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 \( \mathbf{c} = 2\hat{i} - 3\hat{j} + 4\hat{k} \) on the line given by the equation \( \mathbf{r} = (3 + \lambda)\hat{i} + (3 - 2\lambda)\hat{j} + (5 + 6\lambda)\hat{k} \), we can follow these steps:

### Step 1: Identify the direction vector of the line
The line can be expressed in the standard form \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \), where \( \mathbf{a} \) is a point on the line and \( \mathbf{b} \) is the direction vector.

From the equation given:
\[ \mathbf{r} = (3 + \lambda)\hat{i} + (3 - 2\lambda)\hat{j} + (5 + 6\lambda)\hat{k} \]
we can rewrite it as:
\[ \mathbf{r} = 3\hat{i} + 3\hat{j} + 5\hat{k} + \lambda(\hat{i} - 2\hat{j} + 6\hat{k}) \]
Thus, the direction vector \( \mathbf{b} \) is:
\[ \mathbf{b} = \hat{i} - 2\hat{j} + 6\hat{k} \]

### Step 2: Calculate the projection of vector \( \mathbf{c} \) on vector \( \mathbf{b} \)
The formula for the projection of vector \( \mathbf{c} \) onto vector \( \mathbf{b} \) is given by:
\[ \text{proj}_{\mathbf{b}} \mathbf{c} = \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} \]

### Step 3: Compute the dot product \( \mathbf{c} \cdot \mathbf{b} \)
Calculating the dot product:
\[ \mathbf{c} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \mathbf{b} = \hat{i} - 2\hat{j} + 6\hat{k} \]
\[ \mathbf{c} \cdot \mathbf{b} = (2)(1) + (-3)(-2) + (4)(6) = 2 + 6 + 24 = 32 \]

### Step 4: Calculate the magnitude of vector \( \mathbf{b} \)
The magnitude of vector \( \mathbf{b} \) is:
\[ |\mathbf{b}| = \sqrt{(1)^2 + (-2)^2 + (6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41} \]

### Step 5: Calculate the projection
Now substituting back into the projection formula:
\[ \text{proj}_{\mathbf{b}} \mathbf{c} = \frac{32}{41} \mathbf{b} \]
Thus, the projection of \( \mathbf{c} \) on the line is:
\[ \text{proj}_{\mathbf{b}} \mathbf{c} = \frac{32}{41} (\hat{i} - 2\hat{j} + 6\hat{k}) \]

### Final Answer
The projection of the vector \( 2\hat{i} - 3\hat{j} + 4\hat{k} \) on the line is:
\[ \frac{32}{41} \hat{i} - \frac{64}{41} \hat{j} + \frac{192}{41} \hat{k} \]