A vector of length 5 perpendicular to both u = (3, -1, 5) and v =(2, 1, 0) has what coordinates?
Why did the vector go to the circus? Because it wanted to be perpendicular to both "u" and "v"! 🤡
To find a vector that is perpendicular to both "u = (3, -1, 5)" and "v = (2, 1, 0)", let's call the unknown vector "w = (x, y, z)". Now, since "w" is perpendicular to both "u" and "v", the dot product of "w" with "u" should be zero, and the dot product of "w" with "v" should also be zero.
So, let's calculate the dot products:
w · u = x(3) + y(-1) + z(5) = 0
w · v = x(2) + y(1) + z(0) = 0
We have a system of linear equations! Now, solving these equations will give us the coordinates of the vector "w". Let me calculate that for you... 🧮
After solving the system of equations, the coordinates of the vector "w" turn out to be (-5, 11, 1). So, the vector of length 5 perpendicular to both "u" and "v" has the coordinates (-5, 11, 1). Ta-daa! 🎉
To find a vector perpendicular to both u and v, we can take their cross product.
The cross product of two vectors u = (u1, u2, u3) and v = (v1, v2, v3) is given by the following formula:
u x v = (u2*v3 - u3*v2, u3*v1 - u1*v3, u1*v2 - u2*v1)
In this case, u = (3, -1, 5) and v = (2, 1, 0), so we can substitute these values into the formula to find the cross product:
u x v = ( (-1)*(0) - (5)*(1), (5)*(2) - (3)*(0), (3)*(1) - (-1)*(2) )
= ( -5, 10, 5 )
Therefore, a vector perpendicular to both u and v has coordinates (-5, 10, 5).
To find a vector perpendicular to both u = (3, -1, 5) and v = (2, 1, 0), we can use the cross product of the two vectors.
The cross product of two vectors u = (u1, u2, u3) and v = (v1, v2, v3) is calculated as follows:
u x v = (u2*v3 - u3*v2, u3*v1 - u1*v3, u1*v2 - u2*v1)
Let's substitute the values of u = (3, -1, 5) and v = (2, 1, 0) into the formula:
u x v = ( (-1)*0 - 5*1, 5*2 - 3*0, 3*1 - (-1)*2 )
= (-5, 10, 5)
Therefore, a vector of length 5 perpendicular to both u = (3, -1, 5) and v = (2, 1, 0) has the coordinates (-5, 10, 5).
No idea what its coordinates are, but its direction numbers can be found by taking the cross product
u×v = (-5,10,5)