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Question: If \(\mathbf{r}.\mathbf{i} = \mathbf{r}.\mathbf{j} = \mathbf{r}.\mathbf{k}\) and \(|\mathbf{r}| = 3,...

If r.i=r.j=r.k\mathbf{r}.\mathbf{i} = \mathbf{r}.\mathbf{j} = \mathbf{r}.\mathbf{k} and ∣r∣=3,|\mathbf{r}| = 3, then r=\mathbf{r} =

A

±3(i+j+k)\pm 3(\mathbf{i} + \mathbf{j} + \mathbf{k})

B

±13(i+j+k)\pm \frac{1}{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})

C

±13(i+j+k)\pm \frac{1}{\sqrt{3}}(\mathbf{i} + \mathbf{j} + \mathbf{k})

D

±3(i+j+k)\pm \sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})

Answer

±3(i+j+k)\pm \sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})

Explanation

Solution

Let r=xi+yj+zk.\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}. Since r.i=r.j=r.k\mathbf{r}.\mathbf{i} = \mathbf{r}.\mathbf{j} = \mathbf{r}.\mathbf{k}

⇒x=y=z\Rightarrow x = y = z .....(i)

Also ∣r∣=x2+y2+z2=3⇒x=±3|\mathbf{r}| = \sqrt{x^{2} + y^{2} + z^{2}} = 3 \Rightarrow x = \pm \sqrt{3}, {By (i)}

Hence the required vector r=±3(i+j+k).\mathbf{r} = \pm \sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k}).

Trick : As the vector ±3(i+j+k)\pm \sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k}) satisfies both the conditions.