Let `vec(a)` be a vector which is perpendicular to the vector `3 hat(i) + (1)/(2) hat(j) + 2hat(k)` `vec(a) xx (2 hat(i) + hat(k)) = 2hat(i)- 13hat(j)- 4hat(k)`, then the projection of the vector `vec(a)` on the vector `2hat(i) + 2hat(j) + hat(k)` is:

Options

A. `(1)/(3)`
B. 1
C. `(5)/(3)`
D. `(7)/(3)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Define the vector \(\vec{a}\)
Let \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\).

### Step 2: Use the condition of perpendicularity
Since \(\vec{a}\) is perpendicular to the vector \(3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}\), we can use the dot product condition:
\[ \vec{a} \cdot (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}) = 0 \]
This gives us:
\[ 3x + \frac{1}{2}y + 2z = 0 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product condition
We are also given that:
\[ \vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k} \]
Calculating the cross product:
\[ \vec{a} \times (2 \hat{i} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 2 & 0 & 1 \end{vmatrix} \]
Expanding this determinant, we get:
\[ \hat{i}(y \cdot 1 - z \cdot 0) - \hat{j}(x \cdot 1 - z \cdot 2) + \hat{k}(x \cdot 0 - y \cdot 2) = y \hat{i} - (x - 2z) \hat{j} - 2y \hat{k} \]
Setting this equal to \(2 \hat{i} - 13 \hat{j} - 4 \hat{k}\), we have:
\[ y = 2 \quad \text{(Equation 2)} \]
\[ -(x - 2z) = -13 \implies x - 2z = 13 \quad \text{(Equation 3)} \]
\[ -2y = -4 \implies y = 2 \quad \text{(Already known)} \]

### Step 4: Substitute \(y\) into Equation 1
Substituting \(y = 2\) into Equation 1:
\[ 3x + \frac{1}{2}(2) + 2z = 0 \]
This simplifies to:
\[ 3x + 1 + 2z = 0 \implies 3x + 2z = -1 \quad \text{(Equation 4)} \]

### Step 5: Solve Equations 3 and 4
From Equation 3:
\[ x = 13 + 2z \]
Substituting this into Equation 4:
\[ 3(13 + 2z) + 2z = -1 \]
This expands to:
\[ 39 + 6z + 2z = -1 \implies 8z = -40 \implies z = -5 \]
Now substituting \(z = -5\) back into Equation 3:
\[ x - 2(-5) = 13 \implies x + 10 = 13 \implies x = 3 \]

### Step 6: Write the vector \(\vec{a}\)
Now we have:
\[ \vec{a} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k} \]

### Step 7: Find the projection of \(\vec{a}\) on \(2 \hat{i} + 2 \hat{j} + \hat{k}\)
The projection formula is given by:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]
where \(\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}\).

Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (3)(2) + (2)(2) + (-5)(1) = 6 + 4 - 5 = 5 \]

Calculating the magnitude squared of \(\vec{b}\):
\[ |\vec{b}|^2 = 2^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9 \]

### Step 8: Calculate the projection
Thus, the projection is:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{5}{9} \vec{b} = \frac{5}{9}(2 \hat{i} + 2 \hat{j} + \hat{k}) = \frac{10}{9} \hat{i} + \frac{10}{9} \hat{j} + \frac{5}{9} \hat{k} \]

### Final Result
The projection of the vector \(\vec{a}\) on the vector \(2 \hat{i} + 2 \hat{j} + \hat{k}\) is:
\[ \frac{10}{9} \hat{i} + \frac{10}{9} \hat{j} + \frac{5}{9} \hat{k} \]