Q. 263.7( 3 Votes )

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0. Also find the distance of the plane obtained above, from the origin.
OR

Find the distance of the point (2, 12, 5) from the point of intersection of the line and the plane [CBSE 2014]

Answer :

Let P1 be the plane x + y + z = 1 and P2 be the plane 2x + 3y + 4z = 5 and P3 be the plane x - y + z = 0


The plane passing through line of intersection of P1 and P2 is given by P1 + λP2 = 0


x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0


(2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1 = 0


Let this be the plane P which is required


P = (2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1


The direction of normal to a plane with equation ax + by + cz + d = 0 is <a, b, c> or


Hence the direction of normal to plane P is


It is given that the required plane P is perpendicular to plane P3


As the planes P and P3 are perpendiculars, their normal vectors are also perpendicular


The direction of normal to plane P3 is


Take dot product of normal of plane P and normal of plane P3



2λ + 1 – 3λ – 1 + 4λ + 1 = 0


3λ + 1 = 0



Put value of lambda in P


(2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1 = 0





Multiply by 3


x – z + 2 = 0


Distance of plane ax + by + cz + d = 0 from a given point (x1, y1, z1) is given by


Hence distance of plane x – z + 2 = 0 from origin is




distance = √2 units


Hence equation of required plane is x – z + 2 = 0 and its distance from origin is √2 units


OR



To find the distance of the point (2, 12, 5) from the point of intersection of line and plane, we must first find out the point of intersection of line and plane


The equation of a line given


Comparing with the standard form where is point on line and is direction


and


The equation of a line in cartesian form is where (a1, a2, a3) is the point on line and <b1, b2, b3> is the direction


Hence equation of line in cartesian form



x – 2 = 3k, y + 4 = 4k, z – 2 = 2k


x = 3k + 2, y = 4k – 4, z = 2k + 2


Hence general point on line is given by (3k + 2, 4k – 4, 2k + 2)


The equation of plane is


Put to get the cartesian form



x – 2y + z = 0


As the line intersects the plane hence, we will put the general point on the line in the equation of plane so that we will get a specific value of k for which the point will lie on both plane and line which means the intersecting point of plane and line


(3k + 2) – 2(4k – 4) + 2k + 2 = 0


5k + 4 – 8k + 8 = 0


-3k + 12 = 0


3k = 12


k = 4


Hence for k = 4 the general point on line is intersecting point of line and plane


Hence the point is (3(4) + 2, 4(4) – 4, 2(4) + 2) = (14, 12, 10)


Now we have to find the distance between (2, 12, 5) and (14, 12, 10)


Using distance formula




distance = 13 units

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