Given the plane 3x+2y+5z=54 and the points P0(6, 8, 4)[on plane] and P1(13, 18, 5) [not on plane]
A. Find vector n, a vector normal to the plane
B. Find vector v from P0 to P1
C. Find the angle between vector n and vector v
D. Find p, the scalar projection of vector v on vector n
I have:
A. vector n= 3*vector i + 2*vector j + 5* vector k
B. vector v= 7*vector i + 10*vector j + vector k
C. 52.46 degrees
And I'm not quite sure for D...
A. To find a vector normal to the plane, we can use the coefficients of x, y, and z in the equation of the plane. In this case, the equation of the plane is 3x + 2y + 5z = 54. The coefficients of x, y, and z are 3, 2, and 5, respectively. Therefore, a vector normal to the plane is n = (3, 2, 5).
B. To find the vector v from P0 to P1, we subtract the position vectors of the two points. The position vector of P0 is (6, 8, 4), and the position vector of P1 is (13, 18, 5). Therefore, v = (13, 18, 5) - (6, 8, 4) = (7, 10, 1).
C. To find the angle between vector n and vector v, we can use the dot product formula: cos(theta) = (n · v) / (||n|| ||v||), where · represents the dot product, ||n|| represents the magnitude of vector n, and ||v|| represents the magnitude of vector v. The dot product of n and v is (3)(7) + (2)(10) + (5)(1) = 21 + 20 + 5 = 46. The magnitude of vector n is ||n|| = √(3^2 + 2^2 + 5^2) = √(9 + 4 + 25) = √38. The magnitude of vector v is ||v|| = √(7^2 + 10^2 + 1^2) = √(49 + 100 + 1) = √150. Therefore, cos(theta) = (46) / (√38 * √150).
D. To find the scalar projection of vector v on vector n, we can use the formula: p = ||v|| * cos(theta), where p represents the scalar projection. We already calculated cos(theta) in the previous step. Therefore, p = √150 * cos(theta).
Note: Make sure to simplify or rationalize the expressions as needed for the final answers.