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 and Intuition

For two skew (non-intersecting, non-parallel) lines, the shortest distance is the length of the projection of the vector joining any two points on the lines onto the common perpendicular direction $\vec d_1\times\vec d_2$.

Step-by-Step Solution

1. Line 1: point $A=(1,-1,0)$, direction $\vec d_1=(0,1,2)$ (coefficients of $s$).
2. Line 2: point $B=(2,0,1)$, direction $\vec d_2=(1,-1,1)$ (coefficients of $t$).
3. $\vec d_1\times\vec d_2=\begin{vmatrix}\vec i&\vec j&\vec k\\0&1&2\\1&-1&1\end{vmatrix}=\vec i(1\cdot1-2\cdot(-1))-\vec j(0\cdot1-2\cdot1)+\vec k(0\cdot(-1)-1\cdot1)=(3,2,-1)$.
4. $|\vec d_1\times\vec d_2|=\sqrt{9+4+1}=\sqrt{14}$.
5. $B-A=(1,1,1)$; $(B-A)\cdot(3,2,-1)=3+2-1=4$.
6. Shortest distance $=\dfrac{|4|}{\sqrt{14}}=\dfrac{4}{\sqrt{14}}=\sqrt{\dfrac{16}{14}}=\sqrt{\dfrac{8}{7}}$.

Common Mistakes

• Using the wrong base points (must read them off correctly as the constant vector in each line's equation, not the direction vector).
• Forgetting to take absolute value of the scalar triple product before dividing.

ANSWER: D

Concept and Intuition

The shortest distance between two skew lines is measured along the unique common perpendicular to both direction vectors. If $\bar d_1\times\bar d_2$ gives that common perpendicular direction, then projecting the vector joining any two points on the lines onto this direction gives the distance:

Step-by-Step Solution

1. Read off the data: Line 1 passes through $\bar a_1=2\bar i-\bar j=(2,-1,0)$ with direction $\bar d_1=\bar i+2\bar k=(1,0,2)$. Line 2 passes through $\bar a_2=-2\bar i+\bar k=(-2,0,1)$ with direction $\bar d_2=\bar i-\bar j-\bar k=(1,-1,-1)$.
2. Compute $\bar d_1\times\bar d_2 = \begin{vmatrix}\bar i&\bar j&\bar k\\1&0&2\\1&-1&-1\end{vmatrix} = \bar i(0\cdot(-1)-2\cdot(-1)) - \bar j(1\cdot(-1)-2\cdot1) + \bar k(1\cdot(-1)-0\cdot1) = (2,3,-1)$.
3. $|\bar d_1\times\bar d_2| = \sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$.
4. $\bar a_2-\bar a_1 = (-2-2,\,0-(-1),\,1-0) = (-4,1,1)$.
5. Dot product: $(-4)(2)+(1)(3)+(1)(-1) = -8+3-1=-6$.
6. Distance $= \dfrac{|-6|}{\sqrt{14}} = \dfrac{6}{\sqrt{14}}$. Rationalising differently: $\dfrac{6}{\sqrt{14}} = \dfrac{6}{\sqrt2\cdot\sqrt7}=\dfrac{6}{\sqrt2\sqrt7}\cdot\dfrac{\sqrt2}{\sqrt2}=\dfrac{6\sqrt2}{2\sqrt7}=\dfrac{3\sqrt2}{\sqrt7}$, matching option (A) exactly (numerically both are $\approx 1.604$).

Common Mistakes

• Using $\bar a_1-\bar a_2$ instead of $\bar a_2-\bar a_1$ — harmless here since we take the absolute value, but easy to mess up the sign in intermediate steps.
• Mis-expanding the $2\times2$ minors in the cross product (sign errors on the $\bar j$ component are the most common slip).
• Leaving the answer as $6/\sqrt{14}$ and failing to recognise it matches the $3\sqrt2/\sqrt7$ form offered.

ANSWER: A

Concept and Intuition

Two skew lines have a unique common perpendicular direction, given by $\bar d_1\times\bar d_2$. The shortest distance between them is the length of the projection of the vector joining any point on one line to any point on the other, onto this common perpendicular direction.

Step-by-Step Solution

1. Line 1: point $\bar a_1=(-1,-2,-3)$, direction $\bar d_1=(3,-2,-2)$.
2. Line 2: point $\bar a_2=(7,0,4)$, direction $\bar d_2=(1,-2,2)$.
3. $\bar d_1\times\bar d_2 = \begin{vmatrix}\bar i&\bar j&\bar k\\3&-2&-2\\1&-2&2\end{vmatrix} = \bar i(-4-4)-\bar j(6+2)+\bar k(-6+2) = -8\bar i-8\bar j-4\bar k$.
4. $|\bar d_1\times\bar d_2|=\sqrt{64+64+16}=\sqrt{144}=12$.
5. $\bar a_2-\bar a_1=(8,2,7)$. Dot with $(-8,-8,-4)$: $8(-8)+2(-8)+7(-4)=-64-16-28=-108$.
6. Distance $=\dfrac{|-108|}{12}=9$.

Common Mistakes

• Sign errors in the $2\times2$ cofactor expansion of the cross product.
• Using the wrong pair of points (must be one point from each line, not both from the same line).

