Let $\vec a =\hat i+\hat j+\hat k, \vec b=\hat i-\hat j+\hat k$ and $\vec c=\hat i-\hat j-\hat k$ be three vectors. A vector $\vec r$ in the plane of $\vec a$ and $\vec b$, whose projection on $\vec c$ is $\frac{1}{\sqrt{3}}$, is given by
Answer & explanation
Correct answer: option 3
Let $\vec r$ be the required vector. Then,
$\vec r = x\vec a+y\vec b$
$⇒\vec r=(x+y)\hat i+(x-y)\hat j+(x+y)\hat k$
It is given that the projection of $\vec r$ on $\vec c$ is $\frac{1}{\sqrt{3}}$.
$∴\vec r.\frac{1}{\sqrt{3}}(\hat i-\hat j-\hat k)=\frac{1}{\sqrt{3}}$
$⇒x+y-(x-y)-(x + y) = 1$
$⇒-x+y=1$
$⇒y=x+1$
$∴\vec r= (2x+1) \hat i−\hat j + (2x+1)\hat k, x∈ R$
For $x = 1$, we get $\vec r =3\hat i -\hat j+3\hat k$.