1
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Out of Syllabus
If the plane 2x – y + 2z + 3 = 0 has the distances $${1 \over 3}$$ and $${2 \over 3}$$ units from the planes 4x – 2y + 4z + $$\lambda$$ = 0 and 2x – y + 2z + $$\mu$$ = 0, respectively, then the maximum value of $$\lambda$$ + $$\mu$$ is equal to :
A
13
B
9
C
5
D
15
2
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Out of Syllabus
A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :
A
(4, 0, – 1)
B
(2, 0, 1)
C
(1, 0, 2)
D
(– 1, 0, 4)
3
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
If the length of the perpendicular from the point ($$\beta$$, 0, $$\beta$$) ($$\beta$$ $$\ne$$ 0) to the line,
$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$$ is $$\sqrt {{3 \over 2}}$$, then $$\beta$$ is equal to :
A
2
B
1
C
-2
D
-1
4
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
Out of Syllabus
If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the area (in sq. units) of $$\Delta$$PQR is :
A
$${{\sqrt {65} } \over 2}$$
B
$$2\sqrt {13}$$
C
$${{\sqrt {91} } \over 2}$$
D
$${{\sqrt {91} } \over 4}$$
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