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Question : 64 of 70
Marks:
+1,
-0
Solution:
The line of intersection of the planes
r.(3−+)=1 and
r.(−4+2)=2 is common to both the planes.
Therefore, it is perpendicular to normals to the two planes i.e.
n1=3−+ and
n2=+4−2. Hence, it is parallel to the vector
n1×n2=−2+7+13 Thus, we have
w find the equation of the plane passing through
a=+2− and normal to the vector
n=n1×n2.
The equation of the required plane is
(r−a).n or
r.n=a.n or
r.(−2+7+13) =(+2+). (−2+7+13) or
r.(2−7−13)=1
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