Question

the angle between vector p and vector q is cos inverse (-3b/2a),if |p|=|vector a +2bvector|. find vector q . if|p|=|q|

Answer 1

Answer on Question #44747 – Math – Analytic Geometry

the angle between vector $p$ and vector $q$ is cos inverse $(-3b/2a)$, if $|p| = |\text{vector } a + 2b\text{vector}|$. find vector $q$. if $|p| = |q|$

Solution:

$\alpha = \arccos \left(-\frac{3b}{2a}\right)$ – angle between two vectors;

$$|\vec{p}| = |\vec{a} + 2\vec{b}|$$ $$|\vec{p}| = |\vec{q}|$$ $$|\vec{p}|^2 = |\vec{a} + 2\vec{b}|^2$$ $$|\vec{p}|^2 = a^2 + 4\vec{a}\vec{b} + 4b^2$$

The scalar product of two vectors:

$$\vec{p} \cdot \vec{q} = |\vec{p}| \cdot |\vec{q}| \cdot \cos \alpha$$

(1) and (2) in (3):

$$\vec{p} \cdot \vec{q} = |\vec{a} + 2\vec{b}| \cdot |\vec{a} + 2\vec{b}| \cdot \left(-\frac{3b}{2a}\right)$$ $$\vec{p} \cdot \vec{q} = (\vec{a} + 2\vec{b})^2 \cdot \left(-\frac{3b}{2a}\right)$$ $$\vec{p} \cdot \vec{q} = (a^2 + 4\vec{a}\vec{b} + 4b^2) \left(-\frac{3b}{2a}\right)$$

We can't find vector $\vec{q}$ using only vectors $\vec{a}$ and $\vec{b}$, we need more information about vector $\vec{p}$ to find vector $\vec{q}$.

www.AssignmentExpert.com