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If the points with position vectors $60 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}, 40 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}$ and $a \hat{\mathbf{i}}-52 \hat{\mathbf{j}}$ are collinear, then $a$ is equal to
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The correct answer is:
$-40$
Since, the position vectors $60 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}, 40 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}$, and $a \hat{\mathbf{i}}-52 \hat{\mathbf{j}}$ are collinear.
$\therefore \quad\left|\begin{array}{ccc}
60 & 3 & 1 \\
40 & -8 & 1 \\
a & -52 & 1
\end{array}\right|=0$
$\begin{aligned} \Rightarrow & 60(-8+52)-3(40-a) \\ & \quad+1(-2080+8 a)=0 \\ \Rightarrow & 2640-120+3 a-2080+8 a=0 \\ \Rightarrow & 440+11 a=0 \Rightarrow a=-40\end{aligned}$
$\therefore \quad\left|\begin{array}{ccc}
60 & 3 & 1 \\
40 & -8 & 1 \\
a & -52 & 1
\end{array}\right|=0$
$\begin{aligned} \Rightarrow & 60(-8+52)-3(40-a) \\ & \quad+1(-2080+8 a)=0 \\ \Rightarrow & 2640-120+3 a-2080+8 a=0 \\ \Rightarrow & 440+11 a=0 \Rightarrow a=-40\end{aligned}$
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