The product of the d.r's of a line perpendicular to the plane passing through the points (4,0,0), (0,2,0) and (1,0,1) is
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12th
Maths
Three Dimensional Geometry
Planes
The product of the d.r's of...
Maths
The product of the d.r's of a line perpendicular to the plane passing through the points (4,0,0),(0,2,0)and (1,0,1) is
A
6
B
2
C
0
D
1
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Correct option is
A
6
Let three points be A(4,0,0)B(0,2,0)C(1,0,1).
The dr's of AB are (4,−2,0)
The dr;s of BC are (1,−2,1).
Since, both these lines lie on the plane, the normal to the plane will also be a normal to these lines.
Hence, the dr's of the normal can be obtained by the cross product of these 2lines.
n=AB×BC
(−1,2,−3)
Hence, the product is 6.
We have to find the product of direction ratios of a line perpendicular to the plane passing through the points A (4, 0, 0) , B(0, 2, 0) and C(1, 0, 1).
solution : direction ratio of line AB = A(4, 0, 0) - B(0, 2, 0) = (4, -2, 0)
direction ratio of line CB= C(1, 0, 1) - B(0, 2, 0) = (1, -2, 1)
lines AB and BC lie on the same plane. direction ratios of the line is normal to the plane so direction ratios must be perpendicular to lines AB and BC.
so direction ratios of the line :
AB × BC
= (4, - 2, 0) × (1, -2, 1)
= i(-2 × 1 - 0 × -2) + j(4 × 1 - 0 × 1) + k(4 × -2 - (1) × -2)
= i(-2) + j(4) + k(-6)
= -2i + 4j - 6k
find simple ratios of i , j and k
i.e., -1 , 2 and -3 are direction ratios of line.
therefore the product of direction ratios is (-1)(2)(-3) = 6
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