Question

\(\vec{r}\) is a vector perpendicular to the plane determined by \(\vec{r_1}=2\mathbf{i}-\mathbf{j}\) and \(\vec{r_2}=\mathbf{j}+2\mathbf{k}\). If the magnitude of the projection of \(\vec{r}\) on the vector \(2\mathbf{i}+\mathbf{j}+2\mathbf{k}\) is 1, then \(\lvert \vec{r}\rvert =\,?\)

• A. \(\sqrt{6}\)
• B. \(3\sqrt{6}\)
• C. \(\frac{2\sqrt{6}}{3}\)
• D. \(\frac{3\sqrt{6}}{2}\)

Hint:

A vector perpendicular to a plane is parallel to the cross product of two direction vectors in the plane.
- For projection magnitude, use \(\frac{|\mathbf{u}\cdot\mathbf{v}|}{|\mathbf{v}|}\).

Answer 1

The Correct Option is D

Step 1: A vector perpendicular to the plane of \(\vec{r_1}, \vec{r_2}\).<
A suitable vector \(\vec{r}\) is parallel to \(\vec{r_1}\times \vec{r_2}\). First find \[ \vec{r_1}\times \vec{r_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
2 -1 0
0 1 2 \end{vmatrix}. \] Compute: \[ = \mathbf{i}\bigl((-1)\cdot2 -0\cdot1\bigr) -\mathbf{j}\bigl(2\cdot2 -0\cdot0\bigr) +\mathbf{k}\bigl(2\cdot1 -(-1)\cdot0\bigr). \] \[ = \mathbf{i}\,(-2) -\mathbf{j}\,(4) +\mathbf{k}\,(2). \] So \(\vec{r_1}\times \vec{r_2} = -2\,\mathbf{i} -4\,\mathbf{j} + 2\,\mathbf{k}.\) Step 2: Let \(\vec{r}\) be a scalar multiple.
Thus \(\vec{r}=\lambda\,(-2\,\mathbf{i} -4\,\mathbf{j} +2\,\mathbf{k}).\) Step 3: Projection on \(\vec{v}=2\mathbf{i}+\mathbf{j}+2\mathbf{k}\).
The magnitude of the projection of \(\vec{r}\) onto \(\vec{v}\) is \[ \text{proj}_{\vec{v}}(\vec{r}) = \frac{\lvert \vec{r}\cdot \vec{v}\rvert}{\lvert\vec{v}\rvert}. \] We’re given this is \(1\). Now \[ \vec{r}\cdot \vec{v} = \lambda\,(-2\,\mathbf{i} -4\,\mathbf{j} +2\,\mathbf{k}) \,\cdot\,(2\,\mathbf{i}+\mathbf{j}+2\,\mathbf{k}) \] \[ = \lambda\Bigl[ (-2)\cdot2 +(-4)\cdot1 + (2)\cdot2 \Bigr] = \lambda\Bigl[-4 -4 +4\Bigr] = \lambda\cdot(-4). \] So \[ |\vec{r}\cdot \vec{v}|= |\,\lambda\cdot(-4)\,|=4|\lambda|. \] Next, \(\lvert \vec{v}\rvert=\sqrt{2^2 +1^2 +2^2}=\sqrt{4+1+4}= \sqrt{9}=3.\) Hence: \[ \text{proj}_{\vec{v}}(\vec{r}) = \frac{4|\lambda|}{3} =1 \;\;\Longrightarrow\;\; |\lambda| =\frac{3}{4}. \] We can assume \(\lambda >0\) if direction is not specifically reversed. So \(\lambda=\tfrac{3}{4}\). Step 4: \(\lvert \vec{r}\rvert\).
\[ \vec{r}=\tfrac34(-2\,\mathbf{i}-4\,\mathbf{j}+2\,\mathbf{k}). \] Hence \[ \lvert \vec{r}\rvert = \left|\tfrac{3}{4}\right|\;\sqrt{(-2)^2+(-4)^2+(2)^2} = \frac{3}{4}\;\sqrt{4+16+4} = \frac{3}{4}\;\sqrt{24} = \frac{3}{4}\times 2\sqrt{6} = \frac{3\sqrt{6}}{2}. \] Therefore \(\boxed{\tfrac{3\sqrt{6}}{2}}\).