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If \( \vec{A}=2 \hat{i}+3 \hat{j}+8 \hat{k} \) is perpendicular to \( \vec{B}=4 \hat{j}-4 \hat{i}+\alpha \hat{k} \), then the value of ' \( \alpha \) ' is
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The correct answer is:
\( \frac{1}{2} \)
Given, $\vec{A}=2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\vec{B}=4 \hat{j}-4 \hat{i}+\alpha \hat{k}$
Now, $\vec{A} \cdot \vec{B}=|A||B| \cos \theta$
Given, the two vectors are perpendicular. Therefore $\theta=90^{\circ}$
$\Rightarrow \vec{A} \cdot \vec{B}=\left.|A|\right|_{B} \mid \cos 90^{\circ}$
$\Rightarrow \vec{A} \cdot \vec{B}=0$
$\Rightarrow(2 i+3 j+8 k) \cdot(4 j-4 i+\alpha k)=0$
$\Rightarrow 8-12+8 \alpha=0$
$\Rightarrow-4+8 \alpha=0$
$\Rightarrow \alpha=\frac{1}{2}$
Now, $\vec{A} \cdot \vec{B}=|A||B| \cos \theta$
Given, the two vectors are perpendicular. Therefore $\theta=90^{\circ}$
$\Rightarrow \vec{A} \cdot \vec{B}=\left.|A|\right|_{B} \mid \cos 90^{\circ}$
$\Rightarrow \vec{A} \cdot \vec{B}=0$
$\Rightarrow(2 i+3 j+8 k) \cdot(4 j-4 i+\alpha k)=0$
$\Rightarrow 8-12+8 \alpha=0$
$\Rightarrow-4+8 \alpha=0$
$\Rightarrow \alpha=\frac{1}{2}$
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