Question

A unit vector coplanar with \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) is

• A. \(\frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\)
• B. \(\frac{\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{3}}\)
• C. \(\frac{\mathbf{i} + \mathbf{j} + 2\mathbf{k}}{\sqrt{6}}\)
• D. \(\frac{-\mathbf{j} + 2\mathbf{k}}{\sqrt{5}}\)

Hint:

Use linear combination and orthogonality conditions.

Answer 1

The Correct Option is A

To solve the given problem, we need to find a unit vector that is coplanar with the vectors \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), and is perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\).

Step 1: Find the cross product to get the normal vector of the plane.

The cross product of the two vectors \(\mathbf{A} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{B} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}\) will give us a vector perpendicular to the plane containing these vectors.

Calculate the cross product \(\mathbf{A} \times \mathbf{B}\):

\(\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{vmatrix}\)

Using the determinant formula, calculate:

\(\mathbf{A} \times \mathbf{B} = \mathbf{i}(1 \cdot 1 - 2 \cdot 2) - \mathbf{j}(1 \cdot 1 - 2 \cdot 2) + \mathbf{k}(1 \cdot 2 - 1 \cdot 1)\)

This simplifies to:

\(= \mathbf{i}(1 - 4) - \mathbf{j}(1 - 2) + \mathbf{k}(2 - 1)\)

Therefore, the normal vector is:

\(= -3\mathbf{i} + \mathbf{j} + \mathbf{k}\)

Step 2: Find the vector perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) and coplanar with \(\mathbf{A}\) and \(\mathbf{B}\).

Let the required vector be \(\mathbf{X} = a\mathbf{A} + b\mathbf{B}\) such that:

1. \(\mathbf{X}\) is perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\\) This implies \((\mathbf{X} \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0)\).

Express \(\mathbf{X}\) in terms of components:

\(\mathbf{X} = a(1\mathbf{i} + 1\mathbf{j} + 2\mathbf{k}) + b(1\mathbf{i} + 2\mathbf{j} + 1\mathbf{k})\)

Expand and simplify:

\(= (a+b)\mathbf{i} + (a+2b)\mathbf{j} + (2a+b)\mathbf{k}\)

Impose the condition for perpendicularity:

\(((a+b) + (a+2b) + (2a+b) = 0)\)

This yields:

\((4a + 4b = 0 \Rightarrow a + b = 0)\)

Substituting \(b = -a\) in \(\mathbf{X}\):

\(\mathbf{X} = a - a + (a - 2a)\mathbf{j} + (2a - a)\mathbf{k}\)

Simplifying gives:

\(= -a\mathbf{j} + a\mathbf{k}\)

Step 3: Normalize to get the unit vector.

The unit vector in the direction of \(-a\mathbf{j} + a\mathbf{k}\) is

\(\frac{-\mathbf{j} + \mathbf{k}}{\sqrt{1^2 + 1^2}} = \frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\)

The correct option is \(\frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\).