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Question : 62 of 150
Marks:
+1,
-0
Solution:
We have,
a=2+3+4b=−2+ and c=+−Let
r=x+y+zSince,
r×a=b Since,
r×a=b⇒||=−2+ | ⇒4y−3z=1 |
| 4x−2z=2 |
| and 3x−2y=1 |
}....(i)Also,
r⋅c=3⇒(x+y+z)⋅(+−)=3⇒x+y−z=3..... (ii) On solving Eqs. (i) and (ii), we get
x=5,y=7 and z=9∴|r|=√x2+y2+z2=√25+49+81=√155
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