For the equation of the plane 2x + 3y - 5z = 5, find the direction cosines of the
normal to the plane and the distance from the origin.
Answers
Answered by
1
Answer:
Given equation in vector form is
r
⋅(2
i
^
+3
j
^
−
k
^
)=5
∣2
i
^
+3
j
^
−
k
^
∣=
2
2
+3
2
+(−1)
2
=
14
∴
r
⋅(
14
2
i
^
+
14
3
j
^
−
14
1
k
^
)=
14
5
Direction cosines of
n
^
(normal) are
14
2
,
14
3
,
14
−1
and p=
14
5
.
Similar questions