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From the point P(1,2,4) ,a perpendicular is drawn on the plane 2x+y-2z+3=0. Find the eqn , the length and the coordinate of the foot of perpendicular.

Given, Equation of Plane is 2x+y-2x+3=0 ---(1), Hence D.R.S of Normal are 2,1,-2

Also Given Point is (1,2,4)

Hence, Equation of line passing through (1,2,4) and having d.r.s 2,1,-2 (SInce Normal is parallel to line)

(x-1)/2=(y-2)/1=(z-4)/-2= k(say)

ANy coordinate on the line is of the form --> [2k+1,k+2,4-2k]

Since Coordinate also lies on the Plane, Hence

2(2k+1)+k+2-2(4-2k)+3=0 => 4k+2+k+2-8+4k+3=0=> 9k-1=0=> k=1/9

Hence Coordinate of Foot Of Perpendicular= (11/9,19/9,34/9)

ALso Lenght of Perpendicular= 1/3 units (Applying DIstance FOrmula between Points (1,2,4) and (11/9,19/9 ,34/9)

Hope it helps!!!

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Given, Equation of Plane is 2x+y-2x+3=0 ---(1), Hence D.R.S of Normal are 2,1,-2

Also Given Point is (1,2,4)

Hence, Equation of line passing through (1,2,4) and having d.r.s 2,1,-2 (SInce Normal is parallel to line)

(x-1)/2=(y-2)/1=(z-4)/-2= k(say)

ANy coordinate on the line is of the form --> [2k+1,k+2,4-2k]

Since Coordinate also lies on the Plane, Hence

2(2k+1)+k+2-2(4-2k)+3=0 => 4k+2+k+2-8+4k+3=0=> 9k-1=0=> k=1/9

Hence Coordinate of Foot Of Perpendicular= (11/9,19/9,34/9)

ALso Lenght of Perpendicular= 1/3 units (Applying DIstance FOrmula between Points (1,2,4) and (11/9,19/9 ,34/9)
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This is the required solution..

  • 2
From the point P(1,2,4).a Perpendicular is down on the plane 2x + Y - 2 X + 3 is equal to zero Find the equation the length and the co ordinates of The foot Of the perpendicular
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