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Find the equation of a plane passing thr...

Find the equation of a plane passing through origin and which is perpendicular to a normal vector
`2hati+ hatj -hatk`. (Cartesian form )

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Verified by Experts

The correct Answer is:
`2x + y - z = 0`
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Find the Cartesian equation of the plane passing through point A(1, 2, 3) and which is perpendicular to a vector vecn=2hati-3hatj+4hatk

Find the equation of a line passing through the point (1,2,3) and perpendicular to the plane vecr.(2hati-3hatj+4hatk) = 1 .

Find the equation of the plane passing through the point hati-hatj+hatk and perpendicular to the vectro 3hati-hatj-2hatk and show that the point 2hati+4hatj lies on the plane.

Find the equation of a line passing through the point (2,-3,5) and parallel to vector (3hati+2hatj-hatk) .

Find the vector equationof the line passing through the point (3,1,2) and perpendicular to the plane vecr.(2hati-hatj+hatk)=4 . Find also the point of intersection of this line and the plane.

The equation of plane passing through the point hati+hatj+hatk and paralel to the plane vecr.(2hati-hatj+2hatk)=5 is

Find the equation of the line passing through point 2hati-hatj+3hatk and parallel to vector hati+hatj-2hatk in vector form as well as Cartesian form.

Find the equation of the plane through the point 2hati+3hatj-hatk and perpendicular to vector 3hati+3hatj+7hatk. Determine the perpendicular distance of this plane from the origin.

The vector equation of a plane passing through a point having position vector 2hati+3hatj-4hatk and perpendicular to the vector 2hati-hatj+2hatk , is

Find the equation of plane which is at a distance (4)/(sqrt(14)) from the origin and is normal to vector 2hati+hatj-3hatk .

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