The area of the parallelogram whose adjacent sides are given by a = 3i

1.	The area of the parallelogram whose adjacent sides are given by a = 3i - j+2k and 5 =i+49-6 is[Note: Area of the parallelogram is axba) 614 sq unitsb) 614 sq unitsc) 573 unitsd) 32 units
Answer» GIVEN :– ㅤ • $\bf \overrightarrow{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ and $\bf \overrightarrow{b} = \hat{i}+ 4 \hat{j} -6 \hat{k}$ are adjacent SIDES of PARALLELOGRAM . ㅤ TO FIND :– ㅤ • Area of parallelogram = ? ㅤ SOLUTION :– ㅤ • We KNOW that area of parallelogram is – ㅤ $\bf \implies \large{ \boxed{ \bf Area =\| \overrightarrow{a} \times \overrightarrow{b}\|}}$ ㅤ $\bf \implies Area =\|(3 \hat{i} - \hat{j} + 2 \hat{k} )\times ( \hat{i}+ 4 \hat{j} -6 \hat{k})\|$ ㅤ $\bf \implies Area = \left\|\begin{array}{ccc}\bf\hat{i} & \bf\hat{j}& \bf\hat{k} \\ \\\bf3& \bf - 1& \bf2 \\ \\\bf1& \bf 4& \bf - 6\end{array}\right\|$ ㅤ $\bf \implies Area =\|\hat{i} \{ ( - 1)( - 6) - (4)(2)\} -\hat{j} \{ (3)( - 6) - (4)(2)\} + \hat{k} \{ ( 4)(3) - (1)( - 1)\}\|$ ㅤ $\bf \implies Area =\|\hat{i} \{6 - 8\} -\hat{j} \{ - 18- 8\} + \hat{k} \{12 -( - 1)\}\|$ ㅤ $\bf \implies Area =\|\hat{i} \{ - 2\} -\hat{j} \{ -26\} + \hat{k} \{12 + 1\}\|$ ㅤ $\bf \implies Area = \|- 2\hat{i} + 26\hat{j}+13 \hat{k}\|$ ㅤ $\bf \implies Area = \sqrt{ {( - 2)}^{2} + {(26)}^{2} + {(13)}^{2} }$ ㅤ $\bf \implies Area = \sqrt{4+ 676+169}$ ㅤ $\bf \implies Area = \sqrt{849} \:\:sq.\:\: units$ ㅤ $\bf \implies {\boxed{ \bf Area = 29.14 \:\:sq.\:\: units}}$

The area of the parallelogram whose adjacent sides are given by a = 3i - j+2k and 5 =i+49-6 is[Note: Area of the parallelogram is axba) 614 sq unitsb) 614 sq unitsc) 573 unitsd) 32 units

Answer»

GIVEN :–

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• $\bf \overrightarrow{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ and $\bf \overrightarrow{b} = \hat{i}+ 4 \hat{j} -6 \hat{k}$ are adjacent SIDES of PARALLELOGRAM .

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TO FIND :–

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• Area of parallelogram = ?

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SOLUTION :–

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• We KNOW that area of parallelogram is –

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$\bf \implies \large{ \boxed{ \bf Area =| \overrightarrow{a} \times \overrightarrow{b}|}}$

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$\bf \implies Area =|(3 \hat{i} - \hat{j} + 2 \hat{k} )\times ( \hat{i}+ 4 \hat{j} -6 \hat{k})|$

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$\bf \implies Area = \left|\begin{array}{ccc}\bf\hat{i} & \bf\hat{j}& \bf\hat{k} \\ \\\bf3& \bf - 1& \bf2 \\ \\\bf1& \bf 4& \bf - 6\end{array}\right|$