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IB HL Vectors Math Problem (Help!)

Find a unit vector which is perpendictular to both the vectors 4i+2j-3k and 2i-3j+k.

What I understand is I have to first find a vector which is perpendicular to both. If
a = 4i+2j-3k
b = 2i-3j+k
c = xi+yj+zk

a*c = 0
(4x) + (2y) + (-3z) = 0
b*c = 0
(2x) + (-3y) + (1z) = 0

And... then I'm stuck.

Reply 1

user2020user

You might need to use the vector product here. If given two vectors;

\vec{a} = a_x\mathbf{i} + a_y\mathbf{j} + a_z\mathbf{k}

and

\vec{b} = b_x\mathbf{i} + b_y\mathbf{j} + b_z\mathbf{k}

A vector perpendicular to the two,

\vec{n}

, is given by their cross product

\vec{a} \times \vec{b}

, where:

\begin{aligned} \vec{n} = \vec{a} \times \vec{b} & = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \\ & = \mathbf{i} \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} - \mathbf{j} \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} + \mathbf{k} \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} \\ & = \mathbf{i} \left( a_y b_z - a_z b_y \right) - \mathbf{j} \left( a_x b_z - a_z b_x \right) + \mathbf{k} \left( a_x b_y - a_y b_x \right) \end{aligned}

The unit vector,

\hat{n}

, perpendicular to both

\vec{a}

and

\vec{b}

is therefore given by:

\hat{n} = \dfrac{\vec{n}}{|\vec{n}|}

(edited 10 years ago)

Reply 2

the bear

Original post by 2012studious

you have 1 degree of freedom when choosing c... you can set either x or y or z to a number, which could be 1.
then you make two equations with the dot product and can find the 2 missing letters

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