Equation of a Plane Passing through a Point and Parallel to the Two Given Vectors
Let A⃗ be v...
Question
Let →A be vector parallel to line of intersection of planes P1 and P2 through origin. P1 is parallel to the vector 2^j+3^k and 4^j−3^k and P2 is parallel to ^j−^k and 3^i+3^j, then the angle between vector →A and 2^i+^j−2^k is
A
π2,3π4
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B
π4,π2
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C
π6,7π4
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D
π2,3π2
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Solution
The correct options are Aπ2,3π4 Cπ4,π2 The normal vector of plane P1 is (2^j+3^k)×(4^j−3^k), i.e. (−6^i−12^i).
Assuming that point (0,2,3) passes through the first plane we have the plane equation as −18x=0, i.e.x=0.
Similarly for P2, we have the normal vector as (^j−^k)×(3^i+3^j)=(−3^k−3^j+3^i)
The equation of P2 thus becomes 3(x−0)−3(y−1)−3(z+1)=0, i.e.x−y−z=0
The line of intersection of two planes is passing through (0,0,0) and (0,1,−1) , i.e. r=ty−tz.
So the point A can be (0,1,−1) and so cosθ equals 33√2=1√2 implies θ=π4