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1

### 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)
2

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

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are :
A
2, $$-$$1, 1
B
$$2\sqrt 3 ,1, - 1$$
C
$$\sqrt 2 ,1, - 1$$
D
$$\sqrt 2 , - \sqrt 2$$

## Explanation

Let the equation of plane be

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

It passes through (0, 0, 1) then

b + c = 0     . . . . (1)

Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$

$$\Rightarrow$$  a2 $$=$$ $$-$$ 2bc and b $$=$$ $$-$$ c

we get a2 $$=$$ 2c2

$$\Rightarrow$$  a $$=$$ $$\pm$$ $$\sqrt 2$$ c

$$\Rightarrow$$  direction ratio (a, b, c) = ($$\sqrt 2$$, $$-$$1, 1) or ($$\sqrt 2$$, 1, $$-$$ 1)
3

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

Let  $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c$$ is :
A
$$- 10\widehat i - 5\widehat j$$
B
$$- 10\widehat i + 5\widehat j$$
C
$$- 14\widehat i + 5\widehat j$$
D
$$- 14\widehat i - 5\widehat j$$

## Explanation

$$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$$

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

$$\Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$$

$$\Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$$

$$\Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$$

$$\Rightarrow \lambda = 2,3, - 3$$

So,   $$\lambda$$ = 2 (as   $$\overrightarrow a$$ is parallel to $$\overrightarrow c$$ for $$\lambda$$ = $$\pm$$3)

Hence   $$\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$$

$$= - 10\widehat i + 5\widehat j$$
4

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

The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ?
A
($$-$$ 2, 2, 2)
B
(2, 2, 0)
C
(2, 0, $$-$$ 2)
D
(0, $$-$$ 2, 2)

## Explanation

The normal vector of required plane

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

$$= - 8\widehat i + 8\widehat j + 8\widehat k$$

So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$

So required plane is

$$- \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$$

$$\Rightarrow - x + y + z + 4 = 0$$

Which is satisfied by $$\left( {2,0, - 2} \right)$$

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