ANSWER: C

Concept and Intuition

For two lines $\bar r=\bar a+t\bar b$ and $\bar r=\bar p+s\bar q$, the vector $\bar b\times\bar q$ is perpendicular to both direction vectors, so it points along the common perpendicular between the lines. Projecting the vector $(\bar a-\bar p)$ joining a point on each line onto this common-perpendicular direction gives the shortest distance between the lines:

The lines are coplanar exactly when this shortest distance is zero, i.e. when $(\bar a-\bar p)\cdot(\bar b\times\bar q)=0$ — not when it's nonzero.

Step-by-Step Solution

1. Recall the coplanarity criterion: lines are coplanar $\iff (\bar a-\bar p)\cdot(\bar b\times\bar q)=0$.
2. Assertion (A) states the lines are coplanar when this quantity is nonzero — this is the exact opposite of the correct criterion (a nonzero value signals skew lines). So (A) is false.
3. Reason (R) restates the identity $|(\bar a-\bar p)\cdot(\bar b\times\bar q)| = |\bar b\times\bar q|\cdot d$, which follows directly by rearranging the shortest-distance formula above. This identity holds unconditionally (it's even consistent with $d=0$ giving the dot product $=0$), so (R) is true.
4. Since (A) is false and (R) is true, the matching option is (D).

Common Mistakes

• Confusing "nonzero triple product" with "coplanar" instead of "skew".
• Assuming a true Reason automatically validates a false Assertion — in assertion-reason questions each statement is checked independently first.

ANSWER: D

Concept and Intuition

Two lines in space that don't intersect and aren't parallel are "skew," and the shortest segment between them is perpendicular to both simultaneously — i.e. along their common perpendicular direction $\vec d_1\times\vec d_2$. The formula projects the vector connecting any point on line 1 to any point on line 2 onto this common-perpendicular unit vector; the magnitude of that projection is the shortest distance.

Step-by-Step Solution

1. Position vectors: $A=\bar i+\bar j-\bar k$, $B=-\bar i+2\bar j+\bar k$, $C=\bar j+2\bar k$, $D=2\bar i-\bar j+2\bar k$.
2. Direction of line $AB$: $\vec d_1=B-A=(-1-1,\,2-1,\,1-(-1))=(-2,1,2)$.
3. Direction of line $CD$: $\vec d_2=D-C=(2-0,\,-1-1,\,2-2)=(2,-2,0)$.
4. Connecting vector: $C-A=(0-1,\,1-1,\,2-(-1))=(-1,0,3)$.
5. Cross product $\vec d_1\times\vec d_2$:

6. $|\vec d_1\times\vec d_2|=\sqrt{4^2+4^2+2^2}=\sqrt{36}=6$.
7. $(C-A)\cdot(\vec d_1\times\vec d_2)=(-1)(4)+(0)(4)+(3)(2)=-4+0+6=2$.
8. Shortest distance $=\dfrac{|2|}{6}=\dfrac13$.

Common Mistakes

• Using the wrong pair of points for the connecting vector (must connect a point on line 1 to a point on line 2, e.g. $A$ to $C$, not $A$ to $D$).
• Sign errors in the $2\times2$ cofactor expansions of the cross product.

ANSWER: C

Concept and Intuition

Two lines (given in symmetric form) are coplanar exactly when the vector joining a point on each is coplanar with (i.e., has zero scalar triple product with) their direction vectors. This yields a linear equation in $\lambda$. Once $\lambda$ is a concrete number (in radians), $\sin^{-1}(\sin\lambda)$ and $\cos^{-1}(\cos\lambda)$ must be reduced to their principal-value ranges using the periodicity/reflection rules, since $\lambda=4$ radians lies outside $[-\pi/2,\pi/2]$ and $[0,\pi]$ respectively.

Step-by-Step Solution

1. Points: $P_1=(3,2,1)$ on line 1, $P_2=(2,3,2)$ on line 2. Directions: $d_1=(2,3,\lambda)$, $d_2=(3,2,3)$.
2. Coplanarity condition: $(P_2-P_1)\cdot(d_1\times d_2)=0$, where $P_2-P_1=(-1,1,1)$.
3. $d_1\times d_2 = \left(3\cdot3-\lambda\cdot2,\ -(2\cdot3-\lambda\cdot3),\ 2\cdot2-3\cdot3\right) = (9-2\lambda,\ 3\lambda-6,\ -5)$.
4. Dot with $(-1,1,1)$: $-1(9-2\lambda)+1(3\lambda-6)+1(-5) = -9+2\lambda+3\lambda-6-5 = 5\lambda-20$.
5. Set to zero: $5\lambda-20=0 \Rightarrow \lambda=4$ (radians, since it plays the role of an angle argument next).
6. Since $\pi/2\le4\le3\pi/2$ (i.e. $1.57\le4\le4.71$), $\sin^{-1}(\sin4)=\pi-4$.
7. Since $\pi\le4\le2\pi$ (i.e. $3.14\le4\le6.28$), $\cos^{-1}(\cos4)=2\pi-4$.
8. Sum: $(\pi-4)+(2\pi-4)=3\pi-8$.

Common Mistakes

• Treating $\lambda=4$ as already inside the principal domain of $\sin^{-1}$/$\cos^{-1}$ and writing the sum as simply $2\cdot4=8$ or similar — the reduction formulas are essential here since $4$ radians ($\approx229^\circ$) is well outside both principal ranges.
• Sign errors in the cross product components.

ANSWER: C