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The equation of the plane in normal form which passes through the points (-2,1,3),(1,1,1) and (2,3,4) is

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Important Questions on Three Dimensional Geometry

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A plane passes through the point (3,5,7). If the direction ratios of its normal are equal to the intercepts made by the plane x+3y+2z=9 with the coordinate axes, then the equation of that plane is
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The sum of intercepts of the plane 4 x+3 y+2 z=2 on the coordinate axes is
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X intercept of the plane containing the line of intersection of the planes x-2y+z+2=0 and 3x-y-z+1=0 and also passing through (1,1,1) is
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Find the equation of the plane passing through the point (2,1,3) and perpendicular to the planes x-2y+2z+3=0 and 3x-2y+4z-4=0.
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Let a→=i^ and b→=j^ is the point of intersection of the lines r→×a→=b→×a→ and r→×b→=a→×b→ is

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If a1,b1,c1,a2,b2,c2 are the direction cosines of two lines making an angle θ with each other, then cosθ=
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The angle between a line with direction ratios 2,2,1 and the line joining the points (3,1,4) and (7,2,12) is
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Let L1 (respectively L2 ) be the line passing through 2i^-k^ (respectively 2i^+j^-3k^) and parallel to 3i^-j^+2k^ (respectively i^-2j^+k^). Then the shortest distance between the lines L1 and L2 is equal to