1

### JEE Main 2019 (Online) 10th January Evening Slot

If $\overrightarrow \alpha$ = $\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b$  and  $\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b$ be two given vectors $\overrightarrow a$ and $\overrightarrow b$ are non-collinear. The value of $\lambda$ for which vectors $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear, is -
A
4
B
3
C
$-$3
D
$-$4

## Explanation

$\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow \alpha + \overrightarrow b$

$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b$

${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$

$3\lambda - 6 = 4\lambda - 2$

$\lambda = - 4$
2

### JEE Main 2019 (Online) 10th January Evening Slot

The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ?
A
(2, 1, 3)
B
(4, $-$ 1, 2)
C
(4, 1, $-$ 2)
D
($-$ 2, 3, 5)

## Explanation p : 3(x $-$ 0) + 5(y $-$ 2) + 1 (z $-$ 5) = 0

3x + 5y + z = 15
3

### JEE Main 2019 (Online) 11th January Morning Slot

The plane containing the line ${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$ and also containing its projection on the plane 2x + 3y $-$ z = 5, contains which one of the following points ?
A
($-$ 2, 2, 2)
B
(2, 2, 0)
C
(2, 0, $-$ 2)
D
(0, $-$ 2, 2)

## Explanation

The normal vector of required plane

$= \left( {2\widehat i - \widehat j + 3\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$

$= - 8\widehat i + 8\widehat j + 8\widehat k$

So, direction ratio of normal is $\left( { - 1,1,1} \right)$

So required plane is

$- \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$

$\Rightarrow - x + y + z + 4 = 0$

Which is satisfied by $\left( {2,0, - 2} \right)$
4

### JEE Main 2019 (Online) 11th January Morning Slot

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 :
A
$- 10\widehat i - 5\widehat j$
B
$- 10\widehat i + 5\widehat j$
C
$- 14\widehat i + 5\widehat j$
D
$- 14\widehat i - 5\widehat j$

## Explanation

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

$\Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$

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

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

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

$\Rightarrow \lambda = 2,3, - 3$

So,   $\lambda$ = 2 (as   $\overrightarrow a$ is parallel to $\overrightarrow c$ for $\lambda$ = $\pm$3)

Hence   $\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$

$= - 10\widehat i + 5\widehat j$