Let $\vec a=\hat i+2\hat j+\hat k, \vec b =\hat i-\hat j+\hat k, \vec c=\hat i+\hat j-\hat k$. A vector coplanar to $\vec a$ and $\vec b$ has a projection along $\vec c$ of magnitude $1/\sqrt{3}$, then the vector is
Answer & explanation
Correct answer: option 1
A vector $\vec r$ coplanar to $\vec a$ and $\vec b$ is given by
$\vec r=\vec a+λ\vec b$
$⇒\vec r=(\hat i+2\hat j+\hat k)+λ(\hat i-\hat j+\hat k)$
$⇒\vec r=(λ+1)\hat i+(2-λ)\hat j+(λ+1)\hat k$
It is given that the projection of $\vec r$ along $\vec c$ is $\frac{1}{\sqrt{3}}$
$⇒|2-λ|=1⇒2-λ=±1⇒λ=1,3$
For $λ = 1$, we have $\vec r=2\hat i+\hat j+2\hat k$
For $λ = 3$, we have $\vec r=4\hat i -\hat j+4\hat k$