Given the direction vector a of the line L = (1,3), the direction vector of the line L 'is b = (- 1, K). If the line L' passes through the point (0,5) and l ⊥ L ', then what is the equation of the line L' Above is the original title The answer given in the book is ∵a=(1,3) ,b=(-1,k), ∴a·b=-1+3k=0 And because the line L 'passes through the point (0,5) And l ⊥ L ', then the equation of L' is X-3Y+15=0 I don't think the K in B (1, K) should represent the slope? Here's what I did, Let the equation of line L 'be y-0 = K (X-5) That is, the point oblique formula of the column. This should be OK. After multiplying it, it becomes Y = kx-5k, the slope of L 'should be the K I set! Then from the direction vector a + (1,3) of L, we know that the slope of l should be 3 The slope k of L 'should be - 1 / 3 Then we substitute y = kx-5k to get x + 3y-5 = 0 I love to go to the top, thank you very much!

Given the direction vector a of the line L = (1,3), the direction vector of the line L 'is b = (- 1, K). If the line L' passes through the point (0,5) and l ⊥ L ', then what is the equation of the line L' Above is the original title The answer given in the book is ∵a=(1,3) ,b=(-1,k), ∴a·b=-1+3k=0 And because the line L 'passes through the point (0,5) And l ⊥ L ', then the equation of L' is X-3Y+15=0 I don't think the K in B (1, K) should represent the slope? Here's what I did, Let the equation of line L 'be y-0 = K (X-5) That is, the point oblique formula of the column. This should be OK. After multiplying it, it becomes Y = kx-5k, the slope of L 'should be the K I set! Then from the direction vector a + (1,3) of L, we know that the slope of l should be 3 The slope k of L 'should be - 1 / 3 Then we substitute y = kx-5k to get x + 3y-5 = 0 I love to go to the top, thank you very much!

There is no problem in your way of doing it, but the point oblique equation is wrong
You see the coordinates of the point backwards;
The point oblique equation of l-prime should be Y-5 = KX
The final equation should be x + 3y-15 = 0
The answer of the original book is also wrong; its K is indeed not the slope, and others did not say it is the slope. The problem is not here. I think its problem is wrong when substituting the point direction formula of a straight line

Ask a mathematical question about vectors
If there are three points a, B and C on the playground, point B is 50 meters from point a to the West and point C is 50 meters from point B to the south, then the position of point C relative to point a is:

45 degrees south by west, 2 meters from 50 times root

Given the vector a = (1,2), the vector b = (- 3,2), what is the value of K (1). Is Ka + B perpendicular to a-3b? (2). Is Ka + B parallel to a-3b?

1.ka+b=(k-3,2k+2)
If a-3b = (10, - 4) is vertical, 10 (K-3) - 4 (2k + 2) = 0
k=19
2 parallel to integer multiple relationship
ka+b=N(a-3b)
k=-1/3

Two vector math problems in senior one
1. E and F are the middle points of diagonal AC and BD of quadrilateral ABCD respectively. Given vector AB = vector a, vector BC = vector B, vector CD = vector C, vector Da = vector D, find vector EF
2. In the parallelogram ABCD, m and N are the midpoint of DC and BC respectively. The known vectors am = vector C and an = vector D. try to use C and D to represent vectors AB and AD
Need more detailed steps, thank you

1 vector AC = vector a + vector b vector EC = 1 / 2 (vector a + vector b) vector DB = vector D + vector a vector DF = 1 / 2 (vector D + vector a) vector EF = vector EC + vector CD + vector DF = vector a + 1 / 2 vector B + vector C + 1 / 2 vector D 2 link AC vector AC = vector AB + vector Ad vector AB = vector DC vector ad = to