1

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

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$\widehat i$ + $\widehat j$ $-$ $\widehat k$,   $-$$\widehat i$ + 3$\widehat j$ + p$\widehat k$ and 5$\widehat i$ + q$\widehat j$ $-$ 4$\widehat k$, then the point (p, q) lies on a line :
A
parallel to x-axis.
B
parallel to y-axis.
C
making an acute angle with the positive direction of x-axis.
D
making an obtuse angle with the positive direction of x-axis.

## Explanation

Given,

$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$

$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$

$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$

$\therefore$   $\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$

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

$\Delta$ABC is a right angle triangle.

Here $\overrightarrow {AB}$ perpendicular to $\overrightarrow {AC}$

$\therefore$   $\overrightarrow {AB}$  .  $\overrightarrow {AC}$  =  0

$\Rightarrow$   $-$ 8 + 2(q $-$ 1) $-$ 3(p + 1) = 0

$\Rightarrow$   3p $-$ 2q + 13 = 0

$\therefore$   (p, q) lies on the line

3x $-$ 2y + 13 = 0

And slope of the line = ${3 \over 2}$

$\therefore$   line makes an angle less than 90o or acute angle with the positive direction of x-axis.
2

### 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$\sqrt 2$
3

### 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
4

### 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.