Verified Solution • 2025–26 Session moderate

QJEEM23Maefcd: JEE Main 2023 Physics - Class 11 JEEM23Maefcd Solution: Let and . If is a vector such that , and projection of on is 1, …

Let a→=4i^+3j^\overrightarrow a = 4\widehat i + 3\widehat j and b→=3i^−4j^+5k^\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k . If c→\overrightarrow c is a vector such that c→.(a→×b→)+25=0,c→ .(i^+j^+k^)=4\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right) + 25 = 0,\overrightarrow c \,.(\widehat i + \widehat j + \widehat k) = 4 , and projection of c→\overrightarrow c on a→\overrightarrow a is 1, then the projection of c→\overrightarrow c on b→\overrightarrow b equals :

Multiple Choice Options

  • A 32\frac{3}{\sqrt2}
  • B 12\frac{1}{\sqrt2}
  • C 15\frac{1}{5}
  • D 52\frac{5}{\sqrt2}

Step-by-Step Verified Solution

[c→a→b→]=−25

Let c→=li^+nj^+nk^\overrightarrow c = l\widehat i + n\widehat j + n\widehat k

|lmn4303−45|=−25

⇒3l−4m−5n=−5\Rightarrow 3l - 4m - 5n = - 5 ..... (i)

c→ . (i^+j^+k^)=4\overrightarrow c \,.\,(\widehat i + \widehat j + \widehat k) = 4

⇒l+m+n=4\Rightarrow l + m + n = 4 ..... (ii)

c→ . a→∣a→∣=1⇒c→ . a→=5{{\overrightarrow c \,.\,\overrightarrow a } \over {|\overrightarrow a |}} = 1 \Rightarrow \overrightarrow c \,.\,\overrightarrow a = 5

⇒4l+3m=5\Rightarrow 4l + 3m = 5 ...... (iii)

Using (i), (ii) and (iii)

l=2,m=−1,n=3l = 2,m = - 1,n = 3

Now, c→ . b→∣b→∣=2552=52{{\overrightarrow c \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = {{25} \over {5\sqrt 2 }} = {5 \over {\sqrt 2 }}

∴\therefore Option (2) is correct.

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