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For three vectors $\vec{A} = 2 \hat{j} - 3 \hat{k}, \vec{B} = - 2 \hat{i} + \hat{k} a n d \hat{C} = 3 \hat{i} - \hat{j},$ where î, ĵ and k̂ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system, the value of $(\vec{A .} (\vec{B} \times \vec{C}) + 6)$ is

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Correct Answer :

Correct answer is : 6

$\vec{A} = 2 \hat{j} - 3 \hat{k}, \vec{B} = - 2 \hat{i} + \hat{k} a n d \hat{C} = 3 \hat{i} - \hat{j}$

⇒ $\vec{A .} (\vec{B} \times \vec{C}) = | \begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix} |$

⇒ $\vec{A} (\vec{B} \times \vec{C}) = (- 2) (- 3) - 3 (2)$

⇒ $\vec{A .} (\vec{B} \times \vec{C}) = 0$

∴ $\vec{A .} (\vec{B} \times \vec{C}) + 6 = 6$

Solution :

The correct answer is 6.

We are given three vectors in a right-handed rectangular coordinate system:
$\vec{A} = 2 \hat{j} - 3 \hat{k}$
$\vec{B} = - 2 \hat{i} + \hat{k}$
$\vec{C} = 3 \hat{i} - \hat{j}$

We want to find the value of the expression:
$(\vec{A} \cdot (\vec{B} \times \vec{C}) + 6)$

First, let's write out the components of each vector explicitly:
$\vec{A} = 0 \hat{i} + 2 \hat{j} - 3 \hat{k}$
$\vec{B} = - 2 \hat{i} + 0 \hat{j} + \hat{k}$
$\vec{C} = 3 \hat{i} - \hat{j} + 0 \hat{k}$

The scalar triple product $\vec{A} \cdot (\vec{B} \times \vec{C})$ can be calculated as the determinant of a 3×3 matrix formed by the components of vectors $\vec{A}$ , $\vec{B}$ , and $\vec{C}$ :
$\vec{A} \cdot (\vec{B} \times \vec{C}) = | \begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix} |$

We expand this determinant along the first row:
$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \cdot | \begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix} | - 2 \cdot | \begin{matrix} - 2 & 1 \\ 3 & 0 \end{matrix} | + (- 3) \cdot | \begin{matrix} - 2 & 0 \\ 3 & - 1 \end{matrix} |$

Calculating each term:
First term: $0$
Second term: $- 2 \cdot ((- 2) \cdot 0 - 1 \cdot 3) = - 2 \cdot (- 3) = 6$
Third term: $- 3 \cdot ((- 2) \cdot (- 1) - 0 \cdot 3) = - 3 \cdot (2) = - 6$

Adding these terms together:
$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0 + 6 - 6 = 0$

Finally, we substitute this back into the original expression:
$(\vec{A} \cdot (\vec{B} \times \vec{C}) + 6) = 0 + 6 = 6$

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For three vectors $\vec{A} = 2 \hat{j} - 3 \hat{k}, \vec{B} = - 2 \hat{i} + \hat{k} a n d \hat{C} = 3 \hat{i} - \hat{j},$ where î, ĵ and k̂ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system, the value of $(\vec{A .} (\vec{B} \times \vec{C}) + 6)$ is

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For three vectors A → = 2 j ^ − 3 k ^ , B → = − 2 i ^ + k ^ a n d C ^ = 3 i ^ − j ^ , where î, ĵ and k̂ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system, the value of ( A . → ( B → × C → ) + 6 ) is

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For three vectors $\vec{A} = 2 \hat{j} - 3 \hat{k}, \vec{B} = - 2 \hat{i} + \hat{k} a n d \hat{C} = 3 \hat{i} - \hat{j},$ where î, ĵ and k̂ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system, the value of $(\vec{A .} (\vec{B} \times \vec{C}) + 6)$ is