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If the points whose position vectors are $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, 6 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $14 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$ are collinear, then the value of $p$ is
are collinear, then the value of $p$ is
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are collinear, then the value of $p$ is
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Verified Answer
The correct answer is:
$4$
Given vectors $\mathbf{a}=2 \mathbf{i}+\mathbf{j}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=6 \mathbf{i}-\mathbf{j}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=14 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$ are collinear, therefore
$[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=0$
$\begin{array}{cccc} & & \left|\begin{array}{ccc}2 & 1 & 1 \\ 6 & -1 & 2 \\ 14 & -5 & p\end{array}\right|=0 \\ \Rightarrow & 2[-p+10]-1[6 p-28]+1[-30+14]=0 \\ \Rightarrow & -2 p+20-6 p+28-16=0 \\ \Rightarrow & & -8 p+32=0 \\ \Rightarrow & & p=4\end{array}$
$[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=0$
$\begin{array}{cccc} & & \left|\begin{array}{ccc}2 & 1 & 1 \\ 6 & -1 & 2 \\ 14 & -5 & p\end{array}\right|=0 \\ \Rightarrow & 2[-p+10]-1[6 p-28]+1[-30+14]=0 \\ \Rightarrow & -2 p+20-6 p+28-16=0 \\ \Rightarrow & & -8 p+32=0 \\ \Rightarrow & & p=4\end{array}$
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