Given `a=i+j-k, b=-i+2j+k" and "c=-i+2j-k`. A unit vector perpendicular to both a + b and b + c is

Options

A. `(2i+j+k)/sqrt(6)`
B. j
C. k
D. `(i+j+k)/sqrt(3)`

Answer

To find a unit vector that is perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{b} + \mathbf{c} \), we can follow these steps:

### Step 1: Calculate \( \mathbf{a} + \mathbf{b} \)

Given:
\[ \mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k} \]
\[ \mathbf{b} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k} \]

Now, we can add \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{a} + \mathbf{b} = (\mathbf{i} + \mathbf{j} - \mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \]
Combining like terms:
\[ = (\mathbf{i} - \mathbf{i}) + (\mathbf{j} + 2\mathbf{j}) + (-\mathbf{k} + \mathbf{k}) = 0 + 3\mathbf{j} + 0 = 3\mathbf{j} \]

### Step 2: Calculate \( \mathbf{b} + \mathbf{c} \)

Given:
\[ \mathbf{c} = -\mathbf{i} + 2\mathbf{j} - \mathbf{k} \]

Now, we can add \( \mathbf{b} \) and \( \mathbf{c} \):
\[ \mathbf{b} + \mathbf{c} = (-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \]
Combining like terms:
\[ = (-\mathbf{i} - \mathbf{i}) + (2\mathbf{j} + 2\mathbf{j}) + (\mathbf{k} - \mathbf{k}) = -2\mathbf{i} + 4\mathbf{j} + 0 = -2\mathbf{i} + 4\mathbf{j} \]

### Step 3: Find the cross product \( (\mathbf{a} + \mathbf{b}) \times (\mathbf{b} + \mathbf{c}) \)

Now we need to find the cross product of \( \mathbf{u} = 3\mathbf{j} \) and \( \mathbf{v} = -2\mathbf{i} + 4\mathbf{j} \).

Using the determinant form for the cross product:
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 0 \\ -2 & 4 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & 0 \\ -2 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & 3 \\ -2 & 4 \end{vmatrix} \]

Calculating each of these determinants:
1. \( \begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} 0 & 0 \\ -2 & 0 \end{vmatrix} = 0 \)
3. \( \begin{vmatrix} 0 & 3 \\ -2 & 4 \end{vmatrix} = (0 \cdot 4) - (3 \cdot -2) = 6 \)

So, we have:
\[ \mathbf{u} \times \mathbf{v} = 0\mathbf{i} - 0\mathbf{j} + 6\mathbf{k} = 6\mathbf{k} \]

### Step 4: Find the unit vector

To find the unit vector, we divide by the magnitude of the vector \( 6\mathbf{k} \):
\[ \text{Magnitude} = |6\mathbf{k}| = 6 \]

Thus, the unit vector \( \mathbf{p} \) is:
\[ \mathbf{p} = \frac{6\mathbf{k}}{6} = \mathbf{k} \]

### Final Answer
The unit vector perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{b} + \mathbf{c} \) is:
\[ \mathbf{k} \]

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