   Line 1: $\vec{r} = (8+3\lambda)\hat{i} - (9+16\lambda)\hat{j} + (10+7\lambda)\hat{k}$

   This can be rewritten as:

   $\vec{r} = (8\hat{i} - 9\hat{j} + 10\hat{k}) + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k})$

   So point $A = (8, -9, 10)$ and direction $\vec{d}_1 = (3, -16, 7)$.

   Line 2: $\vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$

   So point $B = (15, 29, 5)$ and direction $\vec{d}_2 = (3, 8, -5)$.

2. Find the vector joining a point on each line

   $\overrightarrow{AB} = B - A = (15-8,\; 29-(-9),\; 5-10) = (7, 38, -5)$.

3. Compute the cross product of the direction vectors

   $\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

   $= \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$

   $= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$

   $= \hat{i}(24) - \hat{j}(-36) + \hat{k}(72)$

   $= (24, 36, 72)$.

4. Find the magnitude of the cross product

   $|\vec{d}_1 \times \vec{d}_2| = \sqrt{24^2 + 36^2 + 72^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84$.

5. Compute the scalar projection of $\overrightarrow{AB}$ onto $\vec{d}_1 \times \vec{d}_2$

   The shortest distance $d = \frac{|\overrightarrow{AB} \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$.

   Dot product: $(7, 38, -5) \cdot (24, 36, 72) = 7\cdot24 + 38\cdot36 + (-5)\cdot72$

   $= 168 + 1368 - 360 = 1176$.

   So $d = \frac{|1176|}{84} = 14$.

We have two lines given in vector parametric form. The first line is

$\vec{r} = (1-t)\hat{\imath} + (t-2)\hat{\jmath} + (3-2t)\hat{k}$,

which can be rewritten as

$\vec{r} = \hat{\imath} - 2\hat{\jmath} + 3\hat{k} + t(-\hat{\imath} + \hat{\jmath} - 2\hat{k})$.

So a point on line $L_1$ is $A(1, -2, 3)$ and its direction vector is $\vec{d}_1 = (-1, 1, -2)$.

The second line is

$\vec{r} = (s+1)\hat{\imath} + (2s-1)\hat{\jmath} - (2s+1)\hat{k}$,

which rewrites as

$\vec{r} = \hat{\imath} - \hat{\jmath} - \hat{k} + s(\hat{\imath} + 2\hat{\jmath} - 2\hat{k})$.

So a point on $L_2$ is $B(1, -1, -1)$ and its direction vector is $\vec{d}_2 = (1, 2, -2)$.

The shortest distance between two skew lines is the length of the common perpendicular segment. The key idea: take any vector connecting a point on one line to a point on the other, then project it onto the direction perpendicular to both lines (the cross product of the direction vectors). The magnitude of that projection is the shortest distance.

1. Find a vector joining the two lines.

   Choose $A(1, -2, 3)$ on $L_1$ and $B(1, -1, -1)$ on $L_2$.

   Then $\overrightarrow{AB} = B - A = (1-1, -1-(-2), -1-3) = (0, 1, -4)$.

2. Find the cross product of the direction vectors.

   $\vec{d}_1 = (-1, 1, -2)$, $\vec{d}_2 = (1, 2, -2)$.

   Compute:

Simplify:

• For $\hat{\imath}$: $(-2) - (-4) = 2$
• For $\hat{\jmath}$: $(-1)(-2) = 2$, $(-2)(1) = -2$, so $2 - (-2) = 4$, but with the minus sign: $-4$
• For $\hat{k}$: $(-2) - 1 = -3$ So $\vec{d}_1 \times \vec{d}_2 = (2, -4, -3)$.

3. The shortest distance formula. The distance $d$ is given by:

This is the magnitude of the projection of $\overrightarrow{AB}$ onto the common perpendicular.

4. Compute the numerator.

   $\overrightarrow{AB} = (0, 1, -4)$, cross product = $(2, -4, -3)$.

   Dot product: $0 \cdot 2 + 1 \cdot (-4) + (-4) \cdot (-3) = 0 - 4 + 12 = 8$.

   Absolute value: $|8| = 8$.

5. Compute the denominator.

   $|\vec{d}_1 \times \vec{d}_2| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.

6. Thus the shortest distance is

A general point on each line:

For the lines to intersect the coordinates must match.

z-coordinate: $-1=-1+3\mu\ \Rightarrow\ \mu=0$, so the point on $L_2$ is $(4,-2,-1)$.

x-coordinate: $1+3\lambda=4\ \Rightarrow\ \lambda=1$.

y-coordinate: $1+t\lambda=-2\ \Rightarrow\ 1+t=-2\ \Rightarrow\ t=-3$.

$L_1$: $(1+2t,\,-1+3t,\,1+4t)$.

