1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
Out of Syllabus
The equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is :
A
x $$-$$ 2y + z = 0
B
3x + 2y $$-$$ 3z = 0
C
x + 2y $$-$$ 2z = 0
D
5x + 2y $$-$$ 4z = 0
2
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A
ab'  +  bc'  +  1  =  0
B
cc'  +  a   +  a'  =  0
C
bb'  +  cc'  +  1  =  0
D
aa'  +  c  +  c'  =  0
3
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
Out of Syllabus
The equation of the line passing through (–4, 3, 1), parallel

to the plane x + 2y – z – 5 = 0 and intersecting

the line $${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$$ is :
A
$${{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}$$
B
$${{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}$$
C
$${{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}$$
D
$${{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}$$
4
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
Out of Syllabus
The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :
A
(–3, 0, -1)
B
(3, 2, 1)
C
(3, 3, -1)
D
(–3, 1, 1)
EXAM MAP
Medical
NEET