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1

### JEE Advanced 2017 Paper 2 Offline

Let O be the origin and $$\overrightarrow{OX}$$, $$\overrightarrow{OY}$$, $$\overrightarrow{OZ}$$ be three unit vectors in the directions of the sides $$\overrightarrow{QR}$$, $$\overrightarrow{RP}$$, $$\overrightarrow{PQ}$$ respectively, of a triangle PQR.
|$$\overrightarrow{OX}$$ $$\times$$ $$\overrightarrow{OY}$$| = ?
A
sin(P + Q)
B
sin(P + R)
C
sin(Q + R)
D
sin2R

## Explanation

Now, $$\overrightarrow {OX} = {{\overrightarrow {QR} } \over {QR}}$$

and $$\overrightarrow {OY} = {{\overrightarrow {RP} } \over {RP}}$$

Therefore, $$(\overrightarrow {OX} \times \overrightarrow {OY} ) = {{\overrightarrow {QR} } \over {QR}} \times {{\overrightarrow {RP} } \over {RP}} = {{\overrightarrow {QR} \times \overrightarrow {RP} } \over {PQ}}$$

$$= {{PQ\sin R} \over {PQ}} = \sin R = \sin (\pi - (P + Q) = \sin (P + Q))$$

2

### JEE Advanced 2017 Paper 2 Offline

Let O be the origin and $$\overrightarrow{OX}$$, $$\overrightarrow{OY}$$, $$\overrightarrow{OZ}$$ be three unit vectors in the directions of the sides $$\overrightarrow{QR}$$, $$\overrightarrow{RP}$$, $$\overrightarrow{PQ}$$ respectively, of a triangle PQR.
If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is
A
$$- {3 \over 2}$$
B
$${3 \over 2}$$
C
$${5 \over 3}$$
D
$$- {5 \over 3}$$

## Explanation

cos(P + Q) + cos(Q + R) + cos(R + P)

= $$-$$ (cosR + cosP + cosQ)

Max. of cosP + cosQ + cosR = $${3 \over 2}$$

Min. of cos(P + Q) + cos(Q + R) + cos(R + P) is = $$- {3 \over 2}$$
3

### JEE Advanced 2017 Paper 2 Offline

The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y $$-$$ 2z = 5 and 3x $$-$$ 6y $$-$$ 2z = 7 is
A
14x + 2y $$-$$ 15z = 1
B
$$-$$14x + 2y + 15z = 3
C
14x $$-$$ 2y + 15z = 27
D
14x + 2y + 15z = 31

## Explanation

Let the equation of plane be ax + by + cz = 1. Then

a + b + c = 1

2a + b $$-$$ 2c = 0

3a $$-$$ 6b $$-$$ 2c = 0

$$\Rightarrow$$ a = 7b

c = $${{15b} \over 2}$$

b = $${{2} \over 31}$$, a = $${{14} \over 31}$$, c = $${{15} \over 31}$$

$$\therefore$$ 14x + 2y + 15z = 31
4

### JEE Advanced 2017 Paper 2 Offline

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

$$\overrightarrow{OP}$$ . $$\overrightarrow{OQ}$$ + $$\overrightarrow{OR}$$ . $$\overrightarrow{OS}$$ = $$\overrightarrow{OR}$$ . $$\overrightarrow{OP}$$ + $$\overrightarrow{OQ}$$ . $$\overrightarrow{OS}$$ = $$\overrightarrow{OQ}$$ . $$\overrightarrow{OR}$$ + $$\overrightarrow{OP}$$ . $$\overrightarrow{OS}$$

Then the triangle PQR has S as its
A
centroid
B
orthocentre
C
incentre
D
circumcentre

## Explanation

$$\overrightarrow{OP}$$ . $$\overrightarrow{OQ}$$ + $$\overrightarrow{OR}$$ . $$\overrightarrow{OS}$$ = $$\overrightarrow{OR}$$ . $$\overrightarrow{OP}$$ + $$\overrightarrow{OQ}$$ . $$\overrightarrow{OS}$$

$$\Rightarrow$$ $$\overrightarrow{OP}$$($$\overrightarrow{OQ}$$ $$-$$ $$\overrightarrow{OR}$$) + $$\overrightarrow{OS}$$($$\overrightarrow{OR}$$ $$-$$ $$\overrightarrow{OQ}$$) = 0

$$\Rightarrow$$ ($$\overrightarrow{OP}$$ $$-$$ $$\overrightarrow{OS}$$)($$\overrightarrow{OQ}$$ $$-$$ $$\overrightarrow{OR}$$) = 0

$$\Rightarrow$$ $$\overrightarrow{SP}$$ . $$\overrightarrow{RQ}$$ = 0

Similarly $$\overrightarrow{SR}$$ . $$\overrightarrow{PQ}$$ = 0 and $$\overrightarrow{SQ}$$ . $$\overrightarrow{PR}$$ = 0

$$\therefore \overrightarrow{S}$$ is orthocentre.

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