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Question : 34 of 160
Marks:
+1,
-0
Solution:
Given, normal vector to the plane is
n=2+3+5And direction vector to the line is
b=j−2Since,
b⋅n=(−2)⋅(2+3+5)=3−10=−7≠0⇒ line is not perpendicular to the plane Now,
P(t)=+(−3+t)−2t for minimum distance,
p(t)⋅n=0⇒2+(−9+3t)−10t=0⇒−7t−7=0⇒t=−1‌∴‌‌p=−4+2∴‌‌‌AP=(−4+2)−(−3)‌‌‌=−+2∴ Equation of plane
‌(r−(−4+2))⋅(−+2)=0⇒((x+y+z)−(−4+2))⇒((−+2)=0⇒((x−1)+(y+4)+(z−2)‌(−+2)=0‌⇒‌‌−(y+4)+2(z−2)=0‌⇒‌‌−y−4+2z−4=0‌⇒‌‌−y+2z−8=0‌⇒‌‌−y+2z=8‌⇒r⋅(−+2)=8
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