UPSC NDA I 2021 Mathematics Solved Paper

Concept: The cross product of two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ is given by $\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}=|a| \cdot |b| \sin \theta \hat{1} \;$ and $\left|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}\right|=|a| \cdot |b| \sin \theta$ The scalar product of two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ is given by $\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b}=|a| \times |b| \cos \theta$ If $\overset{\rightarrow}{a}$ is a unit vector then $|a|=1$ Calculations: Statement 1: The cross product of two unit vectors is always a unit vector. Let $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ are two unit vectors. i.e $| a]=1$ and $\left|\overset{\rightarrow}{b}\right|=1$ As we knowthat, the cross product of two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ is given by $\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}=| a] \cdot |b| \sin \theta \hat{f} \;$ and $\left|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}\right|=|a|$ $\Rightarrow \left|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}\right|=|a| \cdot |b| \sin \theta=\sin \theta$ The range of $\sin \theta$ is $[-1,1]$ So, it is notnecessarily true that the cross product of two unit vectors is always a unit vector. Hence, statement 1 is false. Statement 2: The dot product of two unit vectors is always unity. Let $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ are two unit vectors. i.e $|a|=1$ and $|b|=1$ As we know that, the scalar product of two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ is given by $\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b}=\left|\overset{\rightarrow}{a}\right| \times |b| \cos \theta$ $\Rightarrow \left|\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b}\right|=\cos \theta$ The range of $\cos \theta$ is $[-1,1]$ So, it is not necessarily true that the dot product of two unit vectors is always a unit vector. Hence, statement 2 is false. Statement 3: The magnitude of sum of two unit vectors is always greater than the magnitude of their difference. Let $\overset{\rightarrow}{a}=\hat{i}$ and $\overset{\rightarrow}{b}=\hat{j}$ As we can see that, the vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ are two unit vectors $\Rightarrow |\hat{i}+\hat{j}|=\sqrt{2} \; \text{and} \; |\hat{i}-\hat{j}|=\sqrt{2}$ $\Rightarrow \left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right|=\left|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right|$ So, statement 3 is also false. Hence, the correct option is 4 .