Question

4. Does the equation below define a subspace in R -- BE Sa+y+z=0 - 2y + 2z+1=0 What is it geometrically (pont, line, plane, triangle,...)? Find distance from (1,0, 1) to the set defined above.

Answer 1

The answer is given as​

nibhan a The equation define (x+y +z = 0 1x-2y +22+1=0 as- - (1) the equation define we have to a subspace check does in R². Since 10,0,0) is satisfies by equ ci). but does not satisfies by cijs ie 0-0 tot = 0 - which is not possible. as these equations do not contain (0,0,0) hence These equation set does not define a subspace in R3

The equation set S xty +220 - x-2y +22+1=0 which are the planes and we know that in tersection of two plones is a line in. R3 so that equation set define ar line in R3 so geometrically , it is a line in TR3 Now x+y +2 = 0 x-2y+22+1=0 Since Variables are there and egrations are two so taking zat as free Variable. zot x+y = – t 2-2y = -1 -2t substracting, we get 3y = - t +1+2t y – itt x=-t-(t+1) D = – 3t - (t+1) 3

x= -3t-t- 3 x= -4t-/ so the parametrized equation of the required line define by equation set - (dz) = (-44-!, tt, t) Now the distance from (1,0,1) to the set defined by the equations of planes le distance from (1,0,1) to the line. V distance = √(1-647-))+ (0- (44) + (1-4)2 (+4** ) + (++)? + (t-1)2 = √(3 +4€+1 3² + ( ++ 1 ² + (t-u? 3 fut ve (t +1}}+(t+u? +(6712 -J47 (t+1)? + (42

Hence the distance is - distance & 12 17 (t+1 ²+9 (t-1)²
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