# Vectors $2\bar{i}-\bar{j}+2\bar{k}\;and\;\bar{i}+\bar{j}-3\bar{k}$ act along two adjacent sides of a parallelogram. Find the angle between the diagonals of the parallelogram.

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• According to parallelogram law of addition, if $\large\overrightarrow a$ $and$ $\large\overrightarrow b$ are vectors along the adjacent sides of a parallelogram then $\large\overrightarrow a+\overrightarrow b$ $and$ $\large\overrightarrow a-\overrightarrow b$ are vectors along its diagonals.
• Angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ is given by $cos^{-1}\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Given that $\large\overrightarrow a=2\hat i-\hat j+2\hat k$ and
$\large\overrightarrow b=\hat i+\hat j-3\hat k$ are along adjacent sides of a parallelogram.
$\large\overrightarrow a+\overrightarrow b=(2\hat i-\hat j+2\hat k)+(\hat i+\hat j-3\hat k)$
$=\large3\hat i-\hat k$
$\large|\overrightarrow a+\overrightarrow b|=|3\hat i-\hat k|=\sqrt{3^2+(-1)^2}=\sqrt{10}$
$\large\overrightarrow a-\overrightarrow b=(2\hat i-\hat j+2\hat k)-(\hat i+\hat j-3\hat k)$
$=\large\hat i-2\hat j+5\hat k$
$\large|\overrightarrow a-\overrightarrow b|=|\hat i-2\hat j+5\hat k|=\sqrt{1^2+(-2)^2+5^2}=\sqrt{30}$
We know that $\large\overrightarrow a+\overrightarrow b$ and $\large\overrightarrow a-\overrightarrow b$ are diagonals of the parallelogram.
$\large(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)$=$\large(3\hat i-\hat k).(\hat i-2\hat j+5\hat k)$
$=\large\:3+0-5=-2$
We know that angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ is given by $cos^{-1}\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Angle between the diagonals of the parallelogram is given by
$\large cos^{-1}\bigg(\frac{(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)}{|\overrightarrow a+\overrightarrow b||\overrightarrow a-\overrightarrow b|}\bigg)$
$=\large\:cos^{-1}\frac{-2}{\sqrt{10}\sqrt{30}}$
$=\large\:cos^{-1}\bigg(\frac{-1}{5\sqrt3}\bigg)$
answered Apr 19, 2013