Let and be vector such that then is equal to

JEE MainMathematicsModerateVectors

Let
$\overset{\rightarrow}{a} - \overset{\hat{}}{i} - \overset{\hat{}}{j},\overset{\rightarrow}{b} = \overset{\hat{}}{i} + \overset{\hat{}}{j} + \overset{\hat{}}{k}$
and $\overset{\rightarrow}{c}$ be vector such that
$\overset{\rightarrow}{a} \times \overset{\rightarrow}{c} + \overset{\rightarrow}{b} = \overset{\rightarrow}{0}\ and\ \overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c} = 4$
then
$|\overset{\rightarrow}{c}|^{2}$ is equal to

Show answer and solution

\overset{\rightarrow}{a} \times \left( \overset{\rightarrow}{a} \times \overset{\rightarrow}{c} \right) + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0

(\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c})\overset{\rightarrow}{a} - (\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{a})\overset{\rightarrow}{c} + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0

4\overset{\rightarrow}{a} - 2\overset{\rightarrow}{c} + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0

$\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \left| \begin{matrix}
\overset{\hat{}}{i} & \overset{\hat{}}{j} & \overset{\hat{}}{k} \\
1 & - 1 & 0 \\
1 & 1 & 1
\end{matrix} \right|$

= \overset{\hat{}}{i}( - 1) - \overset{\hat{}}{j}(1) + k(2)ww

= - \overset{\hat{}}{i} - \overset{\hat{}}{j} + 2\overset{\hat{}}{k}

2\overset{\rightarrow}{c} = 4(\overset{\hat{}}{i} - \overset{\hat{}}{j}) + ( - \overset{\hat{}}{i} - \overset{\hat{}}{j} + 2\overset{\hat{}}{k})

2\overset{\rightarrow}{c} = 3\overset{\hat{}}{i} - 5\overset{\hat{}}{j} + 2\overset{\hat{}}{k}

\begin{matrix} 2|c| = \sqrt{a + 25 + 4} \\ 2|c| = \sqrt{38} \\ 4|c|^{2} = 38 \\ |c|^{2} = 19/2 \end{matrix}

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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