A vector of magnitude 4 which is equally inclined to $\hat i+\hat j,\hat j +\hat k$ and $\hat k +\hat i$, is
Answer & explanation
Correct answer: option 3
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, $\vec r=±\frac{8}{2\sqrt{3}}(\hat i+\hat j+\hat k)=±\frac{4}{\sqrt{3}}(\hat i+\hat j+\hat k)$