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:
\( \vec{u}=(a_{1}, b_{1}, c_{1}) \)
\(\vec{v}=(a_{2}, b_{2}, c_{2}) \)
\(\vec{u}\cdot\vec{v}=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} \)
Find \((3\vec{a}\cdot\vec{b})\cdot(2\vec{b}-4\vec{a})\text{, if }\vec{a}
=\)\(-\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k }} \text{and } \vec{b}=2\hat{\text{i}}
+4\hat{\text{j}}-5\hat{\text{k}} \)
Step 1
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The three vectors \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \) represents 3 basis vectors for the vector space. We have:
\(\vec{a}=(-1, -3, 1) \)
\(3\vec{a}=(-3, -9, 3) \)
\(4\vec{a}=(-4, -12, 4) \)
\(\vec{b}=(2, 4, -5) \)
\(2\vec{b}=(4, 8, -10) \)
Step 2
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We can now find the dot product:
\((3\vec{a}+\vec{b})\cdot(2\vec{b}-4\vec{a}) \)
\(=(1, -5, -2)\cdot(8, 20, -14) \)
\(=(-1)(8)+-5(20)+-2(-14) \)
\(-8-100+28 \)
\(=-80 \)
Find the angle between the following vectors \(\vec{u}=(-3, 1, 2)\text{ and }\vec{v}=(5, -4,
-1) \)
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\( \vec{u}\cdot\vec{v} = |\vec{u}| |\vec{v}| \cos\theta \)
\((-3, 1, 2)\cdot(5, -4, -1)=\sqrt{3^2+1^2+2^2} \sqrt{5^2+4^2+1^2} \cos{θ} \)
\(-3(5)+1(-4)+2(-1)=\sqrt{14}\sqrt{42}\cos{θ} \)
\(-21=\sqrt{588} \cos{θ} \)
\(\dfrac{-21}{\sqrt{588}}=\cos{θ} \)
\(150°=θ \)
Given \(\vec{a}=(2, 3, 7)\text{ and }\vec{b}=(-4, y ,-14) \), for what value of y are the vectors collinear?
For what value of y are the vectors perpendicular
Step 1
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\(\vec{a}\text{ collinear to } \vec{b}\text{ iff }\vec{a}=k\vec{b} \)
\((2, 3, 7)=k(-4, y, -14) \)
\(2=-4k, 3=ky, 7=-14\)
\(\dfrac{-1}{2}=k \)
\(3=\dfrac{-1}{2}y \)
\(-6=y \)
Step 2
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\(\vec{a}\perp\vec{b}\text{ iff }\vec{a}\cdot\vec{b}=0 \)
\((2, 3, 7)\cdot(-4, y, -14)=0 \)
\(-8+3y-98=0 \)
\(3y=106 \)
\(y=\dfrac{106}{3} \)