Statement-1: Let $\vec r$ be any vector in space. Then, $\vec r=(\vec r.\hat i)\hat i+(\vec r.\hat j)\hat j+(\vec r.\hat k)\hat k$
Statement-2: If $\vec a,\vec b,\vec c$ are three non-coplanar vectors in space and r is vectors in any space, then $\vec r=\left\{\frac{\vec r.\vec a}{|\vec a|^2}\right\}\vec a+\left\{\frac{\vec r.\vec b}{|\vec b|^2}\right\}\vec b+\left\{\frac{\vec r.\vec c}{|\vec c|^2}\right\}\vec c$
Answer & explanation
Correct answer: option 1
We know that any vector in space can be written as a linear combination of three non-coplanar vectors.
$∴\vec r=x\hat a+y\hat b+z\hat c$ ...(i)
where $\hat a,\hat b$ and $\hat c$ are unit vectors along $\vec a,\vec b$ and $\vec c$ respectively.
Clearly, x, y and z are projections of $\vec r$ on $\vec a,\vec b$ and $\vec c$ respectively.
$∴x=\frac{\vec r.\vec a}{|\vec a|},y=\frac{\vec r.\vec b}{|\vec b|}$ and $\frac{\vec r.\vec c}{|\vec c|}$
Substituting the values of x, y, z in (i), we get
$\vec r=\frac{\vec r.\vec a}{|\vec a|}\vec a+\frac{\vec r.\vec b}{|\vec b|}\vec b+\frac{\vec r.\vec c}{|\vec c|}\vec c$
$⇒\vec r=\left\{\frac{\vec r.\vec a}{|\vec a|^2}\right\}\vec a+\left\{\frac{\vec r.\vec b}{|\vec b|^2}\right\}\vec b+\left\{\frac{\vec r.\vec c}{|\vec c|^2}\right\}\vec c$
So, statement-2 is true.
On replacing $\vec a,\vec b$ and $\vec c$ by $\hat i,\hat j$ and $\hat k$, in statement-2, we obtain statement-1.
So, statement-1 is also true and statement-2 is a correct explanation of statement-1.