1
JEE Main 2016 (Offline)
+4
-1
Out of Syllabus
The distance of the point $$(1,-5,9)$$ from the plane $$x-y+z=5$$ measured along the line $$x=y=z$$ is :
A
$${{10} \over {\sqrt 3 }}$$
B
$${20 \over 3}$$
C
$$3\sqrt {10}$$
D
$$10\sqrt {3}$$
2
JEE Main 2015 (Offline)
+4
-1
Out of Syllabus
The equation of the plane containing the line $$2x-5y+z=3; x+y+4z=5,$$ and parallel to the plane, $$x+3y+6z=1,$$ is :
A
$$x+3y+6z=7$$
B
$$2x+6y+12z=-13$$
C
$$2x+6y+12z=13$$
D
$$x+3y+6z=-7$$
3
JEE Main 2015 (Offline)
+4
-1
Out of Syllabus
The distance of the point $$(1, 0, 2)$$ from the point of intersection of the line $${{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}$$ and the plane $$x - y + z = 16,$$ is :
A
$$3\sqrt {21}$$
B
$$13$$
C
$$2\sqrt {14}$$
D
$$8$$
4
JEE Main 2014 (Offline)
+4
-1
Out of Syllabus
The image of the line $${{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,$$ in the plane $$2x-y+z+3=0$$ is the line :
A
$${{x - 3} \over 3} = {{y + 5} \over 1} = {{z - 2} \over { - 5}}$$
B
$${{x - 3} \over { - 3}} = {{y + 5} \over { - 1}} = {{z - 2} \over 5}\,$$
C
$${{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}\,$$
D
$${{x + 3} \over { - 3}} = {{y - 5} \over { - 1}} = {{z + 2} \over 5}$$
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