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: The normal vector to a plane \[ \vec r\cdot \vec n=d \] is \(\vec n\). A vector perpendicular to the normals of two planes must be perpendicular to both normal vectors. Such a vector is obtained using the cross product of the normals. \[ \vec a\times \vec b \] gives a vector perpendicular to both \(\vec a\) and \(\vec b\).

Step 1: Write the normal vectors of the two planes. For the planes \[ \vec r\cdot(2\hat{i}+\hat{j}-\hat{k})=3, \] and \[ \vec r\cdot(6\hat{i}+3\hat{j}+2\hat{k})=4, \] the normal vectors 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: Find a vector perpendicular to both normals. \[ \vec n_1\times \vec n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & 1 & -1 6 & 3 & 2 \end{vmatrix}. \] \[ = \hat{i}(2+3) -\hat{j}(4+6) +\hat{k}(6-6). \] \[ = 5\hat{i}-10\hat{j}. \] \[ = 5(\hat{i}-2\hat{j}). \] Thus a required direction vector is \[ \hat{i}-2\hat{j}. \]

Step 3: Find the corresponding unit vector. The magnitude of \(\hat{i}-2\hat{j}\) is \[ \sqrt{1^2+(-2)^2} = \sqrt5. \] Hence the unit vectors are \[ \pm\frac{\hat{i}-2\hat{j}}{\sqrt5}. \] Therefore, \[ x=\pm\frac1{\sqrt5}, \qquad y=\mp\frac2{\sqrt5}, \qquad z=0. \]

Step 4: Compute \(x+y+z\). \[ x+y+z = \pm\frac1{\sqrt5} \mp\frac2{\sqrt5}. \] \[ = \pm\frac1{\sqrt5}. \]

Step 5: Write the final answer. \[ \boxed{\pm\frac1{\sqrt5}} \]