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Question : 31 of 160
Marks:
+1,
-0
Solution:
Let the points be
A=+2+ and
B=2−− and plane through points are
P1=,P2=2,P3=3Let the point on the line be
r(t)=‌A+t(B−A)⇒r(t)=(+2+)+t[(2−−)‌(+2+)]=‌(+2+)+t(−3−2)=‌(1+t)+(2−3)+(1−2t)Now, vectors in the plane are
‌v1=P2−P1=2−‌v2=P3−P1=3−Normal vectors,
n=v1×v2‌⇒||=(6−0)−(−3−0)+(0+2‌⇒6+3+2Using point
P1= to get plane equation is
‌n⋅(r−)=0‌⇒‌‌⟨6,3,2⟩⋅⟨x−1,y,z⟩=0‌⇒‌‌6x−6+3y+2z=0‌⇒‌‌6x+3y+2z=6Put the value of
r(t)=(1+t,2−3t,1−2t) into the plane Eq. (i), we get
‌6x+3y+2z=6‌⇒6(1+t)+3(2−3t)+2(1−2t)=6‌⇒6+6t+6−9t+2−4t=6‌⇒−7t+14=6⇒7t=8‌⇒‌‌t=‌So,
r(‌)=(1+‌)+(2−3×‌)+(1−2×‌)⇒‌−‌−‌
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