If ABCD is a parallelogram,overrightarrow{AB} | vedclass

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(C) In a parallelogram $ABCD$,by the triangle law of vector addition,we have $\overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AD}$.Therefo

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$\frac{1}{\sqrt{69}}(-i - 2j + 8k)$

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(C) In a parallelogram $ABCD$,by the triangle law of vector addition,we have $\overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AD}$.Therefore,$\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB}$.Substituting the given vectors: $\overrightarrow{BD} = (i + 2j + 3k) - (2i + 4j - 5k) = -i - 2j + 8k$.The magnitude of $\overrightarrow{BD}$ is $|\overrightarrow{BD}| = \sqrt{(-1)^2 + (-2)^2 + 8^2} = \sqrt{1 + 4 + 64} = \sqrt{69}$.The unit vector in the direction of $\overrightarrow{BD}$ is given by $\frac{\overrightarrow{BD}}{|\overrightarrow{BD}|} = \frac{-i - 2j + 8k}{\sqrt{69}} = \frac{1}{\sqrt{69}}(-i - 2j + 8k)$.

If $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is

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