If a vector $$2\hat i + 3\hat j + 8\hat k$$   is perpendicular to the vector $$4\hat j - 4\hat i + \alpha \hat k,$$   then the value of $$\alpha $$ is

Solution :
Concept
If two vectors are perpendicular to each other than their dot product is always equal to zero.
$$\eqalign{ & {\text{Let,}}\,a = 2\hat i + 3\hat j + 8\hat k \cr & b = 4\hat j - 4\hat i + \alpha \hat k \cr & = - 4\hat i + 4\hat j + \alpha \hat k \cr} $$
According to the above hypothesis
$$\eqalign{ & a \bot b \cr & \Rightarrow a \cdot b = 0 \cr & \Rightarrow \left( {2\hat i + 3\hat j + 8\hat k} \right) \cdot \left( { - 4\hat i + 4\hat j + \alpha \hat k} \right) = 0 \cr & \Rightarrow - 8 + 12 + 8\alpha = 0 \cr & \Rightarrow 8\alpha = - 4 \cr & \therefore \alpha = - \frac{4}{8} \cr & = - \frac{1}{2} \cr} $$
NOTE
$$a \cdot b = ab\cos \theta .$$    Here, $$a$$ and $$b$$ are always positive as they are the magnitudes of $$a$$ and $$b.$$