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If $A B C D$ is a parallelogram, $\overrightarrow{A B}=2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k}$ and $\overrightarrow{A D}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$, then the unit vector in the direction of $B D$ is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})$
$\begin{aligned}
& \text {Since } \overrightarrow{A B}+\overrightarrow{B D}=\overrightarrow{A D} \Rightarrow \overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B} \\
& =(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})-(2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k})=-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}
\end{aligned}$
Hence unit vector in the direction of $\overrightarrow{B D}$ is
$\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\left|-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}\right|}=\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\sqrt{69}}$
& \text {Since } \overrightarrow{A B}+\overrightarrow{B D}=\overrightarrow{A D} \Rightarrow \overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B} \\
& =(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})-(2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k})=-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}
\end{aligned}$
Hence unit vector in the direction of $\overrightarrow{B D}$ is
$\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\left|-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}\right|}=\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\sqrt{69}}$
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