The compenent of `vec(A)=hat(i)+hat(j)+5hat(k)` perpendicular to `vec(B)=3hat(i)+4hat(j)` is

Options

A. `-(4)/(25)hat(i)+(3)/(25)hat(j)+5hat(k)`
B. `-(8)/(25)hat(i)-(6)/(25)hat(j)+5hat(k)`
C. `(4)/(25)hat(i)-(3)/(25)hat(j)+5hat(k)`
D. `+(8)/(25)hat(i)-(6)/(25)hat(j)+5hat(k)`

Explanation

To find the component of the vector \(\vec{A} = \hat{i} + \hat{j} + 5\hat{k}\) that is perpendicular to the vector \(\vec{B} = 3\hat{i} + 4\hat{j}\), we can follow these steps:

### Step 1: Identify the vectors
We have:
\[ \vec{A} = \hat{i} + \hat{j} + 5\hat{k} \]
\[ \vec{B} = 3\hat{i} + 4\hat{j} \]

### Step 2: Find the unit vector of \(\vec{B}\)
To find the unit vector \(\hat{b}\) in the direction of \(\vec{B}\), we first calculate the magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Now, the unit vector \(\hat{b}\) is:
\[ \hat{b} = \frac{\vec{B}}{|\vec{B}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \]

### Step 3: Find the projection of \(\vec{A}\) onto \(\vec{B}\)
The projection of \(\vec{A}\) onto \(\vec{B}\) is given by:
\[ \text{proj}_{\vec{B}} \vec{A} = \left(\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\right) \vec{B} \]
First, we calculate the dot product \(\vec{A} \cdot \vec{B}\):
\[ \vec{A} \cdot \vec{B} = (1)(3) + (1)(4) + (5)(0) = 3 + 4 + 0 = 7 \]
Now, substituting this into the projection formula:
\[ \text{proj}_{\vec{B}} \vec{A} = \left(\frac{7}{25}\right)(3\hat{i} + 4\hat{j}) = \frac{21}{25}\hat{i} + \frac{28}{25}\hat{j} \]

### Step 4: Find the component of \(\vec{A}\) perpendicular to \(\vec{B}\)
The component of \(\vec{A}\) that is perpendicular to \(\vec{B}\) can be found by subtracting the projection from \(\vec{A}\):
\[ \vec{A}_{\perp} = \vec{A} - \text{proj}_{\vec{B}} \vec{A} \]
Calculating this gives:
\[ \vec{A}_{\perp} = \left(\hat{i} + \hat{j} + 5\hat{k}\right) - \left(\frac{21}{25}\hat{i} + \frac{28}{25}\hat{j}\right) \]
\[ = \left(1 - \frac{21}{25}\right)\hat{i} + \left(1 - \frac{28}{25}\right)\hat{j} + 5\hat{k} \]
\[ = \left(\frac{25}{25} - \frac{21}{25}\right)\hat{i} + \left(\frac{25}{25} - \frac{28}{25}\right)\hat{j} + 5\hat{k} \]
\[ = \frac{4}{25}\hat{i} - \frac{3}{25}\hat{j} + 5\hat{k} \]

### Final Result
Thus, the component of \(\vec{A}\) that is perpendicular to \(\vec{B}\) is:
\[ \vec{A}_{\perp} = \frac{4}{25}\hat{i} - \frac{3}{25}\hat{j} + 5\hat{k} \]