Question

a vector is in standard position, with its terminal point in the second quadrant and an x - coordinate of - 5. the vector has a magnitude of √81. complete the statements describing the vector. the y - coordinate of the vector to the nearest tenth is 9.3. the direction angle to the nearest whole number is

Explanation:

Step1: Recall vector magnitude formula

The magnitude of a vector $\vec{v}=(x,y)$ is $|\vec{v}|=\sqrt{x^{2}+y^{2}}$. Given $x = - 5$ and $|\vec{v}|=\sqrt{110}$, we have $\sqrt{(-5)^{2}+y^{2}}=\sqrt{110}$. Squaring both sides gives $25 + y^{2}=110$, so $y^{2}=110 - 25=85$ and $y=\sqrt{85}\approx9.2$.

Step2: Recall direction - angle formula

The direction - angle $\theta$ of a vector $\vec{v}=(x,y)$ is given by $\tan\theta=\frac{y}{x}$. Here, $x=-5$ and $y\approx9.2$, so $\tan\theta=\frac{9.2}{-5}=-1.84$. Since the vector is in the second quadrant, $\theta=\pi+\arctan(-1.84)$. $\arctan(-1.84)\approx - 61.5^{\circ}$. In the range $0^{\circ}\leq\theta<360^{\circ}$, $\theta = 180^{\circ}-61.5^{\circ}\approx118^{\circ}$.

Answer:

The direction - angle to the nearest whole number is 118.