$L_2$: $\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}=s\Rightarrow(3+s,\,k+2s,\,s)$.

Match coordinates:

• $z$: $1+4t=s$.
• $x$: $1+2t=3+s\Rightarrow 2t-s=2$. Substitute $s=1+4t$: $2t-(1+4t)=2\Rightarrow-2t-1=2\Rightarrow t=-\tfrac32$, hence $s=1+4(-\tfrac32)=-5$.
• $y$: $-1+3t=k+2s\Rightarrow-1+3(-\tfrac32)=k+2(-5)\Rightarrow-\tfrac{11}{2}=k-10\Rightarrow k=\tfrac92.$

Therefore $2k=9.$

Direction vectors

• Line 1: $\vec d_1=\langle 1,2,3\rangle$ through $(0,2,-3)$.
• Line 2: $\vec d_2=\langle 2,3,4\rangle$ through $(2,6,3)$.

If $\vec d_2=k\vec d_1$, the $x$-components force $k=2$, but then $2\times2=4\ne3$. So the lines are not parallel (and therefore not coincident); they must be either intersecting or skew.

Test for a common point

Parametrise:

From $x$: $\lambda=2+2\mu$.

Substitute into $y$: $2+2(2+2\mu)=6+3\mu\Rightarrow 6+4\mu=6+3\mu\Rightarrow\mu=0,\ \lambda=2$.

Check $z$: $-3+3(2)=3$ and $3+4(0)=3$. The third equation holds.

All three coordinates agree, so the lines meet at $(2,6,3)$.

Concept & Intuition

We have two lines in 3D space. Point P lies on L₁ at a known distance from A, and we need the closest point Q on L₂ to that P. The “closest point” from a point to a line is the foot of the perpendicular — the point on the line that minimizes distance. This is found by setting the vector from a general point on L₂ to P perpendicular to the direction vector of L₂. The key is to first determine P uniquely using the given distance and the x-coordinate condition.

Step-by-step solution

1. Parameterize line L₁ The line is given by

So a general point on L₁ is

2. Use the distance condition from A(2, -1, 1) The vector from A to P is

Its length is

We are told this distance is 6, so

Hence $t = 2$ or $t = -2$.

3. Apply the x-coordinate condition The x-coordinate of P is $2 + 2t$.
   • If $t = 2$, then x = 6 (greater than 2).
   • If $t = -2$, then x = -2 (not greater than 2). So we take $t = 2$. Thus

4. Parameterize line L₂ The line is

So a general point on L₂ is

5. Find the foot of the perpendicular from P to L₂ The direction vector of L₂ is $\mathbf{d} = (1, 2, 2)$. The vector from a point on L₂ to P is

For Q to be the closest point, $\overrightarrow{QP}$ must be perpendicular to $\mathbf{d}$. So their dot product is zero:

Simplify:

Hence $s = 1$.

6. Find Q Substitute $s = 1$ into the parameterization:

[!TLDR]

Set the scalar triple product of the join vector and the two direction vectors to zero; solving gives $\lambda=-2$.

Concept

Lines $\vec r=\vec a_1+s\,\vec d_1$ and $\vec r=\vec a_2+t\,\vec d_2$ are coplanar if and only if $(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0$ (CBSE/NCERT Class 12, Three Dimensional Geometry).

Solution

$L_1$: point $A_1=(1,-3,1)$, direction $\vec d_1=(1,-\lambda,\lambda)$.

$L_2$: point $A_2=(0,1,2)$, direction $\vec d_2=(\tfrac12,1,-1)$.

Join vector $\vec a_2-\vec a_1=(-1,4,1)$.

Coplanarity requires

Expanding along the first row:

[!ANSWER]

(D) $\lambda=-2$

Concept & Intuition

Two lines in 3D intersect if they are not parallel and there exists a point that lies on both. The symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ gives a point $(x_0, y_0, z_0)$ and a direction vector $(a, b, c)$. For intersection, the vector joining a point on one line to a point on the other must lie in the plane spanned by the two direction vectors — equivalently, the scalar triple product of the two direction vectors and the connecting vector must be zero. This is the cleanest algebraic condition.

Step-by-step solution

1. Extract points and direction vectors

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

   Point $P_1 = (1, -1, 1)$, direction $\vec{d}_1 = (2, 3, 4)$.

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

   Point $P_2 = (3, k, 0)$, direction $\vec{d}_2 = (1, 2, 1)$.

2. Form the connecting vector

   $\vec{v} = P_2 - P_1 = (3-1,\; k - (-1),\; 0-1) = (2,\; k+1,\; -1)$.

3. Condition for intersection

   The three vectors $\vec{d}_1, \vec{d}_2, \vec{v}$ must be coplanar. That means their scalar triple product is zero:

4. Compute the cross product $\vec{d}_2 \times \vec{v}$