Q. 133.7( 3 Votes )

# Find the vector equation of the plane through the points (2, 1, – 1) and (– 1, 3, 4) and perpendicular to the plane

x – 2y + 4z = 10.**[CBSE 2013]**

**[CBSE 2013]**

Answer :

Vector equation of a plane is given as

Where is any point on the plane and is a vector perpendicular to the plane.

Now, the given plane x – 2y + 4z = 10 is perpendicular to required plane. So, the normal vector of x – 2y + 4z = 10 will be parallel to the required plane. Hence, is parallel to the required plane.

Points say A(2,1, – 1) and B(– 1,3,4) are on the plane hence the vector is also parallel to the required plane so,

is parallel to the required plane.

Hence as both and are parallel to the plane so the direction of is the cross product of the two vectors.

So, the equation of required plane is,

Hence, the vector equation of required plane is

.

Rate this question :

Find the equation of the plane which contains the line of intersection of the planes

and

and whose intercept on the x-axis is equal to that of on y-axis.

Mathematics - Board PapersFind the coordinates of the point where the line through the points A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane.

Mathematics - Board PapersFind the distance of the point (–1, –5, –10), from the point of intersection of the line and the plane

Mathematics - Board PapersFind the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Find also, the image of the point in the plane.

Mathematics - Board PapersFind the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, - 1, 2) and parallel to the line

Mathematics - Board PapersFind the coordinate of the point P where the line through and crosses the plane passing through three points and Also, find the ratio in which P divides the line segment AB.

Mathematics - Board Papers