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[SOLVED] Find value in vectors

karush

Well-known member
Jan 31, 2012
2,716
(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
Jan 26, 2012
1,403
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...

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

- - - Updated - - -

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
Jan 29, 2012
1,151
(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!