If the projection of $$2\hat{i} + 4\hat{j} - 2\hat{k}$$ on $$\hat{i} + 2\hat{j} + \alpha\hat{k}$$ is zero. Then, the value of $$\alpha$$ will be

We are given two vectors:

$$\vec{A} = 2\hat{i} + 4\hat{j} - 2\hat{k}$$

$$\vec{B} = \hat{i} + 2\hat{j} + \alpha\hat{k}$$

The projection of $$\vec{A}$$ on $$\vec{B}$$ is zero. The projection of $$\vec{A}$$ on $$\vec{B}$$ is given by:

$$\text{Projection} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$$

For this projection to be zero, the dot product must be zero:

$$\vec{A} \cdot \vec{B} = 0$$

Computing the dot product:

$$\vec{A} \cdot \vec{B} = (2)(1) + (4)(2) + (-2)(\alpha)$$

$$= 2 + 8 - 2\alpha$$

$$= 10 - 2\alpha$$

Setting this equal to zero:

$$10 - 2\alpha = 0$$

$$2\alpha = 10$$

$$\alpha = 5$$

Therefore, the value of $$\alpha$$ is 5.