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1

### JEE Main 2019 (Online) 12th January Morning Slot

The sum of the distinct real values of $$\mu$$, for which the vectors, $$\mu \widehat i + \widehat j + \widehat k,$$   $$\widehat i + \mu \widehat j + \widehat k,$$   $$\widehat i + \widehat j + \mu \widehat k$$  are co-planar, is :
A
2
B
$$-$$1
C
0
D
1

## Explanation

$$\left| {\matrix{ \mu & 1 & 1 \cr 1 & \mu & 1 \cr 1 & 1 & \mu \cr } } \right| = 0$$

$$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$$

$${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$$

$${\mu ^3} - 3\mu + 2 = 0$$

$${\mu ^3} - 1 - 3\left( {\mu - 1} \right) = 0$$

$$u = 1,\,\,{\mu ^2} + \mu - 2 = 0$$

$$\mu = 1,\,\,\mu = - 2$$

sum of distinct solutions $$=$$ $$-$$ 1
2

### JEE Main 2019 (Online) 11th January Evening Slot

Let $$\sqrt 3 \widehat i + \widehat j,$$    $$\widehat i + \sqrt 3 \widehat j$$  and   $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta$$ is :
A
4
B
1
C
2
D
3

## Explanation

Angle bisector is x $$-$$ y = 0

$$\Rightarrow$$  $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$

$$\Rightarrow$$  $$\left| {2\beta - 1} \right| = 3$$

$$\Rightarrow$$  $$\beta$$ = 2 or $$-$$ 1
3

### JEE Main 2019 (Online) 11th January Evening Slot

If the point (2, $$\alpha$$, $$\beta$$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha$$ – 3$$\beta$$ is equal to
A
12
B
7
C
17
D
5

## Explanation

Normal vector of plane

$$= \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 5} & 0 \cr 4 & { - 4} & 5 \cr } } \right|$$

$$= - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$$

equation of plane is

5(x $$-$$ 7) + 2y $$-$$ 3(z $$-$$ 6) = 0

5x + 2y $$-$$ 3z = 17

5 $$\times$$ 2 + 2$$\alpha$$ $$-$$ 3$$\beta$$ = 17

$$\therefore$$  2$$\alpha$$ $$-$$ 3$$\beta$$ = 17 $$-$$ 10 = 7
4

### JEE Main 2019 (Online) 11th January Evening Slot

Two lines $${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$$ and $${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
A
(2, 4, 7)
B
(2, $$-$$ 4, $$-$$7)
C
(2, $$-$$ 4, 7)
D
($$-$$ 2, 4, 7)

## Explanation

Point on L1 ($$\lambda$$ + 3, 3$$\lambda$$ $$-$$ 1, $$-$$$$\lambda$$ + 6)

Point on L2 (7$$\mu$$ $$-$$ 5, $$-$$6$$\mu$$ + 2, 4$$\mu$$ + 3

$$\Rightarrow$$  $$\lambda$$ + 3 = 7$$\mu$$ $$-$$ 5      . . . . (i)

3$$\lambda$$ $$-$$ 1 = $$-$$6$$\mu$$ + 2       . . . .(ii)

$$\Rightarrow$$  $$\lambda$$ = $$-$$1, $$\mu$$ = 1

point R(2, $$-$$ 4, 7)

Reflection is (2, $$-$$4, $$-$$ 7)

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