The projection of `2hati-3hatj+4hatk` on the line whose equation is `vecr=(3+lambda)hati+(3-2lambda)hatj+(5+6lambda)hatk`, where `lambda` is a scalar parameter, is

Options

A. `(6)/(sqrt(41))`
B. `(32)/(sqrt(41))`
C. `(16)/(sqrt(41))`
D. `(7)/(5)`

Answer

To find the projection of the vector \( \vec{C} = 2\hat{i} - 3\hat{j} + 4\hat{k} \) onto the line defined by the equation \( \vec{r} = (3 + \lambda)\hat{i} + (3 - 2\lambda)\hat{j} + (5 + 6\lambda)\hat{k} \), we can follow these steps:

### Step 1: Identify the Direction Vector of the Line
The line can be expressed in the form \( \vec{r} = \vec{A} + \lambda \vec{B} \), where \( \vec{A} \) is a point on the line and \( \vec{B} \) is the direction vector.

From the given equation:
- The constant part (point on the line) is \( \vec{A} = 3\hat{i} + 3\hat{j} + 5\hat{k} \).
- The coefficients of \( \lambda \) give us the direction vector \( \vec{B} = \hat{i} - 2\hat{j} + 6\hat{k} \).

### Step 2: Calculate the Dot Product \( \vec{C} \cdot \vec{B} \)
Now, we need to compute the dot product of \( \vec{C} \) and \( \vec{B} \):
\[ \vec{C} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \vec{B} = \hat{i} - 2\hat{j} + 6\hat{k} \]
Calculating the dot product:
\[ \vec{C} \cdot \vec{B} = (2)(1) + (-3)(-2) + (4)(6) \]
\[ = 2 + 6 + 24 = 32 \]

### Step 3: Calculate the Magnitude of \( \vec{B} \)
Next, we find the magnitude of \( \vec{B} \):
\[ |\vec{B}| = \sqrt{(1)^2 + (-2)^2 + (6)^2} \]
\[ = \sqrt{1 + 4 + 36} = \sqrt{41} \]

### Step 4: Calculate the Projection of \( \vec{C} \) onto \( \vec{B} \)
The formula for the projection of \( \vec{C} \) onto \( \vec{B} \) is given by:
\[ \text{Projection of } \vec{C} \text{ on } \vec{B} = \frac{\vec{C} \cdot \vec{B}}{|\vec{B}|} \]
Substituting the values we calculated:
\[ \text{Projection} = \frac{32}{\sqrt{41}} \]

### Final Answer
Thus, the projection of \( \vec{C} \) on the line is:
\[ \frac{32}{\sqrt{41}} \]