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The points with position vectors
$10 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}, 12 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}, \hat{\mathbf{i}}+11 \hat{\mathbf{j}}$ are collinear, if the value of $a$ is
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$10 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}, 12 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}, \hat{\mathbf{i}}+11 \hat{\mathbf{j}}$ are collinear, if the value of $a$ is
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Verified Answer
The correct answer is:
8
Since, the points with position vectors $10 \hat{i}+3 \hat{j}$,
$12 \hat{i}-5 \hat{j}, a \hat{i}+11 \hat{j}$ are collinear.
$\therefore \quad\left|\begin{array}{ccc}10 & 3 & 1 \\ 12 & -5 & 1 \\ a & 11 & 1\end{array}\right|=0$
$\Rightarrow 10(-5-11)-3(12-a)+1(132+5 a)=0$
$\Rightarrow-160-36+3 a+132+5 a=0$
$\Rightarrow-196+132+8 a=0$
$\Rightarrow 8 a=64$
$\Rightarrow a=8$
$12 \hat{i}-5 \hat{j}, a \hat{i}+11 \hat{j}$ are collinear.
$\therefore \quad\left|\begin{array}{ccc}10 & 3 & 1 \\ 12 & -5 & 1 \\ a & 11 & 1\end{array}\right|=0$
$\Rightarrow 10(-5-11)-3(12-a)+1(132+5 a)=0$
$\Rightarrow-160-36+3 a+132+5 a=0$
$\Rightarrow-196+132+8 a=0$
$\Rightarrow 8 a=64$
$\Rightarrow a=8$
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