Question 11
(a) Find an equation of the plane passing through the points
(0, 1, 0), (-1, 3, 2), and (-2, 0, 1)
Answers
Answer:
1
x−1
=
−2
y−1
=
3
z−2
P(1,0,−1),Q(3,2,2)
∴ Direction of
PQ
=2
i
^
+2
j
^
+3
k
^
Plane is parallel to line L & contains
PQ
. (normal to plane ⊥ to PQ & L)
∴
n
=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
2
1
j
^
2
−2
k
^
3
3
∣
∣
∣
∣
∣
∣
∣
∣
=(12)
i
^
−3
j
^
+(−6)
k
^
direction ratios of normal are (4,−1,−2)
∴ Equation plane passing through (1,0,−1) and having directions of normal (4,−1,−2).
4x−y−2z=6
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SIMILAR QUESTIONS
star-struck
The equation of the plane passing through the straight line
2
x−1
=
−1
y+1
=
4
z−3
and perpendicular to plane x+2y+z=12 is:
Medium
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>
Find the equation of a plane which passes through the point (1,2,3.) and which is at the maximum distance from the point (−1,0,2).
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