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I have been asked to solve the following question, and I am not sure where to begin with it :

Given two vectors $a$ and $b$ , where the magnitude of $b$ is double the magnitude of $a$ and the angle between the two vectors is $60°$, we may define two further vectors: $c = a − 2b$ and $d =14a − 2b$ .

Show that the vectors c and d are perpendicular to each other.

I am aware that $$a\cdot b = \vert a\vert\cdot \vert b\vert\cdot cos $$

However im not sure how to apply it in this question.

Any help in solving this question and the methodology to solve it would be greatly appreciated

1 Answers 1

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$$c\cdot d=(a-2b)\cdot (14a-2b)=14|a|^2-30a\cdot b+4|b|^2$$

Use that

$$|b|=2|a| \quad (1)$$ and then

$$a\cdot b=|a||b|\cos 60º \rightarrow a\cdot b=2|a|^2\cdot1/2=|a|^2 \quad (2)$$

Remember that $c \perp d \Leftrightarrow c \cdot d=0$.

Using $(1)$ and $(2)$ we get:

$$c\cdot d=14|a|^2-30a\cdot b+4|b|^2=14|a|^2-30|a|^2+4\cdot 4|a|^2=0$$

  • 0
    how does $$ a* 14a = 14|a|^2$$ not $$a* 14a = 14a^2$$ – 2017-01-09
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    Remember that $a$ is a vector, then $a\cdot a=|a|^2$. For example if $a=(x,y)$ then $(x,y)\cdot (x,y)=x.x+y.y=x^2+y^2=|(x,y)|^2$. is that clear? – 2017-01-09
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    I have always thought that $$|a|= \sqrt {a{_x}^2+a{_y}^2} $$ therefore $$|a|^2= a{_x}^2+a{_y}^2 $$ so how does $$a{_x}^2+a{_y}^2 = a*a$$ – 2017-01-09
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    You are missing the definition of Dot Product. If $u=(u_x,u_y)$ and $v=(v_x,v_y)$ then $$u\cdot v=u_x.v_x+u_y.v_y $$. Now make $u=v=a=(a_x,a_y)$ and conclude that $$a\cdot a=(a_x)^2+(a_y)^2=|a|^2$$. – 2017-01-09
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    Thankyou for this, I am still not sure how to get to the solution from here. – 2017-01-09
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    I finished the solution but I recommend you take a close look in the theory of Dot Product. I mean, definition and properties. – 2017-01-09