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I have been presented with the following question, I cannot solve it:

The vectors $a$ and $b$ are of equal magnitude $k\neq 0$ , the angle between them is $60^{\circ}$. If

$c = (3a-b) $ and $d = (2a-10b)$

a) show that $c$ and $d$ are perpendicular vectors, b)Find the magnitudes and $c$ and $d$ in terms of $k$

The answer to b) is $c= \sqrt{7k}$ and $d= \sqrt{84k}$

If someone could show me the methodology in order to solve this I will be extremely grateful, also my apologies for not putting lines beneath the vectors I am unable to find this on the mathjaxx tutorial.

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    First, express the dot product of $a$ and $b$ in terms of $k$. For (a): Compute the dot product of $c$ and $d$ symbolically. For (b): Compute the dot product of $c$ with itself. The same for $d$. – 2017-01-03
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    To put lines above the symbols, use "\vec{v}" – 2017-01-03
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    Are the $k$'s meant to be under the square roots in your answer for part (b)? – 2017-01-03

3 Answers 3

1

HINT:

Using $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$, we obtain:

$$\vec{a} \cdot \vec{b} = |k| |k| \cos(60^o)$$

$$\vec{a} \cdot \vec{b} = k^2 \cdot \frac{1}{2}$$

We also have:

$$\vec{c} \cdot \vec{d} = (3\vec{a}-\vec{b})(2\vec{a}-10\vec{b})$$

Expanding gives:

$$\vec{c} \cdot \vec{d}=6 \vec{a} \cdot \vec{a} - 32\vec{a} \cdot\vec{b} + 10 \vec{b} \cdot \vec{b}$$

Evaluate $\vec{a}\cdot\vec{a}$ and $\vec{b}\cdot\vec{b}$ in a similar way to $\vec{a}\cdot\vec{b}$ to come to the conclusion that $\vec{c} \cdot \vec{d}=0$, which means the vectors are perpendicular (orthogonal).

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We have $\vec{a}\cdot\vec{b} = ab\cos(60^\circ) = k^2/2$ where $a$ and $b$ are the magnitudes of $\vec{a}$ and $\vec{b}$. So

\begin{align} \vec{c}\cdot\vec{d} &= (3\vec{a}-b)\cdot(2\vec{a}-10\vec{b}) \\ &=6 \vec{a}\cdot\vec{a} - 32\vec{a}\cdot\vec{b} + 10\vec{b}\cdot\vec{b} \\ &= 6k^2 - 32k^2/2 + 10k^2 \\ &= 0 \end{align}

Thus $\vec{c}$ and $\vec{d}$ are orthogonal. In addition we have

$$ c = \|\vec{c}\| =\sqrt{\vec{c}\cdot\vec{c}} = \sqrt{9\vec{a}\cdot\vec{a} - 6\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b}} = \sqrt{9k^2 -3k^2 + k^2} = \sqrt{7} k $$

See if you can do $\vec{d}$ yourself.

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    @FlewittConnor I'm not sure what you mean by that formula – 2017-01-03
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For part (a) you can use the dot-product of $c$ and $d$ to determine if they are orthogonal (they are perpendicular if and only if their dot product is zero). To get you started, we have: $$c\cdot d=(3a-b)\cdot (2a-10b)=6a\cdot a-32a\cdot b+10b\cdot b$$ where you can then use the geometric formula for the dot-product: $u\cdot v=\Vert u\Vert\Vert v\Vert\cos(\theta)$ where $u,v$ are vectors and $\theta$ is the angle between them.

To find the magnitude of a vector, $v$, you can use the square root of the dot-product of $v$ and $v$. That is, $\Vert v\Vert=\sqrt{v\cdot v}$. For part (b), this should get you started on finding the magnitude of $c$: $$\Vert c\Vert=\sqrt{c\cdot c}=\sqrt{(3a-b)\cdot (3a-b)}=\sqrt{9a\cdot a-6a\cdot b+b\cdot b}$$ and again you would use the geometric formula for the dot-product.