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Question : 33 of 160
Marks:
+1,
-0
Solution:
r=a+2b+p(a−2c) r=3a−q(c−b)+k(a−b+c) is
r=xa+yb+zc xa+yb+zc=a+2b+p(a−2c) ‌‌=a(1+p)+b−2pc xa+yb+zc=(3+k)a+(q−k)b+(−q+k)c {| ‌‌x=1+p=3+k⇒k=p−2⇒k=−1 |
| ‌‌y=2=q−k⇒k=q−2⇒q=1. |
z=−2p=−q+k⇒−2p=−q+q−2⇒p=1 ‌‌x=3−1=2,y=2,z=−2 ∴xyz=2×2×(−2=−8
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