[Math] If $u$ is perpendicular to $v$ and $w$, then $u$ is perpendicular to $v + 2 w$

linear algebravectors

True or false (give a reason if true or a counterexample if false):

(a) If $u$ is perpendicular (in three dimensions) to $v$ and $w$, those vectors $v$ and $w$ are parallel.

(b) If $u$ is perpendicular to $v$ and $w$, then $u$ is perpendicular to $v + 2 w$,

(c) If $u$ and $v$ are perpendicular unit vectors then $\lVert u – v \rVert =\sqrt2$

For (a) I think the answer is false because $v$ and $w$ could be going in different directions which means they can't be parallel.

But I have no idea how to firgure out b and c. Any ideas?

Best Answer

Do you know the dot product? That can answer both questions.

(b) We have $u\cdot v=0$ and $u\cdot w=0$. Now expand $u\cdot (v+2w)$.

As for (a), consider $u=(1,0,0),\ v=(0,1,0),\ w=(0,0,1)$.

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[Math] Finding unknowns in basic vectors

$\lambda$ is simply the scalar value of the $\textbf{j}$-component of the vector $\textbf{v}$. The only thing that the question asks of you, is to determine the values of $\lambda$ which will satisfy the conditions in (a) and (b) respectively.

For a):

We know that two vectors are perpendicular if the angle between them are $90\unicode{xb0}$. Where can we use this fact? Clearly the DOT PRODUCT must spring to mind, since we know the dot product of two vectors, $u$ and $v$ are given by:

$$u \cdot v = ||u|| \ ||v|| \ \cos{\theta}$$

Now we know $\theta=$ $90\unicode{xb0}$ and $\cos$($90\unicode{xb0}$)$=0$, so we must have $$u \cdot v = 0$$

We thus need to find $\lambda$ such that \begin{align}u\cdot v &= 0 \\ \therefore (3)(2)+ (2)(\lambda) &=0 \\ \therefore 2\lambda &= -6 \\ \therefore \lambda &= -3 \end{align}

For b):

Two vectors are parallel to one another if their CROSS PRODUCT is zero.

Thus we need to find $\lambda$ such that \begin{align} u \times v &= 0 \\ \therefore \begin{vmatrix}i & j \\ 3 & 2\\ 2 & \lambda\end{vmatrix} &= 0 \\ \therefore 3\lambda - 4 &= 0 \\ \therefore 3\lambda &= 4 \\ \therefore \lambda &= \frac{4}{3}\end{align}

[Math] geometric mean and vectors

You got the geometric mean equation $$ \lvert \bar{u}-\bar{i} \rvert = \sqrt{\lvert \bar{u} \rvert \lvert \bar{u}-2\bar{i} \rvert} \iff \\ \lvert \bar{u}-\bar{i} \rvert^4 = ((\bar{u}-\bar{i})\cdot (\bar{u}-\bar{i}))^2 = \lvert\bar{u} \rvert^2 \lvert \bar{u}-2\bar{i} \rvert^2 = (\bar{u}\cdot \bar{u}) ((\bar{u}-2\bar{i})\cdot (\bar{u}-2\bar{i})) \iff \\ (\bar{u}^2 + \bar{i}^2 - 2 \bar{u}\cdot \bar{i})^2 = \bar{u}^2 (\bar{u}^2 + 4\bar{i}^2 -4 \bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2 + 1 - 2 \bar{u}\cdot \bar{i})^2 = \bar{u}^2 (\bar{u}^2 + 4 -4 \bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2 + 1)^2 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = (\bar{u}^2)^2 + 4\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2)^2 + 1 + 2 \bar{u}^2 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = (\bar{u}^2)^2 + 4\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = 2\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 + 4 (\bar{u}^2-(\bar{u}^2 + 1))(\bar{u}\cdot \bar{i}) = 2\bar{u}^2 \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}\cdot \bar{i}) = 2\bar{u}^2 $$ with $$ \bar{u}\cdot \bar{i} = \lvert \bar{u} \rvert \cos 60^\circ = (1/2) \lvert \bar{u} \rvert $$ we end up with the equation $$ 1 + \lvert \bar{u} \rvert^2 - 2 \lvert \bar{u} \rvert = 2\lvert \bar{u} \rvert^2 \iff \\ \lvert \bar{u} \rvert^2 + 2 \lvert \bar{u} \rvert - 1 = 0 \iff \\ (\lvert \bar{u} \rvert + 1)^2 = 2 \iff \\ \lvert \bar{u} \rvert = \sqrt{2} - 1 $$ I talked with my friend Ruby to check the result:

irb(main):007:0> ul = Math.sqrt(2) - 1
=> 0.41421356237309515
irb(main):008:0> ux = ul / 2
=> 0.20710678118654757
irb(main):009:0> uy = Math.sin(60*Math::PI/180.0) * ul
=> 0.35871946760715046
irb(main):015:0> vx = ux - 1
=> -0.7928932188134524
irb(main):016:0> vy = uy
=> 0.35871946760715046
irb(main):017:0> vl = Math.sqrt(vx**2+vy**2)
=> 0.8702639328851419
irb(main):019:0> wx = ux - 2
=> -1.7928932188134525
irb(main):020:0> wy = uy
=> 0.35871946760715046
irb(main):021:0> wl = Math.sqrt(wx**2 + wy**2)
=> 1.8284271247461903
irb(main):022:0> Math.sqrt(ul*wl)
=> 0.8702639328851421
irb(main):023:0> vl
=> 0.8702639328851419

Best Answer

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