Vector multiplication can be completed using the dot product or cross product. The result of the dot product is a scalar. We can calculate the dot product using geometric or algebraic vectors.
The algebraic dot product of two vectors \(\vec{u}\) and \(\vec{v}\) is defined by:
\(\vec{u}=(a_{1}, b_{1}, c_{1}) \)
\(\vec{v}=(a_{2}, b_{2}, c_{2}) \)
\(\vec{u}\cdot\vec{v}=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} \)
The three vectors \(\hat{i} \), \( \hat{j} \) and \( \hat{k} \) represents 3 basis vectors for the vector space:
First, we can determine the multiples of \(a\):
\(\vec{a}=(-1, -3, 1)\)
\(3\vec{a}=(-3, -9, 3)\)
\(4\vec{a}=(-4, -12, 4)\)
Next, we can determine the multiples of \(b\):
\(\vec{b}=(2, 4, -5)\)
\(2\vec{b}=(4, 8, -10)\)
Then, we can determine the dot product:
\((3\vec{a}+\vec{b})\cdot(2\vec{b}-4\vec{a})\)
\(= (1, -5, -2)\cdot(8, 20, -14) \)
\(=(-1)(8)+-5(20)+-2(-14)\)
\(= -8-100+28\)
\(=-80\)
Therefre, we can determine the dot product is \(\boldsymbol{-80}\).
We can use the following formula to determine the angle between the pair of vectors:
\(\vec{u}\cdot\vec{v} = |\vec{u}| |\vec{v}| \cos\theta\)
Next, we can substitute the appropriate values into the formula and solve:
\((-3, 1, 2)\cdot(5, -4, -1)=\sqrt{3^2+1^2+2^2} \sqrt{5^2+4^2+1^2} \cos{θ} \)
\(-3(5)+1(-4)+2(-1)=\sqrt{14}\sqrt{42}\cos{θ}\)
\(-21=\sqrt{588} \cos{\theta}\)
\(\cfrac{-21}{\sqrt{588}}=\cos{\theta} \)
Then, we can take the inverse of cos to determine the angle:
\(\theta = \cos^{-1}{\left(\cfrac{-21}{\sqrt{588}}\right)}\)
\(\theta = 150°\)
Therefore, we can determine the angle between the pair of vectors is \(\boldsymbol{150°}\).
Given \(\vec{a}=(2, 3, 7)\text{ and }\vec{b}=(-4, y ,-14)\):
i. We can determine the value of \(y\) where the vectors are colliner by determining \(k\), the non-linear scalar, of \(\vec{b}\):
\(\vec{a} = k\vec{b}\)
\((2, 3, 7)=k(-4, y, -14)\)
\(2=-4k, \; 3=ky, \; 7=-14k\)
Next, we can determine \(k\) by finding the ratio for the first components of each vector:
\(-4k = 2\)
\(k = \dfrac{-1}{2}\)
Then, we can substitute \(k\) into the equation for the second components and solve to determine \(y\):
\(ky = 3\)
\(\cfrac{-1}{2}y = 3\)
\(y = -6\)
Therefore, we can determine that if vectors \(\vec{a}\) and \(\vec{b}\) are collinear, \(\boldsymbol{y = -6}\).
ii. We can determine the value of \(y\) where the vectors are perpendicular by setting the dot product of the two vectors equal to \(0\):
\(\vec{a} \cdot \vec{b} = 0\)
We can then substitute the appropriate values into the equation and solve for \(y\):
\((2, 3, 7) \cdot (-4, y, -14) = 0\)
\((2)(-4) + (3)(y) + (7)(-14) = 0\)
\(-8 + 3y - 48 = 0\)
\(3y = 106\)
\(y = \cfrac{106}{3}\)
Therefore, we can determine that if vectors \(\vec{a}\) and \(\vec{b}\) are collinear, \(\boldsymbol{y = \cfrac{106}{3}}\).