1

### 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.
2

### 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
3

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

If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A
ab'  +  bc'  +  1  =  0
B
cc'  +  a   +  a'  =  0
C
bb'  +  cc'  +  1  =  0
D
aa'  +  c  +  c'  =  0

## Explanation

Equation of 1st line is

${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$

Dr's of 1st line = ($a$, 1 , c)

Equation of 2nd line is

${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$

Dr's of 2nd line = ($a'$, c' , 1)

Lines are perpendicular, so the dot product of the Dr's of two lines are zero.

$\therefore$ $aa$' + c + c' = 0
4

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

Let  $\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$   $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$,    $\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$   be three vectors such that the projection vector of $\overrightarrow b$ on $\overrightarrow a$ is $\overrightarrow a$.
If   $\overrightarrow a + \overrightarrow b$   is perpendicular to $\overrightarrow c$ , then $\left| {\overrightarrow b } \right|$ is equal to :
A
$\sqrt {32}$
B
6
C
$\sqrt {22}$
D
4

## Explanation

Projection of $\overrightarrow b$ on $\overrightarrow a$ is $\overrightarrow a$

$\therefore$   ${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$

$\Rightarrow$  ${{{b_1} + {b_2} + 2} \over 2} = 2$

$\Rightarrow$  ${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$

and $\overrightarrow a$ + $\overrightarrow b$ is perpendicular to $\overrightarrow c$

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

$\Rightarrow$  $5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$

$\Rightarrow$  $5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$

solving (1) & (2)

b1 $=$ $-$ 3 and b2 $=$ 5

$\Rightarrow$  $\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$