A vector of magnitude 4 which is equally Inclined to the vectors $\hat i+\hat j,\hat j+\hat k$ and $\hat k + \hat i$, is |
$\frac{4}{\sqrt{3}}(\hat i-\hat j-\hat k)$ $\frac{4}{\sqrt{3}}(\hat i+\hat j-\hat k)$ $\frac{4}{\sqrt{3}}(\hat i+\hat j+\hat k)$ none of these |
$\frac{4}{\sqrt{3}}(\hat i+\hat j+\hat k)$ |
Let the required vector be $\vec r = x\hat i+y\hat j+z\hat k$. Then, $|\vec r|=4⇒ x^2+ y^2+z^2=16$ ...(i) Now, $\vec r$ is equally inclined to the vectors $\hat i+\hat j,\hat j+\hat k$ and $\hat k + \hat i$. $∴\frac{\vec r.(\hat i+\hat j)}{|\vec r|\sqrt{2}}=\frac{\vec r.(\hat j+\hat k)}{|\vec r|\sqrt{2}}=\frac{\vec r.(\hat k+\hat i)}{|\vec r|\sqrt{2}}$ $⇒x + y = y + z = z + x = λ (say)$ $⇒2(x + y + z) = 3λ ⇒ x+y+z=\frac{3λ}{2}$ Now, $x + y = λ$ and $x + y + z=\frac{3λ}{2}⇒z=\frac{λ}{2}$ Similarly, we have $x = y =\frac{λ}{2}$ Substituting these values in (i), we get $λ=±\frac{8}{\sqrt{3}}$ Hence, $\vec r=±\frac{8}{2\sqrt{3}}(\hat i+\hat j+\hat k)=±\frac{4}{\sqrt{3}}(\hat i+\hat j+\hat k)$ |