A) 3
B) 4
C) 5
D) 6
E) 7
Correct Answer: D
Solution :
The unit vector along \[m\hat{i}+2\hat{j}+3\hat{k}\] is \[\frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+4+9}}=\frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+13}}\] Now, \[(\hat{i}+\hat{j}+2\hat{k}).\left( \frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+13}} \right)=2\] \[\Rightarrow \]\[\frac{m}{\sqrt{{{m}^{2}}+13}}+\frac{2}{\sqrt{{{m}^{2}}+13}}+\frac{6}{\sqrt{{{m}^{2}}+13}}=2\] \[\Rightarrow \] \[\frac{m+8}{\sqrt{{{m}^{2}}+13}}=2\] Squaring on both sides, we get \[\Rightarrow \] \[{{(m+8)}^{2}}=4({{m}^{2}}+13)\] \[\Rightarrow \] \[{{m}^{2}}+16m+64=4{{m}^{2}}+52\] \[\Rightarrow \] \[3{{m}^{2}}-16m-12=0\] \[\Rightarrow \] \[m=\frac{16\pm \sqrt{256+144}}{6}=\frac{16\pm 20}{6}\] \[\Rightarrow \] \[m=6,-\frac{2}{3}\] \[\Rightarrow \] \[m=6,\]as \[-\frac{2}{3}\]is not possible.
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