Let `a = i +2j -3k and b = 2i +j-k` then the vector r satisfying `a xx r = a xx b and a.r =0 ` is of length `sqrt(10)`. True or False

Answer

To determine whether the statement is true or false, we need to analyze the vectors given and the conditions provided.

Given:
- \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \)
- \( \mathbf{b} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} \)

We need to find the vector \( \mathbf{r} \) such that:
1. \( \mathbf{a} \times \mathbf{r} = \mathbf{a} \times \mathbf{b} \)
2. \( \mathbf{a} \cdot \mathbf{r} = 0 \)
3. The length of \( \mathbf{r} \) is \( \sqrt{10} \).

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

To find \( \mathbf{a} \times \mathbf{b} \), we use the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{a} \) and \( \mathbf{b} \):

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & -3 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (-3)(1) = -2 + 3 = 1 \)
2. \( \begin{vmatrix} 1 & -3 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (-3)(2) = -1 + 6 = 5 \)
3. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3 \)

Putting it all together:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{i}(1) - \mathbf{j}(5) + \mathbf{k}(-3) = \mathbf{i} - 5\mathbf{j} - 3\mathbf{k} \]

### Step 2: Set up the equation \( \mathbf{a} \times \mathbf{r} = \mathbf{a} \times \mathbf{b} \)

Let \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \). Then:

\[ \mathbf{a} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ x & y & z \end{vmatrix} \]

Calculating this determinant gives:

\[ = \mathbf{i} \begin{vmatrix} 2 & -3 \\ y & z \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ x & z \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ x & y \end{vmatrix} \]

Calculating the 2x2 determinants:

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

Thus:

\[ \mathbf{a} \times \mathbf{r} = (2z + 3y)\mathbf{i} - (z + 3x)\mathbf{j} + (y - 2x)\mathbf{k} \]

Setting this equal to \( \mathbf{i} - 5\mathbf{j} - 3\mathbf{k} \):

\[ 2z + 3y = 1 \quad (1) \]
\[ -(z + 3x) = -5 \quad \Rightarrow \quad z + 3x = 5 \quad (2) \]
\[ y - 2x = -3 \quad (3) \]

### Step 3: Solve the system of equations

From equation (2):

\[ z = 5 - 3x \quad (4) \]

Substituting (4) into (1):

\[ 2(5 - 3x) + 3y = 1 \]
\[ 10 - 6x + 3y = 1 \quad \Rightarrow \quad 3y = 6x - 9 \quad \Rightarrow \quad y = 2x - 3 \quad (5) \]

Substituting (5) into (3):

\[ (2x - 3) - 2x = -3 \]
\[ -3 = -3 \quad \text{(True for all \(x\))} \]

Thus, \(y = 2x - 3\) and \(z = 5 - 3x\) are valid for any \(x\).

### Step 4: Find the length of \( \mathbf{r} \)

The length of \( \mathbf{r} \) is given by:

\[ |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2} \]

Substituting \(y\) and \(z\):

\[ |\mathbf{r}| = \sqrt{x^2 + (2x - 3)^2 + (5 - 3x)^2} \]

Calculating this:

\[ = \sqrt{x^2 + (4x^2 - 12x + 9) + (9 - 30x + 9x^2)} \]
\[ = \sqrt{(1 + 4 + 9)x^2 - (12 + 30)x + (9 + 9)} \]
\[ = \sqrt{14x^2 - 42x + 18} \]

Setting this equal to \( \sqrt{10} \):

\[ 14x^2 - 42x + 18 = 10 \]
\[ 14x^2 - 42x + 8 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 14 \cdot 8}}{2 \cdot 14} \]
\[ = \frac{42 \pm \sqrt{1764 - 448}}{28} = \frac{42 \pm \sqrt{1316}}{28} \]
\[ = \frac{42 \pm 2\sqrt{329}}{28} = \frac{21 \pm \sqrt{329}}{14} \]

### Conclusion

The vector \( \mathbf{r} \) can be found, and its length can indeed be \( \sqrt{10} \) for specific values of \( x \). Therefore, the statement is **True**.