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Question : 101 of 150
Marks:
+1,
-0
Solution:
Given,
‌‌p=ai+bj+ckand
‌‌q=bi+cj+akNow,
cos‌θ=‌=‌| ab+bc+ca |
| √a2+b2+c2√a2+b2+c2 |
We know that,
‌(a+b+c)2=a2+b2+c2+2(ab+bc+ca)‌⇒ab+bc+ca=‌‌∴‌‌cos‌θ=‌| (a+b+c)2−(a2+b2+c2) |
| 2(a2+b2+c2) |
‌Maximum value of
cos‌θ, when
a=b=c=k‌‌ i.e., ‌‌cos‌θ‌=‌=‌=1⇒θ‌=0∘Minimum value of
cos‌θ, when
a+b+c=0‌⇒cos‌θ=‌| 0−(a2+b2+c2) |
| 2(a2+b2+c2) |
=‌‌=−‌‌⇒‌‌θ=‌‌∴‌‌θ∈[0,2π∕3]‌
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