# [SOLVED]Find value in vectors

#### karush

##### Well-known member
(a)
Let $$\displaystyle u=\left[ \begin{array}{c} 2 \\ 3 \\-1 \end{array} \right]$$ and $$\displaystyle w=\left[ \begin{array}{c} 3 \\ -1 \\p \end{array} \right]$$

Given that u is perpendicular to $$\displaystyle w$$, find the value of $$\displaystyle p$$

so by Dot Product $$\displaystyle u \bullet w = 0$$ then $$\displaystyle u \perp w$$

using TI-Nspire solve(dotP(u,w)=0,p) $$\displaystyle p=3$$

(b)

Let $$\displaystyle v=\left[ \begin{array}{c} 1 \\ q \\5 \end{array} \right]$$ Given that $$\displaystyle |v|=\sqrt{42}$$ , find the possible values of $$\displaystyle q$$

does this mean

$$\displaystyle |\sqrt{1^2+q^2+5^2}|=\sqrt{42}$$ if so $$\displaystyle q=\pm 4$$

#### Prove It

##### Well-known member
MHB Math Helper
Re: find value in vectors

(a)
Let $$\displaystyle u=\left[ \begin{array}{c} 2 \\ 3 \\-1 \end{array} \right]$$ and $$\displaystyle w=\left[ \begin{array}{c} 3 \\ -1 \\p \end{array} \right]$$

Given that u is perpendicular to $$\displaystyle w$$, find the value of $$\displaystyle p$$

so by Dot Product $$\displaystyle u \bullet w = 0$$ then $$\displaystyle u \perp w$$

using TI-Nspire solve(dotP(u,w)=0,p) $$\displaystyle p=3$$

(b)

Let $$\displaystyle v=\left[ \begin{array}{c} 1 \\ q \\5 \end{array} \right]$$ Given that $$\displaystyle |v|=\sqrt{42}$$ , find the possible values of $$\displaystyle q$$

does this mean

$$\displaystyle |\sqrt{1^2+q^2+5^2}|=\sqrt{42}$$ if so $$\displaystyle q=\pm 4$$
You shouldn't need a CAS to solve the first equation, it is really simple.

\displaystyle \displaystyle \begin{align*} \mathbf{u} = \left[ \begin{matrix} \phantom{-}2 \\ \phantom{-}3 \\ -1 \end{matrix} \right] \end{align*} and \displaystyle \displaystyle \begin{align*} \mathbf{w} = \left[ \begin{matrix} \phantom{-}3 \\ -1 \\ \phantom{-} p \end{matrix} \right] \end{align*}, and since the two vectors are orthogonal...

\displaystyle \displaystyle \begin{align*} \mathbf{u} \cdot \mathbf{w} &= 0 \\ 2(3) + 3(-1) + (-1)p &= 0 \\ 6 - 3 - p &= 0 \\ 3 - p &= 0 \\ 3 &= p \end{align*}

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Let $$\displaystyle v=\left[ \begin{array}{c} 1 \\ q \\5 \end{array} \right]$$ Given that $$\displaystyle |v|=\sqrt{42}$$ , find the possible values of $$\displaystyle q$$

does this mean

$$\displaystyle |\sqrt{1^2+q^2+5^2}|=\sqrt{42}$$ if so $$\displaystyle q=\pm 4$$
Well, it's \displaystyle \displaystyle \begin{align*} \sqrt{1^2 + q^2 + 5^2 } = \sqrt{42} \end{align*}, not \displaystyle \displaystyle \begin{align*} \left| \sqrt{1^2 + q^2 + 5^2} \right| \end{align*}. Otherwise you are correct.

#### HallsofIvy

##### Well-known member
MHB Math Helper
(a)
Let $$\displaystyle u=\left[ \begin{array}{c} 2 \\ 3 \\-1 \end{array} \right]$$ and $$\displaystyle w=\left[ \begin{array}{c} 3 \\ -1 \\p \end{array} \right]$$

Given that u is perpendicular to $$\displaystyle w$$, find the value of $$\displaystyle p$$

so by Dot Product $$\displaystyle u \bullet w = 0$$ then $$\displaystyle u \perp w$$
More to the point is the other way around: if $$\displaystyle u\perp w$$, then $$\displaystyle [u\bullet w= 0$$

using TI-Nspire solve(dotP(u,w)=0,p) $$\displaystyle p=3$$
That's really sad! $$\displaystyle u\bullet w= 6- 3- p= 0$$ so 3- p= 0.
You should be able to solve that faster yourself than the time it takes you to turn a calculator on!

(b)

Let $$\displaystyle v=\left[ \begin{array}{c} 1 \\ q \\5 \end{array} \right]$$ Given that $$\displaystyle |v|=\sqrt{42}$$ , find the possible values of $$\displaystyle q$$

does this mean

$$\displaystyle |\sqrt{1^2+q^2+5^2}|=\sqrt{42}$$ if so $$\displaystyle q=\pm 4$$
Yes, that is correct!