Find `lambda` if the scalar projection of `veca=lambda hati+hat j+4hatk" on "vecb=2hati+6hatj+3hatk` is 4 units.

Answer

To find the value of \( \lambda \) such that the scalar projection of the vector \( \vec{a} = \lambda \hat{i} + \hat{j} + 4 \hat{k} \) on the vector \( \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \) is 4 units, we can follow these steps:

### Step 1: Understand the formula for scalar projection
The scalar projection of vector \( \vec{a} \) on vector \( \vec{b} \) is given by the formula:
\[ \text{Scalar Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

### Step 2: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
We first need to compute the dot product of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (\lambda \hat{i} + \hat{j} + 4 \hat{k}) \cdot (2 \hat{i} + 6 \hat{j} + 3 \hat{k}) \]
Calculating the dot product:
\[ = \lambda \cdot 2 + 1 \cdot 6 + 4 \cdot 3 = 2\lambda + 6 + 12 = 2\lambda + 18 \]

### Step 3: Calculate the magnitude of \( \vec{b} \)
Next, we calculate the magnitude of vector \( \vec{b} \):
\[ |\vec{b}| = \sqrt{(2)^2 + (6)^2 + (3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

### Step 4: Set up the equation for the scalar projection
According to the problem, the scalar projection is given as 4 units:
\[ \frac{2\lambda + 18}{7} = 4 \]

### Step 5: Solve for \( \lambda \)
Now we can solve for \( \lambda \):
\[ 2\lambda + 18 = 4 \cdot 7 \]
\[ 2\lambda + 18 = 28 \]
Subtract 18 from both sides:
\[ 2\lambda = 28 - 18 \]
\[ 2\lambda = 10 \]
Divide both sides by 2:
\[ \lambda = 5 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 5 \]