Given the vectors $x = (1, −4, 3)^T$, $y = (0, 1, 1)^T$ and $z = (2, 0, −1)^T$, find the values of both $\alpha$ and $\beta$ such that the vector $x + \alpha y + \beta z$ is orthogonal to both $y$ and $z$.
I'm not totally sure where to begin? I think I'm suppose to look at the Null Space of the transpose of a matrix formed of vector y and vector z? If that's the case thought, I see no point to be given an x vector?
Thanks for the help!
On
If you don't know the cross product:
You want to find $a$ and $b$ such that $V = x+ay+bz$ is orthogonal to both $y$ and $z$.
Then, using "$\cdot$" for dot product, you want $V\cdot y =V\cdot z = 0 $.
$V\cdot y =(x+ay+bz)\cdot y = x\cdot y+a (y\cdot y) + b (z\cdot y) $ and $V\cdot z =(x+ay+bz)\cdot z = x\cdot z+a (y\cdot z) + b (z\cdot z) $.
You therefore have two equations in the two unknowns $a$ and $b$: $x\cdot y+a (y\cdot y) + b (z\cdot y) = 0$ and $x\cdot z+a (y\cdot z) + b (z\cdot z)=0$.
Solve these and you are done.
The cross product of the vectors y and z will give a vector that is perpendicular to both y and z.
The cross product of <0,1,1> and <2,0,-1) is <1,-2,2>. Setting this equal to the vector x+αy+βz gives $\alpha = -2$ and $\beta = 2$.