Find a unit vector perpendicular to each of the vector a→ and b→ where a→=5ı^+6ȷ^−2k^ and b→=7ı^+6ȷ^+2k^ORFind the volume of the parallelopiped whose adjacent edges are represented by 2a→,−b→ and 3c→ where a→=ı^−ȷ^+2k^b→=3ı^+4ȷ^−5k^, and c→=2ı^−ȷ^+3k^

First, we will find a→×b→∴a→×b→=ı^ȷ^k^56−2762=(12+12)ı^−(10+14)j+(30−42)k^=24ı^−24j−12k^Now for unit vector, |a→×b→|=242+(−24)2+(−12)2=1296=36Therefore, unit vector perpendicular to both a→ and b→ is, a→×b→|a→×b→|=24i^−24j^−12k^36=2436ı^−2436ȷ^−1236k^=23ı^−23ȷ^−13k^ORGiven, a→=ı^−ȷ^+2k^⇒2a→=2(ı^−ȷ^+2k^)=2ı^−2ȷ^+4k^Similarly, −b→=−3ı^−4ȷ^+5k^ and 3c→=6ı^−3ȷ^+9k^Now, Volume of the parallelopiped =2−24−3−456−39=2(−36−(−15))+2(−27−30)+4(9+24)=2(−21)+2(−57)+4(33)=−42−114+132=|−24|=24Therefore, the volume of the parallelopiped is 24.

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Find a unit vector perpendicular to each of the vector $\vec{a} and \vec{b} where \vec{a} = 5 \hat{ı} + 6 \hat{ȷ} -$ $2 \hat{k} and \vec{b} = 7 \hat{ı} + 6 \hat{ȷ} + 2 \hat{k}$
OR
Find the volume of the parallelopiped whose adjacent edges are represented by $2 \vec{a}, - \vec{b} and 3 \vec{c}$ where
$\vec{a} = \hat{ı} - \hat{ȷ} + 2 \hat{k}$
$\begin{array}{l} \vec{b} = 3 \hat{ı} + 4 \hat{ȷ} - 5 \hat{k}, and \\ \vec{c} = 2 \hat{ı} - \hat{ȷ} + 3 \hat{k} \end{array}$

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Detailed Solution

First, we will find $\vec{a} \times \vec{b}$
$∴ \vec{a} \times \vec{b} = |\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 5 & 6 & - 2 \\ 7 & 6 & 2 \end{array}| = (12 + 12) \hat{ı} - (10 + 14) j + (30 - 42) \hat{k} = 24 \hat{ı} - 24 j - 12 \hat{k}$
Now for unit vector, $| \vec{a} \times \vec{b} | = \sqrt{24^{2} + (- 24)^{2} + (- 12)^{2}} = \sqrt{1296} = 36$
Therefore, unit vector perpendicular to both $\vec{a} and \vec{b} is,$
$\frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |} = \frac{24 \hat{i} - 24 j^{^} - 12 \hat{k}}{36} = \frac{24}{36} \hat{ı} - \frac{24}{36} \hat{ȷ} - \frac{12}{36} \hat{k} = \frac{2}{3} \hat{ı} - \frac{2}{3} \hat{ȷ} - \frac{1}{3} \hat{k}$

Given,
$\vec{a} = \hat{ı} - \hat{ȷ} + 2 \hat{k} \Rightarrow 2 \vec{a} = 2 (\hat{ı} - \hat{ȷ} + 2 \hat{k}) = 2 \hat{ı} - 2 \hat{ȷ} + 4 \hat{k}$
Similarly, $- \vec{b} = - 3 \hat{ı} - 4 \hat{ȷ} + 5 \hat{k} and 3 \vec{c} = 6 \hat{ı} - 3 \hat{ȷ} + 9 \hat{k}$
Now, Volume of the parallelopiped $= |\begin{array}{ccc} 2 & - 2 & 4 \\ - 3 & - 4 & 5 \\ 6 & - 3 & 9 \end{array}| = 2 (- 36 - (- 15)) + 2 (- 27 - 30) +$
$4 (9 + 24) = 2 (- 21) + 2 (- 57) + 4 (33) = - 42 - 114 + 132 = | - 24 | = 24$
Therefore, the volume of the parallelopiped is 24.