Practicing Success
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 |
$\hat i -3\hat j+ 3\hat k$ $-3\hat i -3\hat j-\hat k$ $3\hat i -\hat j+3\hat k$ $\hat i +3\hat j-3\hat k$ |
$3\hat i -\hat j+3\hat k$ |
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 $\vec r =3\hat i -\hat j+3\hat k$. |