1

### 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
2

### 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.
3

### 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$\overrightarrow G$ $= 2\overrightarrow C + \overrightarrow R$

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

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

=   ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$
4

### JEE Main 2017 (Offline)

If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,

${x \over 1} = {y \over 4} = {z \over 5}$ is Q, then PQ is equal to:
A
$2\sqrt {42}$
B
$\sqrt {42}$
C
$6\sqrt 5$
D
$3\sqrt 5$

## Explanation

Equation of line PQ is ${{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}$

Let F be ($\lambda$ + 1, 4$\lambda$ $-$ 2, 5$\lambda$ + 3)

Since F lies on the plane

$\therefore$   2($\lambda$ + 1) + 3(4$\lambda$ $-$ 2) $-$ 4(5$\lambda$ + 3) + 22 $=$ 0

$\Rightarrow$   $-$ 6$\lambda$ + 6 = 0 $\Rightarrow$  $\lambda$ = 1

$\therefore$   F is (2, 2, 8)

PQ = 2 PF = 2$\sqrt {{1^2} + {4^2} + {5^2}}$ = 2$\sqrt {42}$