If `vec(a) = 2hat(i) + 2hat(j) + 3hat(k), vec(b)= - hat(i) + 2hat(j) + 3 and vec(c ) = 3hat(i) + hat(j)` are three vectors such that `vec(a) + t vec(b)` is perpendicular to c. then what is t equal to?

Options

A. 8
B. 6
C. 4
D. 2

Answer

To solve the problem, we need to find the value of \( t \) such that the vector \( \vec{a} + t \vec{b} \) is perpendicular to \( \vec{c} \). This means that the dot product of \( \vec{a} + t \vec{b} \) and \( \vec{c} \) must equal zero.

### Step 1: Write down the vectors
We have:
\[ \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \vec{b} = -\hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{c} = 3\hat{i} + \hat{j} \]

### Step 2: Set up the equation for perpendicularity
The condition for perpendicularity is:
\[ (\vec{a} + t\vec{b}) \cdot \vec{c} = 0 \]

### Step 3: Expand the dot product
Expanding the left-hand side:
\[ (\vec{a} + t\vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + t(\vec{b} \cdot \vec{c}) \]

### Step 4: Calculate \( \vec{a} \cdot \vec{c} \)
Calculating the dot product \( \vec{a} \cdot \vec{c} \):
\[ \vec{a} \cdot \vec{c} = (2\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + \hat{j}) \]
\[ = 2 \cdot 3 + 2 \cdot 1 + 3 \cdot 0 = 6 + 2 + 0 = 8 \]

### Step 5: Calculate \( \vec{b} \cdot \vec{c} \)
Now, calculating the dot product \( \vec{b} \cdot \vec{c} \):
\[ \vec{b} \cdot \vec{c} = (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j}) \]
\[ = -1 \cdot 3 + 2 \cdot 1 + 0 = -3 + 2 + 0 = -1 \]

### Step 6: Substitute back into the equation
Now substituting back into the equation:
\[ 8 + t(-1) = 0 \]
\[ 8 - t = 0 \]

### Step 7: Solve for \( t \)
Rearranging gives:
\[ t = 8 \]

Thus, the value of \( t \) is \( 8 \).