1

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

The plane passing through the point (4, –1, 2) and parallel to the lines  ${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$  and  ${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$ also passes through the point -
A
(1, 1, $-$ 1)
B
(1, 1, 1)
C
($-$ 1, $-$ 1, $-$1)
D
($-$ 1, $-$ 1, 1)

## Explanation

Let $\overrightarrow n$ be the normal vector to the plane passing through (4, $-$1, 2) and parallel to the lines L1 & L2

then $\overrightarrow n$ = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$

$\therefore$  $\overrightarrow n$ = $- 7\widehat i - 7\widehat j + 7\widehat k$

$\therefore$  Equation of plane is

$-$ 1(x $-$ 4) $-$ 1(y + 1) + 1(z $-$ 2) = 0

$\therefore$  x + y $-$ z $-$ 1 = 0

Now check options
2

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

Let A be a point on the line $\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$ and B(3, 2, 6) be a point in the space. Then the value of $\mu$ for which the vector $\overrightarrow {AB}$  is parallel to the plane x $-$ 4y + 3z = 1 is -
A
${1 \over 8}$
B
${1 \over 2}$
C
${1 \over 4}$
D
$-$ ${1 \over 4}$

## Explanation

Let point A is

$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$

and point B is (3, 2, 6)

then $\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$

which is parallel to the the plane x $-$ 4y + 3z = 1

$\therefore$  2 + 3$\mu$ $-$ 12 + 4$\mu$ + 12 $-$ 15$\mu$ = 0

8$\mu$ = 2

$\mu$ = ${1 \over 4}$
3

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

On which of the following lines lies the point of intersection of the line,   ${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$  and the plane, x + y + z = 2 ?
A
${{x - 4} \over 1} = {{y - 5} \over 1} = {{z - 5} \over { - 1}}$
B
${{x - 2} \over 2} = {{y - 3} \over 2} = {{z + 3} \over 3}$
C
${{x - 1} \over 1} = {{y - 3} \over 2} = {{z + 4} \over { - 5}}$
D
${{x + 3} \over 3} = {{4 - y} \over 3} = {{z + 1} \over { - 2}}$

## Explanation

General point on the given line is

x = 2$\lambda$ + 4

y = 2$\lambda$ + 5

z = $\lambda$ + 3

Solving with plane,

2$\lambda$ + 4 + 2$\lambda$ + 5 + $\lambda$ + 3 = 2

5$\lambda$ + 12 = 2

5$\lambda$ = $-$ 10

$\lambda$ = $-$ 2
4

### 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

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

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

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

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

$\lambda = - 4$