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Question : 34 of 160
Marks:
+1,
-0
Solution:
Let
d=x+y+zGiven,
|d|=1⇒x2+y2+z2=1‌‌... (i)
d⋅c=0⇒−x+2z=0⇒x=2z‌‌...‌ (ii) ‌a,b and
d are coplanar
‌||=0‌2(2z+y)+x(l)=0‌4z+2y+2z=0‌6z+2y=0⇒y=−3z‌[∵x=2z]From Eq. (i),
z2+4z2+9z2=114z2=1z2=‌⇒z=‌If
z=‌,x=‌,y=‌If
z=‌,x=‌,y=‌|d⋅b|=(‌+‌)=‌=√‌
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