1

### JEE Main 2016 (Online) 9th April Morning Slot

The distance of the point (1, − 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is :
A
$2\sqrt 2$
B
2
C
$\sqrt 2$
D
${1 \over {\sqrt 2 }}$

## Explanation

Equation of plane passing through point (1, 2, 2) is,

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

This plane is perpendicular to the plane

x $-$ y + 2z = 3   and  2x $-$ 2y + z + 12 = 0

When two planes,

a1x + b1y + c1z + d1 = 0

and a2x + b2y + c2z + d2 = 0

are perpendicular to each other then

a1a2 + b1b2 + c1c2 = 0

So, we can say,

a $\times$ 1 + b $\times$ ($-$ 1) + c $\times$ 2 = 0

$\Rightarrow$   a $-$ b + 2c = 0           . . . (2)

and, a $\times$ 2 + b($-$2) + c(1) = 0

$\Rightarrow$   2a $-$ 2b + c = 0           . . .(3)

Solving (2) and (3)

${a \over { - 1 + 4}}$ = ${b \over {4 - 1}}$ = ${c \over { - 2 + 2}}$ = $\lambda$

$\Rightarrow$    a = 3$\lambda$, b = 3$\lambda$, c = 0

Putting the values of a, b, c in equation (1) we get,

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

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

$\Rightarrow$   3x + 3y $-$ 9 = 0

$\Rightarrow$   x + y $-$ 3 = 0

$\therefore$   Distance of point (1, $-$2, 4) from plane x + y $-$ 3 = 0 is,

= $\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$ = ${4 \over {\sqrt 2 }}$ = 2$\sqrt 2$
2

### JEE Main 2016 (Online) 10th April Morning Slot

ABC is a triangle in a plane with vertices
A(2, 3, 5), B(−1, 3, 2) and C($\lambda$, 5, $\mu$).

If the median through A is equally inclined to the coordinate axes, then the value of ($\lambda$3 + $\mu$3 + 5) is :
A
1130
B
1348
C
676
D
1077

## Explanation

${{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5$

i.e.  ${{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}$

As medium is making equal angles with coordinate axes,

$\therefore$   ${{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}$

$\Rightarrow$   $\lambda$ = 7,   $\mu$ = 10

$\therefore$   $\lambda$3 + $\mu$3 + 5 = 73 + 103 + 5 = 1348
3

### JEE Main 2016 (Online) 10th April Morning Slot

The number of distinct real values of $\lambda$ for which the lines

${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}}$ and ${{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2}$ are coplanar is :
A
4
B
1
C
2
D
3

## Explanation

As planes are coplanar, so

$\left| {\matrix{ {3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right|$ = 0

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

$\Rightarrow$   2(4 $-$ $\lambda$4) + 4($\lambda$2 $-$ 2) = 0

$\Rightarrow$   4 $-$ $\lambda$4 + 2$\lambda$2 $-$ 4 = 0

$\Rightarrow$   $\lambda$2($\lambda$2 $-$ 2) = 0

$\Rightarrow$   $\lambda$ = 0, $\sqrt 2 , - \sqrt 2$

$\therefore$   3 distinct real values are possible.
4

### JEE Main 2016 (Online) 10th April Morning Slot

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ and ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$ respectively, then the position vector of the orthocentre of this triangle, is :
A
${\overrightarrow a + \overrightarrow b + \overrightarrow c }$
B
$- \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$
C
$\overrightarrow 0$
D
$\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$

## Explanation

Given,

Position vector of circumcentre, $\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$

We know, position vector of centroid, $\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$

Now, let $\overrightarrow R$ be the orthocentre of the triangle.

We know, $\overrightarrow G$ $= {{2\overrightarrow C + \overrightarrow R } \over 3}$

$\Rightarrow$   3$\overrightarrow G$ $= 2\overrightarrow C + \overrightarrow R$

$\Rightarrow$   $\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C$

=   $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$

=   ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$

NEET