Let $\vec a=2\hat i-\hat j+\hat k,\vec b=\hat i+2\hat j-\hat k$ and $\vec c=\hat i+\hat j-2\hat k$. A vector in the plane of $\vec b$ and $\vec c$ whose projection on $\vec a$ has the magnitude $\sqrt{\frac{2}{3}}$ is
Answer & explanation
Correct answer: option 3
Let $\vec r$ be the required vector. Then, $\vec r =x\vec b+y\vec c$.
We have,
Projection of $\vec r$ on $\vec a =\sqrt{\frac{2}{3}}$
$⇒\vec r.\hat a=\sqrt{\frac{2}{3}}$
$⇒(x\vec b+y\vec c).\hat a=\sqrt{\frac{2}{3}}$
$⇒x(\vec b.\hat a)+y(\vec c.\hat a)=\sqrt{\frac{2}{3}}$
$⇒-\frac{x}{\sqrt{6}}-\frac{y}{\sqrt{6}}=\sqrt{\frac{2}{3}}⇒x+y=-2⇒y= -x-2$
Now,
$\vec r=x\vec b+y\vec c$
$⇒\vec r= (x + y)\hat i + (2x + y)\hat j -(x+2y)\hat k$
$⇒\vec r=-2\hat i+(x-2)\hat j+(x+4)\hat k$
Clearly, option (3) satisfies this for x = 1.