Two vectors lie with their tails at the same point.When the angle between them is increased by 20° their scalar product has the same magnitude but changes from positive to negative.The original angle between them was:
A)0°
B)60°
C)70°
D)80°
E)90°

Question

Quizplus · Accepted Answer

Let the two vectors be $\vec{a}$ and $\vec{b}$ with initial angle $	heta$ between them. We know that $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos	heta$.

When the angle between them is increased by $20^\circ$, the new angle between them is $	heta+20^\circ$. We are told that the scalar product has the same magnitude but changes sign, meaning $\vec{a}\cdot\vec{b}=-|\vec{a}||\vec{b}|\cos(	heta+20^\circ)$.

Setting these two expressions equal to each other, we have $|\vec{a}||\vec{b}|\cos	heta=-|\vec{a}||\vec{b}|\cos(	heta+20^\circ)$. Dividing both sides by $|\vec{a}||\vec{b}|$ and simplifying, we get $\cos	heta=-\cos(	heta+20^\circ)$.

Using the identity $\cos(-x)=\cos x$, we can rewrite this as $\cos	heta+\cos(	heta+20^\circ)=0$. Using the sum-to-product identity, we can simplify this to $2\cos\frac{2	heta+20^\circ}{2}\cos 10^\circ=0$.

Since $\cos 10^\circ$ is not zero, we must have $\cos\frac{2	heta+20^\circ}{2}=0$, which means $\frac{2	heta+20^\circ}{2}=90^\circ+k\cdot 180^\circ$ for some integer $k$. Solving for $	heta$, we get $	heta=35^\circ+k\cdot 180^\circ$.

Since $	heta$ is an angle between two vectors, we know it must be between $0^\circ$ and $180^\circ$. Therefore, $	heta=35^\circ$ is the only valid solution, and the initial angle between the vectors was $\boxed{	extbf{(D)}\ 80^\circ}$.

Two Vectors Lie with Their Tails at the Same Point