1

### JEE Main 2019 (Online) 9th January Morning Slot

Let $\overrightarrow a$ = $\widehat i - \widehat j$, $\overrightarrow b$ = $\widehat i + \widehat j + \widehat k$ and $\overrightarrow c$

be a vector such that $\overrightarrow a$ × $\overrightarrow c$ + $\overrightarrow b$ = $\overrightarrow 0$

and $\overrightarrow a$.$\overrightarrow c$ = 4, then |$\overrightarrow C$|2 is equal to :
A
8
B
$19 \over 2$
C
9
D
$17 \over 2$

## Explanation

Given that,

$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0$

$\Rightarrow$  $\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow a \times b = \overrightarrow 0$

$\Rightarrow$  $\left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a \cdot \overrightarrow a } \right)\overrightarrow c + \overrightarrow a \times \overrightarrow b = \overrightarrow 0$

given that

$\overrightarrow a \cdot \overrightarrow c = 4$

and $\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$

$\Rightarrow$  $4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$

Now  $\overrightarrow a \times \overrightarrow b$

$= \left| {\matrix{ i & j & k \cr 1 & { - 1} & 0 \cr 1 & 1 & 1 \cr } } \right|$

$= - \widehat i - \widehat j + 2\widehat k$

$\therefore$  $2\overrightarrow c = 4\left( {\widehat i - \widehat j} \right) + \left( { - \widehat i - \widehat j + \widehat k} \right)$

$= 4\widehat i - 4\widehat j - \widehat i - \widehat j + \widehat k$

$= 3\widehat i - 5\widehat j + \widehat k$

$\therefore$  $\overrightarrow c = {3 \over 2}\widehat i - {5 \over 2}\widehat j + \widehat k$

$\therefore$  $\left| {\overrightarrow c } \right| = \sqrt {{9 \over 4} + {{25} \over 4} + 1}$

$= \sqrt {{{38} \over 4}}$

$= \sqrt {{{19} \over 2}}$

$\therefore$  ${\left| {\overrightarrow c } \right|^2} = {{19} \over 2}$
2

### JEE Main 2019 (Online) 9th January Morning Slot

The equation of the line passing through (–4, 3, 1), parallel
to the plane x + 2y – z – 5 = 0 and intersecting
the line ${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$ is
A
${{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}$
B
${{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}$
C
${{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}$
D
${{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}$

## Explanation

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x1, y1, z1) and it passes through L2.

${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$

$\therefore$   x1 = $-$1 $-$ 3t, y1 = 3 + 2t, z1 = 2 $-$ t

$\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$

$\overrightarrow n = \widehat i + 2\widehat j - \widehat k$

As $\overrightarrow {AR}$ and $\overrightarrow n$ are perpendicular to each other, So

$\overrightarrow {AR}$ $\cdot$ $\overrightarrow n$ = 0

$\Rightarrow$  (3 $-$ 3t) 1 + (2t)2 + (1 $-$ t) ($-$ 1) = 0

$\Rightarrow$  3 $-$ 3t + 4t $-$ 1 + t = 0

$\Rightarrow$  2 + 2t = 0

$\Rightarrow$  t = $-$ 1

$\therefore$   point R = (2, 1, 3)

$\therefore$  DR of line L is

= (2 $-$ ($-$ 4), $1$ $-$ 3, 3 $-$ 1)

= (6, $-$ 2, 2)

$\therefore$  Equation of line is

${{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}$

or ${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$
3

### JEE Main 2019 (Online) 9th January Morning Slot

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :
A
(–3, 0, -1)
B
(3, 2, 1)
C
(3, 3, -1)
D
(–3, 1, 1)

## Explanation

The equation of plane

(x + y + z $-$ 1) + $\lambda$ (2x + 3y $-$ z + 4) = 0

$\Rightarrow$  (1 + 2$\lambda$)x + (1 + 3$\lambda$)y + (1 $-$ $\lambda$)z + 4$\lambda$ $-$ 1 = 0

As plane is parallel to y axis so the normal vector of plane and dot product of $\widehat j$ is zero.

$\therefore$  1 + 3$\lambda$ = 0

$\Rightarrow$  $\lambda$ = $-$ ${1 \over 3}$

$\therefore$  So the equation of the plane is

x(1 $-$ ${2 \over 3}$) + (1 $-$ ${3 \over 3}$) y + (1 + ${1 \over 3}$) $-$ ${4 \over 3}$ $-$ 1 = 0

$\Rightarrow$  x (${1 \over 3}$) + z(${4 \over 3}$) $-$ ${7 \over 3}$ = 0

$\Rightarrow$  x + 4z $-$ 7 = 0

By checking each options you can see only point (3, 2, 1) lies on the plane.
4

### JEE Main 2019 (Online) 9th January Evening Slot

The equation of the plane containing the straight line ${x \over 2} = {y \over 3} = {z \over 4}$ and perpendicular to the plane containing the straight lines ${x \over 3} = {y \over 4} = {z \over 2}$ and ${x \over 4} = {y \over 2} = {z \over 3}$ is :
A
x $-$ 2y + z = 0
B
3x + 2y $-$ 3z = 0
C
x + 2y $-$ 2z = 0
D
5x + 2y $-$ 4z = 0

## Explanation

Vector $\bot$ to given plane = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 4 & 2 \cr 4 & 2 & 3 \cr } } \right|$

= $\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$

= $8\widehat i - \widehat j - 10\widehat k\,$      . . . . (1)

Vector parallel to given line

= $2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,$      . . . (2)

Vector $\bot \,\,\,$ to both (1) & (2) vector

= $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 1} & { - 10} \cr 2 & 3 & 4 \cr } } \right|$

= $\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)$

= $26\widehat i - 52\widehat j + 26\widehat k$

Dr's of normal of required plane is

(26, $-$52, 26) $\Rightarrow$ (1, $-$2, 1)

Equation of plane whose Dr's of Normal is (1, $-$2, 1) and passes through origin

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

x $-$ 2y + z = 0