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Question

Let the plane ax+by+cz+d=0 bisect the line joining the points 4,-3,1 and 2,3,-5 at the right angles. If a, b, c, d are integers, then the minimum value of a2+b2+c2+d2 is


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Solution

Explanation for the correct option:

Step-1 Using mid point concept:

Given that the equation of the plane is ax+by+cz+d=0 and the line passed the plane having the coordinates as P4,-3,1 and Q2,3,-5.

Let Mx,y,z be the mid-point of the line passing through the plane.

Using the mid-point formula find the coordinate of Mx,y,z.

x=x1+x22⇒x=4+22⇒x=62⇒x=3

And

y=y1+y22⇒y=-3+32⇒y=02⇒y=0

And

z=z1+z22⇒z=1-52⇒z=-42⇒z=-2

Therefore, the mid-point Mx,y,z is obtained as 3,0,-2.

Step-2 : Normal vector to the plane:

The normal of the plane PQ is determined as,

PQ=-2i+6j-6k[wherecoefficientofi=2-4,coefficientofj=3-(-3),coefficientofk=-5-1]

Then the value of a is -2, b is 6 and c is -6.

Now the equation of plane is -2x+6y-6z+d=0

Using the mid-point coordinate 3,0,-2 finding the value of d.

-23+60-6-2+d=0⇒-6+0+12+d=0⇒6+d=0⇒d=-6

Therefore, the equation of the plane is -2x+6y-6z+-6=0.

⇒x-3y+3z+3=0[equationisoftheformax+by+cz+d=0]

Now substitute the values of a as 1, b as -3, c as 3 and d as 3 in a2+b2+c2+d2.

a2+b2+c2+d2=12+-32+32+32=1+9+9+9=28

Therefore, the minimum value of a2+b2+c2+d2 is 28.


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