The vectors $\vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} - 2 \hat{k}$ are the adjacent sides of a parallelogram. The acture angle between its diagonals is _____________.

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Solution

Given:
vectors $\vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} - 2 \hat{k}$ are the adjacent sides of a parallelogram

$Let θ be the acute angle between the diagonals . The diagonals of the parallelogram are \vec{a} + \vec{b} = (3 \hat{i} - 2 \hat{j} + 2 \hat{k}) + (- \hat{i} - 2 \hat{k}) = 2 \hat{i} - 2 \hat{j} . . . (1) and \vec{a} - \vec{b} = (3 \hat{i} - 2 \hat{j} + 2 \hat{k}) - (- \hat{i} - 2 \hat{k}) = 4 \hat{i} - 2 \hat{j} + 4 \hat{k} . . . (2) The angle between the diagonals is : \cos θ = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})}{|\vec{a} + \vec{b}| |\vec{a} - \vec{b}|} = \frac{(2) (4) + (- 2) (- 2) + (0) (4)}{\sqrt{{(2)}^{2} + {(- 2)}^{2}} \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}}} = \frac{8 + 4}{\sqrt{4 + 4} \sqrt{16 + 4 + 16}} (From (1) and (2)) = \frac{12}{\sqrt{8} \sqrt{36}} = \frac{12}{6 \times 2 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}$

Hence, the acute angle between its diagonals is $\frac{π}{4}$ .

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¯ ¯¯¯¯¯¯ ¯ A B = ¯ ¯¯¯ ¯ 3 i - ¯ ¯¯¯ ¯ 2 j + ¯ ¯¯¯¯ ¯ 2 k

¯ ¯¯¯¯¯¯ ¯ B C = - ¯ i - ¯ ¯¯¯¯ ¯ 2 k

\vec{a} = \hat{i} + \hat{j} - \hat{k} and \vec{b} = - 2 \hat{i} + \hat{j} + 2 \hat{k} .

A = 3 i - k, B = i + 2 j

\vec{a} + \vec{b} and \vec{a} - \vec{b}, where \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} .

\to a = ˆ i + 2 ˆ j + 3 ˆ k, a n d \to b = - 3 ˆ i - 2 ˆ j + ˆ k

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