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If $\overline{\mathrm{a}}=2 \hat{\imath}-\hat{\mathrm{j}}+\widehat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\imath}+2 \hat{\mathrm{\jmath}}-3 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=3 \hat{\imath}+\lambda \hat{\mathrm{\jmath}}+5 \hat{\mathrm{k}}$ are coplanar,
then $\lambda$ is the root of the equation
Options:
then $\lambda$ is the root of the equation
Solution:
2518 Upvotes
Verified Answer
The correct answer is:
$x^{2}+3 x=4$
Since given vectors are coplanar, we write
$$
\begin{aligned}
&\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & -3 \\
3 & \lambda & 5
\end{array}\right|=0 \\
\therefore & 2(10+3 \lambda)+1(5+9)+1(\lambda-6)=0 \\
\therefore & 20+6 \lambda+14+\lambda-6=0 \Rightarrow 7 \lambda+28=0 \Rightarrow \lambda=-4
\end{aligned}
$$
Put $x=-4$ in all options.
(A) $8-12=-4 \neq 6$
(B) $16-8=8 \neq 4$
(C) $16-12=4$
(D) $16-8=8 \neq 6$
$$
\begin{aligned}
&\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & -3 \\
3 & \lambda & 5
\end{array}\right|=0 \\
\therefore & 2(10+3 \lambda)+1(5+9)+1(\lambda-6)=0 \\
\therefore & 20+6 \lambda+14+\lambda-6=0 \Rightarrow 7 \lambda+28=0 \Rightarrow \lambda=-4
\end{aligned}
$$
Put $x=-4$ in all options.
(A) $8-12=-4 \neq 6$
(B) $16-8=8 \neq 4$
(C) $16-12=4$
(D) $16-8=8 \neq 6$
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