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Find the distance of the point 3i-2j+k from the plane 3x+y-z+2=0 measured parallel to the line x-1/2=y+2/-3=z-1/1. Also find the foot of the perpendicular from the given point upon the given plane.

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Please find below the solution to the asked query:

Direction ratio of line x-12=y+2-3=z-11 are a,b,c=2,-3,1Given point is 3i^-2j^+k^ i.e. 3,-2,1Hence we have to measure distance along the line:x-1a=y+2b=z-1c=t, where t is parameterx-32=y+2-3=z-11=tHence any general point on line will be:x,y,z=2t+3,-3t-2,t+1It will satisfy plane 3x+y-z+2=0 for some value of t,⇒32t+3+-3t-2-t+1+2=0⇒6t+9-3t-2-t-1+2=0⇒2t=-8⇒t=-4⇒2t+3,-3t-2,t+1=-8+3,12-2,-4+1=-5,10,-3Required distance between 3,-2,1 and -5,10,-3=3+52+-2-102+1+32=64+144+16=224To find foot of perpendicular from point 3,-2,1 to 3x+y-z+2=0 apply:x-x13=y-y11=z-z1-1=-3x1+y1-z1+232+12+12⇒x-x13=y-y11=z-z1-1=-3x1+y1-z1+211⇒x-33=y+21=z-1-1=-9-2-1+211=-811⇒x-33=-811⇒x=-2411+3=911y+21=-811⇒y=-811-2=-3011z-1-1=-811⇒z=811+1=1911Hence foot of perpendicular is 911,-3011,1911
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listen first of all u did not give me the point u gave me the plane but i am taking the point as ( 3i , -2j , k ) and telling u the concept

first as u know the point and the direction ratio of the line ( as this line is parallel to the given line ) u can form a equation of required line , 

now when u form the line find the point of line which lie on the plane by letting the line = lambda  and getting x , y , z co-ordinates then u pu these co-ordinates in the plane and get lambda , now at last put the lambda in the points and get the required points that lie on plane now ues the distance formula to find the distance between the plane and line .

for the perpendicular distance u the formula  as given in the ncert 

CHEERS..!!

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