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If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number then $\left[\lambda(\vec{a}+\vec{b}) \lambda^2 \vec{b} \lambda \vec{c}\right]=[\vec{a} \vec{b}+\vec{c} \vec{b}]$ for
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Verified Answer
The correct answer is:
no value of $\lambda$
no value of $\lambda$
$$
\begin{aligned}
& {\left[\begin{array}{lll}
\lambda(\vec{a}+\vec{b}) & \lambda^2 \vec{b} & \lambda \vec{c}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{b}
\end{array}\right]} \\
& \left|\begin{array}{ccc}
\lambda & \lambda & 0 \\
0 & \lambda^2 & 0 \\
0 & 0 & \lambda
\end{array}\right|=\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 0
\end{array}\right| \\
& \Rightarrow \lambda^4=-1 \\
&
\end{aligned}
$$
Hence no real value of $\lambda$.
\begin{aligned}
& {\left[\begin{array}{lll}
\lambda(\vec{a}+\vec{b}) & \lambda^2 \vec{b} & \lambda \vec{c}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{b}
\end{array}\right]} \\
& \left|\begin{array}{ccc}
\lambda & \lambda & 0 \\
0 & \lambda^2 & 0 \\
0 & 0 & \lambda
\end{array}\right|=\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 0
\end{array}\right| \\
& \Rightarrow \lambda^4=-1 \\
&
\end{aligned}
$$
Hence no real value of $\lambda$.
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