# Math Help - Vector question

1. ## Vector question

Im hoping this is in right place?
Ok, am very stuck with this =(.
Find a vector perpendicular to both i + 2k - k and 3i -j + k.
Im sure its very simple..i know it is but i just can't do it =/...the answer is i - 4j - 7k.

I would greatly appreciate a poke in the right direction

Thankyou!!

2. Say that $\vec{c}$ is the vector that results from taking the cross product of $\vec{u}$ and $\vec{v}$.

You should recall that $\vec{c}$ is perpendicular to both $\vec{u}$ and $\vec{v}$.

3. You can find the cross product by:

$\begin{vmatrix}
{i}&{j}&{k}\\
{1}&{2}&{-1}\\
{3}&{-1}&{1}
\end{vmatrix}$

You can interchange the 2nd and 3rd rows, but you'll get the negative of this answer.

4. Hello, AshleyT!

Even if you didn't know about the cross product,
. . we can still answer the question.

Of course, it takes longer . . .

Find a vector perpendicular to both $i + 2j - k$ and $3i -j + k$
We are given two vectors: . $\begin{array}{ccc} \vec u &=& \langle1,2,\text{-}1\rangle \\ \vec v &=& \langle 3,\text{-}1,1\rangle \end{array}$

We want a vector, $\vec w \:=\:\langle a,b,c\rangle$, perpendicular to both $\vec u$ and $\vec v.$

Recall that two vectors are perpendicular if their dot product is zero.

We have:

. . $\vec u\perp\vec w \quad\Rightarrow\quad \vec u\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 1,2,\text{-}1\rangle \cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad a + 2b - c \:=\:0\;\;{\color{blue}[1]}
$

. . $\vec v\perp\vec w \quad\Rightarrow\quad \vec v\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 3,\text{-}1,1\rangle\cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad 3a - b + c \:=\:0\;\;{\color{blue}[2]}$

Add [1] and [2]: . $4a + b \:=\:0 \quad\Rightarrow\quad b \:=\:\text{-}4a$

Substitute into [2]: . $3a + 4a + c \:=\:0 \quad\Rightarrow\quad c \:=\:\text{-}7a$

We have: . $\begin{array}{ccc}a &=& a \\ b &=& \text{-}4a \\ c &=& \text{-}7a \end{array}$

On the right, replace $a$ with a parameter $t$: . $\begin{array}{ccc}a &=& t \\ b &=& \text{-}4t \\ c &=& \text{-}7t\end{array}$

This represents all vectors perpendicular to both $\vec u$ and $\vec v.$

Let $t = 1$ and we have: . $\langle 1,\text{-}4,\text{-}7\rangle \:=\:i - 4j - 7k$

5. Originally Posted by Soroban
Hello, AshleyT!

Even if you didn't know about the cross product,
. . we can still answer the question.

Of course, it takes longer . . .

We are given two vectors: . $\begin{array}{ccc} \vec u &=& \langle1,2,\text{-}1\rangle \\ \vec v &=& \langle 3,\text{-}1,1\rangle \end{array}$

We want a vector, $\vec w \:=\:\langle a,b,c\rangle$, perpendicular to both $\vec u$ and $\vec v.$

Recall that two vectors are perpendicular if their dot product is zero.

We have:

. . $\vec u\perp\vec w \quad\Rightarrow\quad \vec u\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 1,2,\text{-}1\rangle \cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad a + 2b - c \:=\:0\;\;{\color{blue}[1]}
$

. . $\vec v\perp\vec w \quad\Rightarrow\quad \vec v\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 3,\text{-}1,1\rangle\cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad 3a - b + c \:=\:0\;\;{\color{blue}[2]}$

Add [1] and [2]: . $4a + b \:=\:0 \quad\Rightarrow\quad b \:=\:\text{-}4a$

Substitute into [2]: . $3a + 4a + c \:=\:0 \quad\Rightarrow\quad c \:=\:\text{-}7a$

We have: . $\begin{array}{ccc}a &=& a \\ b &=& \text{-}4a \\ c &=& \text{-}7a \end{array}$

On the right, replace $a$ with a parameter $t$: . $\begin{array}{ccc}a &=& t \\ b &=& \text{-}4t \\ c &=& \text{-}7t\end{array}$

This represents all vectors perpendicular to both $\vec u$ and $\vec v.$

Let $t = 1$ and we have: . $\langle 1,\text{-}4,\text{-}7\rangle \:=\:i - 4j - 7k$

Perfect thank-you! We havn't been shown the cross product yet so we have to do things the long way =(.
Thank-you to everyone for your replies too =).