If \(O\) is the origin and \(P\) is the point \((2, -4, 6)\), then what is the equation of the plane through \(P\) and perpendicular to \(OP\)?
\(x - 2y + 3z - 28 = 0\)
Since the plane is perpendicular to \(OP\), the vector \(\vec{OP}=(2,-4,6)\) serves as its normal. Using point \(P(2,-4,6)\), the plane is \(2(x-2)-4(y+4)+6(z-6)=0\), which simplifies to \(2x-4y+6z-56=0\), that is, \(x-2y+3z-28=0\).
Similar Questions
The equation of the plane passing through the intersection of the planes 2x + y + 2z = 9, 4x – 5y – 4z = 1 and the point (3, 2, 1) is
The locus of a point P(x, y, z) which moves in such a way that z = 7 is a
A variable plane passes through a fixed point (a, b, c) and cuts the axes in A, B and C respectively. The locus of the center of the sphere OABC, O being the origin, is
What is the equation of the plane which cuts an intercept 5 units on the z-axis and is parallel to xy-plane?
With reference to the plane \(Ax + By + Cz + D = 0\), the image of the point \(P(1, 3, 4)\) is \((-3, 5, 2)\). If \(A + B + C = 2\), then what is \(D\) equal to?
Important Questions from Equation of a Plane
If the foot of the perpendicular drawn from (-2, 1, 0) on a plane is (1, -2, 1), then the equation of the plane is
The equation of the plane which contain the points (0, 6, 0) and (-2, -3, 4) and which is parallel to the ray with direction ratios (2, 3, -2) is:
The image of the point (–3, 8, 4) in the plane 6x – 3y – 2z + 1 = 0, is -
The equation of the plane through the point (1, 2, –3) and normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1) is-
Determine the vector equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\) , and the point (1, 1, 1)?
