find the intercepts cut off by the plane 3x-4y+3z=0 find the vector equation of the plane in the standard form
Answers
Answered by
0
Answer:
Given, 2x+y−z=5
⇒
5
2
x+
5
y
−
5
z
=1
⇒
5/2
x
+
5
y
+
−5
z
=1
It is known that the equation of a plane in intercept from is
a
x
+
b
y
+
c
z
=1, where a,b,c are the intercepts cut off by the plane at x,y and z axes respectively.
Therefore, for the given equation,
a=
2
5
,b=5 and c=−5
Thus, the intercepts cut off by the given plane are
2
5
,5 and −5.
Similar questions