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