NIMCET 2014 — Mathematics PYQ
NIMCET | Mathematics | 2014For the vectors a=−4i^+2j^, b=2i^+j^ and c=2i^+3j^, if c=ma+nb, then the value of m+n is:
Choose the correct answer:
- A.
21
- B.
23
- C.
25
25
Explanation
Solution
Concept:
If two vectors a=a1i^+a2j^+a3k^ and b=b1i^+b2j^+b3k^ are equal, then a1=b1, a2=b2 and c1=c2.
Calculation:
We have c=ma+nb.
⇒2i^+3j^=m(−4i^+2j^)+n(2i^+j^)
⇒2i^+3j^=(−4m+2n)i^+(2m+n)j^
Equating the scalar coefficients, we get:
−4m+2n=2…(1)
2m+n=3…(2)
Multiplying equation (2) by 2 and adding to equation (1), we get:
4n=8
⇒n=2
Using either of the equations above, we also get:
m=21
Therefore:
∴m+n=2+21=25
Correct Option: 3. (25)
Explanation
Solution
Concept:
If two vectors a=a1i^+a2j^+a3k^ and b=b1i^+b2j^+b3k^ are equal, then a1=b1, a2=b2 and c1=c2.
Calculation:
We have c=ma+nb.
⇒2i^+3j^=m(−4i^+2j^)+n(2i^+j^)
⇒2i^+3j^=(−4m+2n)i^+(2m+n)j^
Equating the scalar coefficients, we get:
−4m+2n=2…(1)
2m+n=3…(2)
Multiplying equation (2) by 2 and adding to equation (1), we get:
4n=8
⇒n=2
Using either of the equations above, we also get:
m=21
Therefore:
∴m+n=2+21=25
Correct Option: 3. (25)

