Correct Answer - Option 1 :
\(\vec{p}\ +\ \vec{r}\ =\ k\vec{q}\)
Concept:
Let \(\rm\vec {a}\) and \(\rm \vec {b}\) be the two vectors,
The dot product of two vectors is given by:
\(\rm \vec{a}.\vec{b} = |\vec{a}|.|\vec{b}| cos θ\)
Cross product of two vectors is given by:
\(\rm \vec{a}× \vec{b} = |\vec{a}|.|\vec{b}| \sin θ\; ̂{n}\)
\(\rm \vec{a}× \vec{b} = -\ \vec{b}× \vec{a}\)
If \({\rm{\vec a\;and\;\vec b}}\) are two vectors parallel to each other then
\({\rm\vec{a} = λ \vec{b}}\) or \(\rm \vec{a} × \vec{b} =0\)
Calculation:
Given that,
\(\vec{p}\times\vec{q}\ =\ \vec{q}\times \vec{r}\)
\(⇒\ \vec{p}\times\vec{q}\ -\ \vec{q}\times \vec{r}\ =\ 0\)
We know that,
\(\rm \vec{a}× \vec{b} = -\ \vec{b}× \vec{a}\)
\(⇒\ \vec{p}\times\vec{q}\ +\ \vec{r}\times \vec{q}\ =\ 0\)
\(⇒\ (\vec{p}\ \ +\ \vec{r})\times \vec{q}\ =\ 0\)
\(⇒\ (\vec{p}\ \ +\ \vec{r})\parallel \vec{q}\ \)
\(\Rightarrow \ \vec{p}\ +\ \vec{r}\ =\ k\vec{q}\)
Where k is any scaler.
Hence, option 1 is correct.
