If the points with position vectors i ^ + 10 j ^ + 13 k ^ , 6 i ^ + 11 j ^ + 11 k ^ , 9 2 i ^ + j ^ - 8 k ^ are collinear, then ( 19 - 6 ) 2 is equal to [2023]

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Solution

Q.
If the points with position vectors $α \hat{i} + 10 \hat{j} + 13 \hat{k}$ , $6 \hat{i} + 11 \hat{j} + 11 \hat{k}$ , $\frac{9}{2} \hat{i} + β \hat{j} - 8 \hat{k}$ are collinear, then ${(19 α - 6 β)}^{2}$ is equal to [2023]

1 36

2 25

3 16

4 49

Ans.
(1)

$Let \vec{a} = α \hat{i} + 10 \hat{j} + 13 \hat{k},$

$\vec{b} = 6 \hat{i} + 11 \hat{j} + 11 \hat{k} and \vec{c} = \frac{9}{2} \hat{i} + β \hat{j} - 8 \hat{k}$

$For collinear, \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0}$

$Now, \vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 10 & 13 \\ 6 & 11 & 11 \end{matrix} |$

$= - 33 \hat{i} - (11 α - 78) \hat{j} + (11 α - 60) \hat{k}$

$\vec{b} \times \vec{c} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 11 & 11 \\ \frac{9}{2} & β & - 8 \end{matrix} |$

$= (- 88 - 11 β) \hat{i} + \frac{195}{2} \hat{j} + (6 β - \frac{99}{2}) \hat{k}$

$\vec{c} \times \vec{a} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{9}{2} & β & - 8 \\ α & 10 & 13 \end{matrix} |$

$= (13 β + 80) \hat{i} - (\frac{117}{2} + 8 α) \hat{j} + (45 - α β) \hat{k}$

$∴$ $(- 41 + 2 β) \hat{i} + (117 - 19 α) \hat{j} + (11 α + 6 β - α β - \frac{129}{2}) \hat{k} = \vec{0}$

$\Rightarrow β = \frac{41}{2}, α = \frac{117}{19}$

$∴$ ${(19 α - 6 β)}^{2} = {(117 - 123)}^{2} = 36$

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