Question

If the unit vector which is perpendicular to the normals drawn to the planes \[ \vec r\cdot(2\hat{i}+\hat{j}-\hat{k})=3 \] and \[ \vec r\cdot(6\hat{i}+3\hat{j}+2\hat{k})=4 \] is \[ x\hat{i}+y\hat{j}+z\hat{k}, \] then \(x+y+z=\)

• A. \( \pm\dfrac{3}{\sqrt5} \)
• B. \(0\)
• C. \( \pm\dfrac{1}{\sqrt5} \)
• D. \( \pm1 \) \bigskip

Hint:

For planes of the form \[ \vec r\cdot\vec n=d, \] the vector \(\vec n\) is the normal. A vector perpendicular to two normals is obtained using the cross product: \[ \vec n_1\times\vec n_2. \] To get a unit vector, divide by its magnitude.

Answer 1

The Correct Option is C

Concept: A line common to both planes must be perpendicular to both their normal vectors. Find the cross product of the normals to get a direction vector, then convert to a unit vector and sum its components.

Step 1: Extract normals from the plane equations. For \(\vec r\cdot(2\hat{i}+\hat{j}-\hat{k})=3\) and \(\vec r\cdot(6\hat{i}+3\hat{j}+2\hat{k})=4\), the normals are \(\vec n_1=2\hat{i}+\hat{j}-\hat{k}\), \(\vec n_2=6\hat{i}+3\hat{j}+2\hat{k}\).

Step 2: Compute the cross product for a direction vector. \(\vec n_1\times\vec n_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\6&3&2\end{vmatrix} = \hat{i}(2+3) - \hat{j}(4+6) + \hat{k}(6-6) = 5\hat{i}-10\hat{j} = 5(\hat{i}-2\hat{j})\). A direction vector is \(\hat{i}-2\hat{j}\).

Step 3: Find the unit vector and its components. Magnitude = \(\sqrt{1^2+(-2)^2}=\sqrt5\). Unit vectors: \(\pm\frac{\hat{i}-2\hat{j}}{\sqrt5}\). So \(x=\pm\frac1{\sqrt5}, y=\mp\frac2{\sqrt5}, z=0\).

Step 4: Sum the components. \(x+y+z = \pm\frac1{\sqrt5} \mp\frac2{\sqrt5} = \pm\frac1{\sqrt5}\).

Step 5: Write the final answer. \(\boxed{\pm\frac1{\sqrt5}}\)