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Q.
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
Two vectors must be perpendicular if their dot product is zero.
Let $\vec{ a } =2 \hat{ i }+3 \hat{ j }+8 k$
$\vec{ b } =4 \hat{ j }-4 \hat{ i }+\alpha \hat{ k } $
$=-4 \hat{ i }+4 \hat{ j }+ \alpha \hat{ k }$
According to the above hypothesis
$\vec{ a } \perp \vec{ b } $
$\Rightarrow \vec{ a } \cdot \vec{ b }=0 $
$\Rightarrow (2 \hat{ i }+3 \hat{ j }+ 8 \hat{ k }) .(-4 \hat{ i }+4 \hat{ j }+\alpha \hat{ k })=0$
$\Rightarrow -8+12+8 \alpha=0 $
$\Rightarrow 8 \alpha=-4 $
$\therefore \alpha=-\frac{4}{8}=-\frac{1}{2}$ NOTE: $\vec{ a } \cdot \vec{ b }=a b \cos \theta$. Here, $a$ and $b$ are always positive as they are the magnitudes $a \vec{ a }$ and $\vec{ b }$.