QJEEM19Maaa85: JEE Main 2019 Physics - Class 11 JEEM19Maaa85 Solution: Let      and   be coplanar…

Question

Verified Solution • 2025–26 Session moderate

Let

\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,

\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k

and

\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k

be coplanar vectors. Then the non-zero vector

\overrightarrow a \times \overrightarrow c

is :

Accepted Answer

\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0

\Rightarrow | \begin{matrix} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & λ^{2} - 1 \end{matrix} | = 0

\Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0

\Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0

\Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0

\Rightarrow \lambda = 2,3, - 3

So,

\lambda

= 2 (as

\overrightarrow a

is parallel to

\overrightarrow c

for

\lambda

=

\pm

3)

Hence

\vec{a} \times \vec{c} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{matrix} |

= - 10\widehat i + 5\widehat j

QJEEM19Maaa85: JEE Main 2019 Physics - Class 11 JEEM19Maaa85 Solution: Let&nbsp;&nbsp; &nbsp; &nbsp;and&nbsp; &nbsp;be coplanar…