find the D.C's of the normal to the plane x+2y+2z-4=0
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Step-by-step explanation:
Correct option is A)
Directional cosines of normal to a plane ax+by+cz+d=0
is ±(
a
2
+b
2
+c
2
a
,
a
2
+b
2
+c
2
b
,
a
2
+b
2
+c
2
c
)
∴ directional cosines of normal to plane x+2y-3z+4=0 are
a =1 , b = 2, c = -3 , d = 4
dc's ⇒±[
1+4+9
1
,
1+4+9
2
,
1+4+9
−3
]
∴ The required dc's are (
14
1
,
14
2
,
14
−3
) and (
14
−1
,
14
−2
,
14
3
)
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