Question

If vector A = 3 i + j + 2 k and vector B = 2 i - 2 j + 4 k then find ⏐A x B⏐

Answer 1

$\huge{\mathfrak{\underline{\red{Answer!}}}}$

Answer:

ans of 1=-(ans of 2)

I use the determination process for cross product.

and Happy janmastami ☺️

Answer 2

${\underline{\underline{\bf{\pink{GivEn:-}}}}}$

• $\sf \overrightarrow{a} = 3\hat{i} + \hat{j} + 2 \hat{k}$

• $\sf \overrightarrow{b} = 2\hat{i} - 2\hat{j} + 4 \hat{k}$

${\underline{\underline{\bf{\red{To\;find:-}}}}}$

• $\sf \overrightarrow{a} \times \overrightarrow{b}$

${\underline{\underline{\bf{\blue{SoluTion:-}}}}}$

As we know that if,

$\sf \overrightarrow{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$

$\sf \overrightarrow{b} = \lambda \hat{i} - \mu \hat{j} + \delta \hat{k}$

So, $\sf \overrightarrow{a} \times \overrightarrow{b}$ is,

$\left|\begin{array}{c c c} \hat{i} & \hat{j} & \hat{k} \\ \alpha & \beta & \gamma \\ \lambda & \mu & \delta \end{array} \right|$

therefore,

$\dashrightarrow\sf \overrightarrow{a} \times \overrightarrow{b} = \hat{i} (\beta \delta - \gamma \mu) - \hat{j} (\alpha \delta - \gamma \lambda) + \hat{k}(\alpha \mu - \beta \lambda)$

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Here,

• $\sf \overrightarrow{a} = 3\hat{i} + \hat{j} + 2 \hat{k}$

• $\sf \overrightarrow{b} = 2\hat{i} - 2\hat{j} + 4 \hat{k}$

☯ Therefore, $\sf \overrightarrow{a} \times \overrightarrow{b}$ is,

$\left|\begin{array}{c c c} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 2 & 4 \end{array} \right|$

$\dashrightarrow\sf \hat{i} (3 \times 4 - 1 \times 2) - \hat{j} (2 \times 4 - 1 \times 3) + \hat{k}(2 \times 2 - 3 \times 3)$

$\dashrightarrow\sf \hat{i} (12 - 2) - \hat{j} (8 - 3) + \hat{k}(4 - 9)$

$\dashrightarrow\sf \purple{ 10 \hat{i} - 5 \hat{j} - 5 \hat{k}}$ ★

Similarly, we can say that,

$\sf (\overrightarrow{a} \times \overrightarrow{b}) = - ( \overrightarrow{a} \times \overrightarrow{b})$

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