Answer :

Given :


Equation of plane : 3x – y – z = 7


A = (2, 3, 7)


To Find :


i) Length of perpendicular = d


ii) coordinates of the foot of the perpendicular


Formulae :


1) Unit Vector :


Let be any vector


Then unit vector of is



Where,


2) Length of perpendicular :


The length of the perpendicular from point A with position vector to the plane is given by,



Note :


If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then



Given equation of the plane is


3x – y – z = 7 ………..eq(1)


Therefore direction ratios of normal vector of the plane are


(3, -1, -1)


Therefore normal vector of the plane is






From eq(1), p = 7


Given point A = (2, 3, 7)


Position vector of A is



Now,



= (2×3) + (3×(-1)) + (7×(-1))


= 6 – 3 – 7


= -4


Length of the perpendicular from point A to the plane is






Let P be the foot of perpendicular drawn from point A to the given plane,


Let P = (x, y, z)



As normal vector and are parallel



x = 3k+2, y = - k+3, z = -k+7


As point P lies on the plane, we can write


3(3k+2) - (- k+3) - (-k+7) = 7


9k + 6 + k – 3 + k – 7 = 7


11k = 11



,




Therefore co-ordinates of the foot of perpendicular are


P(x, y, z) =


P = (5, 2, 6)


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