MHB Find Value of p & q in Vectors

[karush]

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

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

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

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

(b)

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

does this mean

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

 

[Prove It]

Re: find value in vectors

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

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

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

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

(b)

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

does this mean

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

You shouldn't need a CAS to solve the first equation, it is really simple.

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

[math]\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*}[/math]

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

does this mean

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

Well, it's [math]\displaystyle \begin{align*} \sqrt{1^2 + q^2 + 5^2 } = \sqrt{42} \end{align*}[/math], not [math]\displaystyle \begin{align*} \left| \sqrt{1^2 + q^2 + 5^2} \right| \end{align*}[/math]. Otherwise you are correct.

 

[HallsofIvy]

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

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

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

More to the point is the other way around: if [math]u\perp w[/math], then [math][u\bullet w= 0[/math]

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

That's really sad! [math]u\bullet w= 6- 3- p= 0[/math] 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 $$v=\left[ \begin{array}{c} 1 \\ q \\5 \end{array} \right] $$ Given that $$|v|=\sqrt{42}$$ , find the possible values of $$q$$

does this mean

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

Yes, that is correct!