Consider three vectors `p=hat(i)+hat(j)+hat(k), q=2hat(i)+4hat(j)-hat(k) and r=hat(i)+hat(j)+3hat(k)` and let s be a unit vector, then
Q. If `(ptimesq)timesr=up+vq+wr`, then (u+v+w) is equal to

Options

A. `8`
B. `2`
C. `-2`
D. `4`

Answer

To solve the problem, we need to compute the expression \((\mathbf{p} \times \mathbf{q}) \times \mathbf{r}\) and express it in the form \(u\mathbf{p} + v\mathbf{q} + w\mathbf{r}\). We will then find \(u + v + w\).

### Step 1: Define the vectors
Given:
\[ \mathbf{p} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{q} = 2\hat{i} + 4\hat{j} - \hat{k} \]
\[ \mathbf{r} = \hat{i} + \hat{j} + 3\hat{k} \]

### Step 2: Compute \(\mathbf{p} \times \mathbf{q}\)
Using the determinant method to find the cross product:
\[ \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot -1 - 1 \cdot 4) - \hat{j} (1 \cdot -1 - 1 \cdot 2) + \hat{k} (1 \cdot 4 - 1 \cdot 2) \]
\[ = \hat{i} (-1 - 4) - \hat{j} (-1 - 2) + \hat{k} (4 - 2) \]
\[ = -5\hat{i} + 3\hat{j} + 2\hat{k} \]

Thus,
\[ \mathbf{p} \times \mathbf{q} = -5\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 3: Compute \((\mathbf{p} \times \mathbf{q}) \times \mathbf{r}\)
Now we need to compute:
\[ (\mathbf{p} \times \mathbf{q}) \times \mathbf{r} = (-5\hat{i} + 3\hat{j} + 2\hat{k}) \times (\hat{i} + \hat{j} + 3\hat{k}) \]

Using the determinant method again:
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 3 & 2 \\ 1 & 1 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -5 & 2 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -5 & 3 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i} (3 \cdot 3 - 2 \cdot 1) - \hat{j} (-5 \cdot 3 - 2 \cdot 1) + \hat{k} (-5 \cdot 1 - 3 \cdot 1) \]
\[ = \hat{i} (9 - 2) - \hat{j} (-15 - 2) + \hat{k} (-5 - 3) \]
\[ = 7\hat{i} + 17\hat{j} - 8\hat{k} \]

### Step 4: Express in terms of \(\mathbf{p}, \mathbf{q}, \mathbf{r}\)
We want to express \(7\hat{i} + 17\hat{j} - 8\hat{k}\) in terms of \(\mathbf{p}, \mathbf{q}, \mathbf{r}\):
\[ \mathbf{p} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{q} = 2\hat{i} + 4\hat{j} - \hat{k} \]
\[ \mathbf{r} = \hat{i} + \hat{j} + 3\hat{k} \]

We need to find \(u\), \(v\), and \(w\) such that:
\[ 7\hat{i} + 17\hat{j} - 8\hat{k} = u(\hat{i} + \hat{j} + \hat{k}) + v(2\hat{i} + 4\hat{j} - \hat{k}) + w(\hat{i} + \hat{j} + 3\hat{k}) \]

### Step 5: Set up the equations
This gives us a system of equations:
1. \(u + 2v + w = 7\) (coefficient of \(\hat{i}\))
2. \(u + 4v + w = 17\) (coefficient of \(\hat{j}\))
3. \(u - v + 3w = -8\) (coefficient of \(\hat{k}\))

### Step 6: Solve the equations
From equations 1 and 2, we can eliminate \(w\):
Subtract equation 1 from equation 2:
\[ (4v + w) - (2v + w) = 17 - 7 \]
\[ 2v = 10 \implies v = 5 \]

Substituting \(v = 5\) into equation 1:
\[ u + 2(5) + w = 7 \implies u + 10 + w = 7 \implies u + w = -3 \quad \text{(Equation 4)} \]

Now substitute \(v = 5\) into equation 3:
\[ u - 5 + 3w = -8 \implies u + 3w = -3 \quad \text{(Equation 5)} \]

### Step 7: Solve Equations 4 and 5
From Equation 4:
\[ u + w = -3 \implies u = -3 - w \]
Substituting into Equation 5:
\[ (-3 - w) + 3w = -3 \]
\[ -3 + 2w = -3 \implies 2w = 0 \implies w = 0 \]
Substituting \(w = 0\) back into Equation 4:
\[ u + 0 = -3 \implies u = -3 \]
And we already have \(v = 5\).

### Step 8: Find \(u + v + w\)
Now we have:
\[ u = -3, \quad v = 5, \quad w = 0 \]
Thus,
\[ u + v + w = -3 + 5 + 0 = 2 \]

### Final Answer
The value of \(u + v + w\) is \(2\).