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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:
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Verified Answer
The correct answer is:
$- \frac{1}{2}$
$\vec{a}=2 \hat{i}+3 \hat{i}+8 \hat{k}, b=4 \hat{j}+3 \hat{i}+\alpha \hat{k}$
$\begin{aligned}
& \vec{a} \cdot \vec{b}=0, \text { if } a^{-1} \cdot b^{-1} \\
& \therefore(2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(-4 i+4 j+\alpha k)=\frac{1}{2} \\
& \Rightarrow \alpha=-\frac{1}{2}
\end{aligned}$
$\begin{aligned}
& \vec{a} \cdot \vec{b}=0, \text { if } a^{-1} \cdot b^{-1} \\
& \therefore(2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(-4 i+4 j+\alpha k)=\frac{1}{2} \\
& \Rightarrow \alpha=-\frac{1}{2}
\end{aligned}$
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