What is the projection of the vector `hat(i)-2 hat(j) + hat(k)` on the vector `4hat(i) - 4hat(j)+ 7hat(k)`?

Options

A. `(sqrt5)/(2)`
B. `(19)/(9)`
C. `(sqrt5)/(4)`
D. `(11)/(3)`

Answer

To find the projection of the vector **A** = \( \hat{i} - 2\hat{j} + \hat{k} \) on the vector **B** = \( 4\hat{i} - 4\hat{j} + 7\hat{k} \), we can use the formula for the projection of vector **A** onto vector **B**:

\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{\mathbf{B} \cdot \mathbf{B}} \mathbf{B} \]

### Step 1: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)

\[ \mathbf{A} \cdot \mathbf{B} = (1)(4) + (-2)(-4) + (1)(7) \]
\[ = 4 + 8 + 7 = 19 \]

### Step 2: Calculate the dot product \( \mathbf{B} \cdot \mathbf{B} \)

\[ \mathbf{B} \cdot \mathbf{B} = (4)(4) + (-4)(-4) + (7)(7) \]
\[ = 16 + 16 + 49 = 81 \]

### Step 3: Substitute the dot products into the projection formula

\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{19}{81} \mathbf{B} \]

### Step 4: Write the projection vector

Now, substitute **B** back into the equation:

\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{19}{81} (4\hat{i} - 4\hat{j} + 7\hat{k}) \]
\[ = \left(\frac{76}{81}\hat{i} - \frac{76}{81}\hat{j} + \frac{133}{81}\hat{k}\right) \]

### Step 5: Calculate the magnitude of the projection

To find the magnitude of the projection, we can calculate:

\[ \|\text{proj}_{\mathbf{B}} \mathbf{A}\| = \sqrt{\left(\frac{76}{81}\right)^2 + \left(-\frac{76}{81}\right)^2 + \left(\frac{133}{81}\right)^2} \]

Calculating each term:

\[ = \sqrt{\frac{5776}{6561} + \frac{5776}{6561} + \frac{17689}{6561}} \]
\[ = \sqrt{\frac{28841}{6561}} = \frac{\sqrt{28841}}{81} \]

### Step 6: Simplify the square root

Finding the square root of 28841:

\[ \sqrt{28841} = 169 \]

Thus, the magnitude of the projection is:

\[ \|\text{proj}_{\mathbf{B}} \mathbf{A}\| = \frac{169}{81} \]

### Final Answer

The projection of the vector \( \hat{i} - 2\hat{j} + \hat{k} \) on the vector \( 4\hat{i} - 4\hat{j} + 7\hat{k} \) is:

\[ \frac{169}{81} \text{ or approximately } 2.0864 \]