why is the line <x,y,z>=<8,9,10>+t<3,4,5> perpendicular to the plane 3x=4y+5z=10

3x=4y+5z=10

3x + 4y
or
3x - 4y ???

3x+4y

To determine if the line and the plane are perpendicular, we need to check if the direction vector of the line is perpendicular to the normal vector of the plane.

First, let's find the direction vector of the line. The direction vector can be obtained from the coefficients of the parameter 't' in the equation of the line. In this case, the direction vector of the line is <3, 4, 5>.

Next, we need to find the normal vector of the plane. To find the normal vector, we need to rewrite the equation of the plane in the form Ax + By + Cz + D = 0.

Given the equation of the plane, 3x - 4y + 5z - 10 = 0, we can see that A = 3, B = -4, C = 5.

Therefore, the normal vector of the plane is <3, -4, 5>.

Now, to determine if the line and the plane are perpendicular, we can take the dot product of the direction vector of the line and the normal vector of the plane. If the dot product is equal to zero, then the line and the plane are perpendicular.

Calculating the dot product: <3, 4, 5> · <3, -4, 5> = (3)(3) + (4)(-4) + (5)(5) = 9 - 16 + 25 = 18.

Since the dot product of the line's direction vector and the plane's normal vector is not equal to zero (it is equal to 18), we can conclude that the line <x, y, z> = <8, 9, 10> + t<3, 4, 5> is not perpendicular to the plane 3x - 4y + 5z - 10 = 0.