Question

Consider the following
Assertion (A): The two lines $\vec{r} = \vec{a}+t(\vec{b})$ and $\vec{r}=\vec{b}+s(\vec{a})$ intersect each other.
Reason (R): The shortest distance between the lines $\vec{r}=\vec{p}+t(\vec{q})$ and $\vec{r}=\vec{c}+s(\vec{d})$ is equal to the length of projection of the vector $(\vec{p}-\vec{c})$ on $(\vec{q}\times\vec{d})$.
The correct answer is

• A. Both (A) and (R) are true and (R) is the correct explanation of (A)
• B. Both (A) and (R) are true and (R) is not the correct explanation of (A)
• C. (A) is true, but (R) is false
• D. (A) is false, but (R) is true

Hint:

Two lines $\vec{r}=\vec{a}+t\vec{b}$ and $\vec{r}=\vec{c}+s\vec{d}$ intersect if the shortest distance between them is zero. This is equivalent to the condition that the vectors $(\vec{a}-\vec{c})$, $\vec{b}$, and $\vec{d}$ are coplanar, which means their scalar triple product is zero: $[(\vec{a}-\vec{c}) \ \vec{b} \ \vec{d}] = 0$.

Answer 1

The Correct Option is A

Step 1: Analyze Reason (R).
The formula for the shortest distance between two skew lines $\vec{r}=\vec{p}+t(\vec{q})$ and $\vec{r}=\vec{c}+s(\vec{d})$ is given by $SD = \frac{|(\vec{p}-\vec{c})\cdot(\vec{q}\times\vec{d})|}{|\vec{q}\times\vec{d}|}$.
The expression represents the magnitude of the projection of the vector $(\vec{p}-\vec{c})$ onto the vector $(\vec{q}\times\vec{d})$ (which is the common normal to both lines).
So, Reason (R) is a correct statement of the formula for the shortest distance between two skew lines.

Step 2: Analyze Assertion (A) using the concept from Reason (R).
For two lines to intersect, the shortest distance between them must be zero.
Let the two lines be $L_1: \vec{r} = \vec{a}+t(\vec{b})$ and $L_2: \vec{r}=\vec{b}+s(\vec{a})$.
Here, $\vec{p}=\vec{a}, \vec{q}=\vec{b}, \vec{c}=\vec{b}, \vec{d}=\vec{a}$.

Step 3: Calculate the shortest distance for the lines in Assertion (A).
The condition for intersection is that the numerator of the shortest distance formula is zero.
We need to check if $(\vec{p}-\vec{c})\cdot(\vec{q}\times\vec{d}) = 0$.
Substituting the vectors: $(\vec{a}-\vec{b})\cdot(\vec{b}\times\vec{a})$.
This is a scalar triple product. We know that the scalar triple product $[\vec{u} \ \vec{v} \ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$ is zero if any two vectors are the same or if the vectors are coplanar.
Let $\vec{u} = \vec{a}-\vec{b}, \vec{v}=\vec{b}, \vec{w}=\vec{a}$.
The scalar triple product is $(\vec{a}-\vec{b})\cdot(\vec{b}\times\vec{a}) = \vec{a}\cdot(\vec{b}\times\vec{a}) - \vec{b}\cdot(\vec{b}\times\vec{a})$.
The term $\vec{a}\cdot(\vec{b}\times\vec{a})$ is the scalar triple product $[\vec{a} \ \vec{b} \ \vec{a}]$, which is zero because a vector is repeated.
The term $\vec{b}\cdot(\vec{b}\times\vec{a})$ is the scalar triple product $[\vec{b} \ \vec{b} \ \vec{a}]$, which is zero because a vector is repeated.
So, the result is $0 - 0 = 0$.

Step 4: Conclude about Assertion (A) and its relation to Reason (R).
Since the shortest distance between the lines is zero, the lines are coplanar and must intersect (unless they are parallel, which they are not, as $\vec{a}$ and $\vec{b}$ are not stated to be parallel).
Thus, Assertion (A) is true.
Reason (R) provides the formula and concept for the shortest distance. The condition for intersection (A) is that this shortest distance is zero. We used the formula from (R) to prove (A).
Therefore, both (A) and (R) are true, and (R) is the correct explanation of (A).