If `veca=hati+2hatj+3hatk, vecb=-2hati+hatj+hatk, vecc=10hatj-hatk` and `vecaxx(vecbxxvecc)=uveca+v vecb+wvecv`, then

Options

A. w=0
B. w=-3
C. u=0
D. v=17

Answer

To solve the problem, we need to find the values of \( u \), \( v \), and \( w \) in the expression \( \vec{a} \times (\vec{b} \times \vec{c}) = u \vec{a} + v \vec{b} + w \vec{c} \).

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \vec{b} = -2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{c} = 10\hat{j} - \hat{k} \]

### Step 2: Calculate \( \vec{b} \times \vec{c} \)
Using the determinant method for the cross product:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 0 & 10 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left( 1 \cdot (-1) - 1 \cdot 10 \right) - \hat{j} \left( -2 \cdot (-1) - 1 \cdot 0 \right) + \hat{k} \left( -2 \cdot 10 - 1 \cdot 0 \right) \]
\[ = \hat{i} (-1 - 10) - \hat{j} (2) + \hat{k} (-20) \]
\[ = -11\hat{i} - 2\hat{j} - 20\hat{k} \]

### Step 3: Calculate \( \vec{a} \times (\vec{b} \times \vec{c}) \)
Now we need to compute \( \vec{a} \times (\vec{b} \times \vec{c}) \):
\[ \vec{a} \times (-11\hat{i} - 2\hat{j} - 20\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -11 & -2 & -20 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left( 2 \cdot (-20) - 3 \cdot (-2) \right) - \hat{j} \left( 1 \cdot (-20) - 3 \cdot (-11) \right) + \hat{k} \left( 1 \cdot (-2) - 2 \cdot (-11) \right) \]
\[ = \hat{i} (-40 + 6) - \hat{j} (-20 + 33) + \hat{k} (-2 + 22) \]
\[ = \hat{i} (-34) - \hat{j} (13) + \hat{k} (20) \]
\[ = -34\hat{i} - 13\hat{j} + 20\hat{k} \]

### Step 4: Compare with \( u\vec{a} + v\vec{b} + w\vec{c} \)
We need to express \( -34\hat{i} - 13\hat{j} + 20\hat{k} \) in terms of \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \):
\[ u \vec{a} + v \vec{b} + w \vec{c} = u(\hat{i} + 2\hat{j} + 3\hat{k}) + v(-2\hat{i} + \hat{j} + \hat{k}) + w(0\hat{i} + 10\hat{j} - \hat{k}) \]
\[ = (u - 2v)\hat{i} + (2u + v + 10w)\hat{j} + (3u + v - w)\hat{k} \]

### Step 5: Set up equations
From the comparison:
1. \( u - 2v = -34 \)
2. \( 2u + v + 10w = -13 \)
3. \( 3u + v - w = 20 \)

### Step 6: Solve the equations
From equation 1:
\[ u = -34 + 2v \]

Substituting \( u \) into equation 2:
\[ 2(-34 + 2v) + v + 10w = -13 \]
\[ -68 + 4v + v + 10w = -13 \]
\[ 5v + 10w = 55 \quad \text{(Equation 4)} \]
\[ v + 2w = 11 \quad \text{(Equation 5)} \]

Substituting \( u \) into equation 3:
\[ 3(-34 + 2v) + v - w = 20 \]
\[ -102 + 6v + v - w = 20 \]
\[ 7v - w = 122 \quad \text{(Equation 6)} \]

### Step 7: Solve Equations 5 and 6
From equation 5:
\[ w = \frac{11 - v}{2} \]

Substituting into equation 6:
\[ 7v - \frac{11 - v}{2} = 122 \]
Multiplying through by 2 to eliminate the fraction:
\[ 14v - (11 - v) = 244 \]
\[ 14v + v - 11 = 244 \]
\[ 15v = 255 \quad \Rightarrow \quad v = 17 \]

Now substituting \( v \) back to find \( w \):
\[ w = \frac{11 - 17}{2} = \frac{-6}{2} = -3 \]

Now substituting \( v \) back to find \( u \):
\[ u = -34 + 2(17) = -34 + 34 = 0 \]

### Final Answers
Thus, the values are:
\[ u = 0, \quad v = 17, \quad w = -3 \]