x2+y2+z2 = 5, x+y-z=3 হলে yz+zx-xy এর মান কত?
Solution
Correct Answer: Option D
দেওয়া আছে, x2+y2+z2=5 .....(1)
x+y-z=3......(2)
সমীকরণ (2) নং হতে পাই (x+y-z)2=32 [বর্গ করে]
⇒ {(x+y)-z}2=9
⇒ (x+y)2-2(x+y) z+(z)2=9
⇒ x2+2xy+y2-2xz-2yz+z2=9
⇒ x2+y2+z2-2xz-2yz+2xy=9
⇒ x2+y2+z2-2(yz+zx-xy)=9
⇒ 5-2(xy+zx-xy)=9
⇒ -2(yz+zx-xy)=9-5
⇒ yz+zx-xy=4/-2
অতএব, yz+zx-xy=-2
x+y-z=3......(2)
সমীকরণ (2) নং হতে পাই (x+y-z)2=32 [বর্গ করে]
⇒ {(x+y)-z}2=9
⇒ (x+y)2-2(x+y) z+(z)2=9
⇒ x2+2xy+y2-2xz-2yz+z2=9
⇒ x2+y2+z2-2xz-2yz+2xy=9
⇒ x2+y2+z2-2(yz+zx-xy)=9
⇒ 5-2(xy+zx-xy)=9
⇒ -2(yz+zx-xy)=9-5
⇒ yz+zx-xy=4/-2
অতএব, yz+zx-xy=-2
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