If two vectors $\vec{A}$ and $\vec{B}$ are at an angle $\theta \n

If two vectors $\vec{A}$ and $\vec{B}$ are at an angle $\theta \neq 0^\circ$, then

”<tex>$| ec{A}|+| ec{B}|=| ec{A}+ ec{B}|$</tex>”

”<tex>$| ec{A}|+| ec{B}|>| ec{A}+ ec{B}|$</tex>”

”<tex>$| ec{A}|+| ec{B}|<| ec{A}+ ec{B}|$</tex>”

”<tex>$| ec{A}|+| ec{B}|=| ec{A}- ec{B}|$</tex>”

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC CDS-1 – 2019

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The correct option is B.

For any two vectors $\vec{A}$ and $\vec{B}$, the triangle inequality states that the magnitude of their sum is less than or equal to the sum of their magnitudes: $|\vec{A}+\vec{B}| \le |\vec{A}|+|\vec{B}|$. Equality holds only when the vectors are collinear and point in the same direction, which corresponds to the angle between them $\theta = 0^\circ$. If $\theta \neq 0^\circ$, the vectors form two sides of a triangle, and the sum vector forms the third side. In a triangle, the length of any side is strictly less than the sum of the lengths of the other two sides. Therefore, if $\theta \neq 0^\circ$, $|\vec{A}+\vec{B}| < |\vec{A}|+|\vec{B}|$, or equivalently, $|\vec{A}|+|\vec{B}| > |\vec{A}+\vec{B}|$.

The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$. Since $|\vec{A}|+|\vec{B}|$ is always positive, we can compare their squares: $(|\vec{A}|+|\vec{B}|)^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|$. Comparing this with $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$, we see that $(|\vec{A}|+|\vec{B}|)^2 > |\vec{A}+\vec{B}|^2$ when $\cos\theta < 1$, which is true for $\theta \neq 0^\circ$.

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