12. Let $\vec { a } = \hat { i } + 2 \hat { j } + \hat { k } , \vec { b } = \hat { i } - \hat { j } + \hat { k }$ and $\vec { c } = \hat { i } - \hat { j } - \hat { k }$. A vector in the plane of $\vec { a }$ and $\vec { b }$ whose projection on $\vec { c }$ is $\frac { 1 } { \sqrt { 3 } }$, is\\
(A) $4 \hat { i } - \hat { j } + 4 \hat { k }$\\
(B) $3 \hat { i } + \hat { j } - 3 \hat { k }$\\
(C) $2 \hat { i } + \hat { j } - 2 \hat { k }$\\
(D) $4 \hat { i } + \hat { j } - 4 \hat { k }$

\section*{Sol. (A)}
Vector lying in the plane of $\vec { a }$ and $\vec { b }$ is $\vec { r } = \lambda _ { 1 } \vec { a } + \lambda _ { 2 } \vec { b }$ and its projection on $\vec { c }$ is $\frac { 1 } { \sqrt { 3 } }$\\
$\Rightarrow \quad \left[ \left( \lambda _ { 1 } + \lambda _ { 2 } \right) \hat { \mathrm { i } } - \left( 2 \lambda _ { 1 } - \lambda _ { 2 } \right) \hat { \mathrm { j } } + \left( \lambda _ { 1 } + \lambda _ { 2 } \right) \hat { \mathrm { k } } \right] \cdot \frac { [ \hat { \mathrm { i } } - \hat { \mathrm { j } } - \hat { \mathrm { k } } ] } { \sqrt { 3 } } = \frac { 1 } { \sqrt { 3 } }$\\
$\Rightarrow \quad 2 \lambda _ { 1 } - \lambda _ { 2 } = - 1 \Rightarrow \overrightarrow { \mathrm { r } } = \left( 3 \lambda _ { 1 } + 1 \right) \hat { \mathrm { i } } - \hat { \mathrm { j } } + \left( 3 \lambda _ { 1 } + 1 \right) \hat { \mathrm { k } }$\\
Hence the required vector is $4 \hat { i } - \hat { j } + 4 \hat { k }$.

\section*{Alternate:}
Vector lying in the plane of $\vec { a }$ and $\vec { b }$ is $\vec { a } + \lambda \vec { b }$, and its projection on $C$ is $\frac { 1 } { \sqrt { 3 } }$.\\
$\Rightarrow \left( ( 1 + \lambda ) \hat { \mathrm { i } } + ( 2 - \lambda ) \hat { \mathrm { j } } + ( 1 + \lambda ) \hat { \mathrm { k } } \cdot \frac { ( \hat { \mathrm { i } } - \hat { \mathrm { j } } - \hat { \mathrm { k } } ) } { \sqrt { 3 } } \right) = \frac { 1 } { \sqrt { 3 } }$\\
$\Rightarrow \lambda = 3$.\\
Hence the required vector is $4 \hat { i } - \hat { j } + 4 \hat { k }$.

\section*{Section - B (May have more than one option correct)}
\begin{enumerate}
  \setcounter{enumi}{12}
  \item The equations of the common tangents to the parabola $y = x ^ { 2 }$ and $y = - ( x - 2 ) ^ { 2 }$ is/are\\
(A) $\mathrm { y } = 4 ( \mathrm { x } - 1 )$\\
(B) $\mathrm { y } = 0$\\
(C) $y = - 4 ( x - 1 )$\\
(D) $y = - 30 x - 50$
\end{enumerate}

Sol. (A), (B)\\
Equation of tangent to $x ^ { 2 } = y$ is

$$\mathrm { y } = \mathrm { mx } - \frac { 1 } { 4 } \mathrm {~m} ^ { 2 }$$

Equation of tangent to $( x - 2 ) ^ { 2 } = - y$ is

$$\mathrm { y } = \mathrm { m } ( \mathrm { x } - 2 ) + \frac { 1 } { 4 } \mathrm {~m} ^ { 2 }$$

(1) and (2) are identical.\\
$\Rightarrow \mathrm { m } = 0$ or 4\\
$\therefore \quad$ Common tangents are $\mathrm { y } = 0$ and $\mathrm { y } = 4 \mathrm { x } - 4$.\\