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1
JEE Main 2021 (Online) 17th March Morning Shift
+4
-1
The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is :
A
x + 3z = 0
B
3x $$-$$ z = 0
C
x + 3z = 10
D
3x + z = 6
2
JEE Main 2021 (Online) 16th March Evening Shift
+4
-1
Let $$\overrightarrow a$$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r$$ $$\times$$ $$\overrightarrow a$$ = $$\overrightarrow b$$ $$\times$$ $$\overrightarrow r$$,

$$\overrightarrow r$$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the

value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :
A
13
B
11
C
9
D
15
3
JEE Main 2021 (Online) 16th March Evening Shift
+4
-1
If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, l $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L1 and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to :
A
$${1 \over {\sqrt 6 }}$$
B
$${1 \over 2}$$
C
$${1 \over {\sqrt 3 }}$$
D
$$\sqrt {{2 \over 3}}$$
4
JEE Main 2021 (Online) 16th March Evening Shift
+4
-1
If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
$$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$$ is equal to :
A
3
B
39
C
$$-$$45
D
0
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