If `a = 2i+j+k, b =i+2j+k,c=2i-3j+4k` and r is a vector such that `r xx b = c xxb` and r.a = 0 then r is equal to

Options

A. `-2,2,2`
B. `-2,1,3`
C. `-3,2,4`
D. `1,-5,3`

Answer

To solve the problem, we need to find the vector \( r \) given the conditions \( r \times b = c \times b \) and \( r \cdot a = 0 \). Let's break down the solution step by step.

### Step 1: Define the vectors
We have the following vectors:
- \( a = 2i + j + k \)
- \( b = i + 2j + k \)
- \( c = 2i - 3j + 4k \)

### Step 2: Calculate \( c \times b \)
We need to compute the cross product \( c \times b \).

\[ c \times b = \begin{vmatrix} i & j & k \\ 2 & -3 & 4 \\ 1 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = i \begin{vmatrix} -3 & 4 \\ 2 & 1 \end{vmatrix} - j \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} + k \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -3 & 4 \\ 2 & 1 \end{vmatrix} = (-3)(1) - (4)(2) = -3 - 8 = -11 \)

2. \( \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} = (2)(1) - (4)(1) = 2 - 4 = -2 \)

3. \( \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = (2)(2) - (-3)(1) = 4 + 3 = 7 \)

Putting it all together:

\[ c \times b = -11i + 2j + 7k \]

### Step 3: Set up the equation \( r \times b = c \times b \)
Let \( r = xi + yj + zk \). Then we have:

\[ r \times b = \begin{vmatrix} i & j & k \\ x & y & z \\ 1 & 2 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = i \begin{vmatrix} y & z \\ 2 & 1 \end{vmatrix} - j \begin{vmatrix} x & z \\ 1 & 1 \end{vmatrix} + k \begin{vmatrix} x & y \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} y & z \\ 2 & 1 \end{vmatrix} = y(1) - z(2) = y - 2z \)

2. \( \begin{vmatrix} x & z \\ 1 & 1 \end{vmatrix} = x(1) - z(1) = x - z \)

3. \( \begin{vmatrix} x & y \\ 1 & 2 \end{vmatrix} = x(2) - y(1) = 2x - y \)

Thus,

\[ r \times b = (y - 2z)i - (x - z)j + (2x - y)k \]

### Step 4: Set the components equal
Setting \( r \times b = c \times b \):

\[ (y - 2z)i - (x - z)j + (2x - y)k = -11i + 2j + 7k \]

This gives us the following equations:

1. \( y - 2z = -11 \)
2. \( -(x - z) = 2 \) → \( x - z = -2 \) → \( x = z - 2 \)
3. \( 2x - y = 7 \)

### Step 5: Solve the system of equations
Substituting \( x = z - 2 \) into the third equation:

\[ 2(z - 2) - y = 7 \\ 2z - 4 - y = 7 \\ y = 2z - 11 \]

Now substitute \( y = 2z - 11 \) into the first equation:

\[ (2z - 11) - 2z = -11 \\ -11 = -11 \text{ (this is always true)} \]

Now substitute \( y \) into the second equation:

\[ x = z - 2 \]

### Step 6: Choose a value for \( z \)
Let’s choose \( z = 3 \):

Then,

\[ y = 2(3) - 11 = 6 - 11 = -5 \\ x = 3 - 2 = 1 \]

Thus, we have:

\[ r = 1i - 5j + 3k \]

### Final Answer
The vector \( r \) is:

\[ \boxed{i - 5j + 3k} \]