1
WB JEE 2024
+2
-0.5

Angle between two diagonals of a cube will be

A
$$\cos ^{-1}\left(\frac{1}{3}\right)$$
B
$$\sin ^{-1}\left(\frac{1}{3}\right)$$
C
$$\frac{\pi}{2}-\cos ^{-1}\left(\frac{1}{3}\right)$$
D
$$\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)$$
2
WB JEE 2023
+1
-0.25

If the distance between the plane $$\alpha x - 2y + z = k$$ and the plane containing the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}$$ is $$\sqrt 6$$, then $$|k|$$ is

A
36
B
12
C
6
D
$$2\sqrt 3$$
3
WB JEE 2023
+1
-0.25

The angle between a normal to the plane $$2x - y + 2z - 1 = 0$$ and the X-axis is

A
$${\cos ^{ - 1}}{2 \over 3}$$
B
$${\cos ^{ - 1}}{1 \over 5}$$
C
$${\cos ^{ - 1}}{3 \over 4}$$
D
$${\cos ^{ - 1}}{1 \over 3}$$
4
WB JEE 2022
+1
-0.25

The equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y $$-$$ z + 4 = 0 and parallel to the x-axis is

A
y + 3z + 6 = 0
B
y + 3z $$-$$ 6 = 0
C
y $$-$$ 3z + 6 = 0
D
y $$-$$ 3z $$-$$ 6 = 0
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