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BITSAT 2005 Solved Question Paper

Section: Mathematics

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Question : 45 of 150

Marks: +1, -0

If

\vec{\alpha}=2 \hat{i}+3 \hat{j}-\hat{k}, \vec{\beta}=-\hat{i}+2 \hat{j}-4 \hat{k},

\vec{\gamma} = \hat{i}+\hat{j}+\hat{k}

, then what is the value of

(\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma})?

47
74
$-74$
none of the above

Solution:

Given that

\vec{\alpha}=2\hat{i}+3\hat{j}-\hat{k},\vec{\beta}=-\hat{i}+2\hat{j}-4\hat{k}

and

\vec{\gamma}=\hat{i}+\hat{j}+\hat{k}

Now,

\vec{\alpha} \times \vec{\beta} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{vmatrix}

= \hat{i}(-12+2) - \hat{j}(-8-1) + \hat{k}(4+3)

=-10\hat{i}+9\hat{j}+7\hat{k}

and

(\vec{\alpha} \times \vec{\gamma}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix}

= \hat{i}(3+1) - \hat{j}(2+1) + \hat{k}(2-3)

=-4 \hat{i} -3 \hat{j} -\hat{k}

Now,

(\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma})

=(-10 \hat{i}+9 \hat{j}+7 \hat{k}) \cdot (4 \hat{i}-3 \hat{j}-\hat{k})

=-40-27-7

=-74

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