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Question : 142 of 150
Marks:
+1,
-0
Solution:
3lm−4nl+mn=0 ..........(i)
and
l+2m+3n=0 ............(ii)
l=−2m−3n Put
l=−2m−3n in eq. (i),
3(−2m−3n)m−4(−2m−3n)n+mn=0 ⇒
−6m2−9mn+8mn+12n2+mn=0 ⇒
−6m2+12n2=0 ⇒
12n2=6m2⇒2n2=m2 ⇒
m2−2n2⇒m=±√2n ∵ If
m=√2n⇒l=−2(2√2+2)n ⇒
m=−√2n⇒l=−(−2√2+3)n∴ Directions ratios are
−(2√2+3)n,√2n,nand−(−2√2+3)n,−√2n,n) ⇒
−(2√2+3),√2,1and−(−2√2+3),−√2,1 ∵ We know that
cosθ=a1a2+b1b2+c1c2 =(3+2√2)(3−2√2)+(√2)(−√2)+(1)(1) =9−8−2+1=0 ∴
cosθ=0,θ=
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